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## Maximizing Sales and Efficiency with an Integrated Ecommerce Solution

In today’s digital age, businesses are constantly seeking ways to increase sales and improve operational efficiency. One effective solution that has gained popularity is an integrated ecommerce platform. An integrated ecommerce solution seamlessly connects various aspects of online sales, such as inventory management, order processing, and customer relationship management (CRM). In this article, we will explore the benefits of implementing an integrated ecommerce solution and how it can help businesses maximize their sales and efficiency.

## Streamlined Operations for Increased Efficiency

A key advantage of an integrated ecommerce solution is its ability to streamline operations. With all aspects of online sales connected in one centralized platform, businesses can eliminate the need for manual data entry across different systems. This not only saves time but also reduces the risk of errors that may occur when transferring data between multiple platforms.

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By streamlining operations through integration, businesses can focus on what they do best – providing excellent products or services – rather than getting bogged down by administrative tasks.

## Enhanced Customer Experience for Increased Sales

An integrated ecommerce solution also helps businesses enhance their customer experience, leading to increased sales. With a centralized CRM system integrated into the platform, businesses have access to comprehensive customer data that enables personalized marketing efforts.

For example, when a customer makes a purchase on the website, their information is automatically stored in the CRM database. This data can then be used to tailor marketing campaigns based on customers’ preferences and purchase history. By sending targeted promotions or personalized recommendations via email or social media channels, businesses can effectively engage with customers and drive repeat purchases.

Moreover, an integrated ecommerce solution allows for seamless customer support. Customer inquiries and support tickets can be managed within the platform, ensuring timely responses and efficient issue resolution. A positive customer experience not only leads to increased sales but also fosters customer loyalty and advocacy.

## Real-time Analytics for Informed Decision Making

Another benefit of an integrated ecommerce solution is the availability of real-time analytics. By having access to comprehensive sales data, businesses can make informed decisions regarding inventory management, pricing strategies, and product assortment.

For instance, with real-time sales data, businesses can identify top-selling products and adjust inventory levels accordingly. They can also analyze customer behavior patterns to determine which marketing channels are most effective in driving conversions. This information allows businesses to allocate their resources effectively and optimize their marketing efforts.

Furthermore, an integrated ecommerce solution provides insights into customer demographics, allowing businesses to better understand their target audience. This knowledge is invaluable when creating targeted marketing campaigns or launching new products tailored specifically to customers’ needs and preferences.

## Scalability for Future Growth

Lastly, an integrated ecommerce solution offers scalability for future growth. As a business expands its online presence or diversifies its product offerings, the platform can easily accommodate these changes without requiring significant modifications or additional integrations.

With the ability to seamlessly integrate with third-party applications such as accounting software or email marketing platforms, businesses can easily adapt to evolving market trends and technologies. This scalability ensures that the integrated ecommerce solution remains a valuable asset as the business continues to grow.

In conclusion, implementing an integrated ecommerce solution offers numerous benefits for businesses looking to maximize sales and operational efficiency. By streamlining operations, enhancing the customer experience, providing real-time analytics, and offering scalability for future growth, businesses can stay ahead of the competition in today’s fast-paced digital landscape.

This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability.

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## 7.6E: Exercises for Numerical Integration

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In exercises 1 - 5, approximate the following integrals using either the midpoint rule, trapezoidal rule, or Simpson’s rule as indicated. (Round answers to three decimal places.)

1) \( \displaystyle ∫^2_1\frac{dx}{x};\) trapezoidal rule; \( n=5\)

2) \( \displaystyle ∫^3_0\sqrt{4+x^3}\;dx;\) trapezoidal rule; \( n=6\)

3) \( \displaystyle ∫^3_0\sqrt{4+x^3}\;dx;\) Simpson’s rule; \( n=6\)

4) \( \displaystyle ∫^{12}_0x^2\;dx;\) midpoint rule; \( n=6\)

5) \( \displaystyle ∫^1_0\sin^2(\pi x)\;dx;\) midpoint rule; \( n=3\)

6) Use the midpoint rule with eight subdivisions to estimate \( \displaystyle ∫^4_2x^2\;dx.\)

7) Use the trapezoidal rule with four subdivisions to estimate \( \displaystyle ∫^4_2x^2\;dx.\)

8) Find the exact value of \( \displaystyle ∫^4_2x^2\;dx.\) Find the error of approximation between the exact value and the value calculated using the trapezoidal rule with four subdivisions. Draw a graph to illustrate.

Approximate the integral to four decimal places using the indicated rule.

9) \( \displaystyle ∫^1_0\sin^2(\pi x)\;dx;\) trapezoidal rule; \( n=6\)

10) \( \displaystyle ∫^3_0\frac{1}{1+x^3}\;dx;\) trapezoidal rule; \( n=6\)

11) \( \displaystyle ∫^3_0\frac{1}{1+x^3}\;dx;\) Simpson’s rule; \( n=6\)

12) \( \displaystyle ∫^{0.8}_0e^{−x^2}\;dx;\) trapezoidal rule; \( n=4\)

13) \( \displaystyle ∫^{0.8}_0e^{−x^2}\;dx;\) Simpson’s rule; \( n=4\)

14) \(\displaystyle ∫^{0.4}_0\sin(x^2)\;dx;\) trapezoidal rule; \( n=4\)

15) \(\displaystyle ∫^{0.4}_0\sin(x^2)\;dx;\) Simpson’s rule; \( n=4\)

16) \( \displaystyle ∫^{0.5}_{0.1}\frac{\cos x}{x}\;dx;\) trapezoidal rule; \(n=4\)

17) \( \displaystyle ∫^{0.5}_{0.1}\frac{\cos x}{x}\;dx;\) Simpson’s rule; \(n=4\)

18) Evaluate \( \displaystyle ∫^1_0\frac{dx}{1+x^2}\) exactly and show that the result is \( π/4\). Then, find the approximate value of the integral using the trapezoidal rule with \( n=4\) subdivisions. Use the result to approximate the value of \( π\).

19) Approximate \( \displaystyle ∫^4_2\frac{1}{\ln x}\;dx\) using the midpoint rule with four subdivisions to four decimal places.

20) Approximate \( \displaystyle ∫^4_2\frac{1}{\ln x}\;dx\) using the trapezoidal rule with eight subdivisions to four decimal places.

21) Use the trapezoidal rule with four subdivisions to estimate \( \displaystyle ∫^{0.8}_0x^3\;dx\) to four decimal places.

22) Use the trapezoidal rule with four subdivisions to estimate \( \displaystyle ∫^{0.8}_0x^3\;dx.\) Compare this value with the exact value and find the error estimate.

23) Using Simpson’s rule with four subdivisions, find \( \displaystyle ∫^{π/2}_0\cos(x)\;dx.\)

24) Show that the exact value of \( \displaystyle ∫^1_0xe^{−x}\;dx=1−\frac{2}{e}\). Find the absolute error if you approximate the integral using the midpoint rule with 16 subdivisions.

25) Given \( \displaystyle ∫^1_0xe^{−x}\;dx=1−\frac{2}{e},\) use the trapezoidal rule with 16 subdivisions to approximate the integral and find the absolute error.

26) Find an upper bound for the error in estimating \( \displaystyle ∫^3_0(5x+4)\;dx\) using the trapezoidal rule with six steps.

27) Find an upper bound for the error in estimating \( \displaystyle ∫^5_4\frac{1}{(x−1)^2}\;dx\) using the trapezoidal rule with seven subdivisions.

28) Find an upper bound for the error in estimating \( \displaystyle ∫^3_0(6x^2−1)\;dx\) using Simpson’s rule with \( n=10\) steps.

29) Find an upper bound for the error in estimating \( \displaystyle ∫^5_2\frac{1}{x−1}\;dx\) using Simpson’s rule with \( n=10\) steps.

30) Find an upper bound for the error in estimating \( \displaystyle ∫^π_02x\cos(x)\;dx\) using Simpson’s rule with four steps.

31) Estimate the minimum number of subintervals needed to approximate the integral \( \displaystyle ∫^4_1(5x^2+8)\;dx\) with an error magnitude of less than 0.0001 using the trapezoidal rule.

32) Determine a value of n such that the trapezoidal rule will approximate \( \displaystyle ∫^1_0\sqrt{1+x^2}\;dx\) with an error of no more than 0.01.

33) Estimate the minimum number of subintervals needed to approximate the integral \( \displaystyle ∫^3_2(2x^3+4x)\;dx\) with an error of magnitude less than 0.0001 using the trapezoidal rule.

34) Estimate the minimum number of subintervals needed to approximate the integral \( \displaystyle ∫^4_3\frac{1}{(x−1)^2}\;dx\) with an error magnitude of less than 0.0001 using the trapezoidal rule.

35) Use Simpson’s rule with four subdivisions to approximate the area under the probability density function \( y=\frac{1}{\sqrt{2π}}e^{−x^2/2}\) from \( x=0\) to \( x=0.4\).

36) Use Simpson’s rule with \( n=14\) to approximate (to three decimal places) the area of the region bounded by the graphs of \( y=0, x=0,\) and \( x=π/2.\)

37) The length of one arch of the curve \( y=3\sin(2x)\) is given by \( L=∫^{π/2}_0\sqrt{1+36\cos^2(2x)}\;dx.\) Estimate L using the trapezoidal rule with \( n=6\).

38) The length of the ellipse \( x=a\cos(t),y=b\sin(t),0≤t≤2π\) is given by \( L=4a∫^{π/2}_0\sqrt{1−e^2\cos^2(t)}dt\), where e is the eccentricity of the ellipse. Use Simpson’s rule with \( n=6\) subdivisions to estimate the length of the ellipse when \( a=2\) and \( e=1/3.\)

39) Estimate the area of the surface generated by revolving the curve \( y=\cos(2x),0≤x≤\frac{π}{4}\) about the x-axis. Use the trapezoidal rule with six subdivisions.

40) Estimate the area of the surface generated by revolving the curve \( y=2x^2, 0≤x≤3\) about the x-axis. Use Simpson’s rule with \( n=6.\)

41) The growth rate of a certain tree (in feet) is given by \( y=\dfrac{2}{t+1}+e^{−t^2/2},\) where t is time in years. Estimate the growth of the tree through the end of the second year by using Simpson’s rule, using two subintervals. (Round the answer to the nearest hundredth.)

42) [T] Use a calculator to approximate \( \displaystyle ∫^1_0\sin(πx)\;dx\) using the midpoint rule with 25 subdivisions. Compute the relative error of approximation.

43) [T] Given \( \displaystyle ∫^5_1(3x^2−2x)\;dx=100,\) approximate the value of this integral using the midpoint rule with 16 subdivisions and determine the absolute error.

44) Given that we know the Fundamental Theorem of Calculus, why would we want to develop numerical methods for definite integrals?

45) The table represents the coordinates \( (x,y)\) that give the boundary of a lot. The units of measurement are meters. Use the trapezoidal rule to estimate the number of square meters of land that is in this lot.

46) Choose the correct answer. When Simpson’s rule is used to approximate the definite integral, it is necessary that the number of partitions be____

a. an even number

b. odd number

c. either an even or an odd number

d. a multiple of 4

47) The “Simpson” sum is based on the area under a ____.

48) The error formula for Simpson’s rule depends on___.

a. \( f(x)\)

b. \( f′(x)\)

c. \( f^{(4)}(x)\)

d. the number of steps

## Collapse menu

Introduction, 1 analytic geometry.

- 2. Distance Between Two Points; Circles
- 3. Functions
- 4. Shifts and Dilations

## 2 Instantaneous Rate of Change: The Derivative

- 1. The slope of a function
- 2. An example
- 4. The Derivative Function
- 5. Properties of Functions

## 3 Rules for Finding Derivatives

- 1. The Power Rule
- 2. Linearity of the Derivative
- 3. The Product Rule
- 4. The Quotient Rule
- 5. The Chain Rule

## 4 Trigonometric Functions

- 1. Trigonometric Functions
- 2. The Derivative of $\sin x$
- 3. A hard limit
- 4. The Derivative of $\sin x$, continued
- 5. Derivatives of the Trigonometric Functions
- 6. Implicit Differentiation
- 7. Limits revisited

## 5 Curve Sketching

- 1. Maxima and Minima
- 2. The first derivative test
- 3. The second derivative test
- 4. Concavity and inflection points
- 5. Asymptotes and Other Things to Look For

## 6 Applications of the Derivative

- 1. Optimization
- 2. Related Rates
- 3. Newton's Method
- 4. Linear Approximations
- 5. The Mean Value Theorem

## 7 Integration

- 1. Two examples
- 2. The Fundamental Theorem of Calculus
- 3. Some Properties of Integrals
- 4. Substitution

## 8 Applications of Integration

- 1. Area between curves
- 2. Distance, Velocity, Acceleration
- 4. Average value of a function

## 9 Transcendental Functions

- 1. Inverse functions
- 2. The natural logarithm
- 3. The exponential function
- 4. Other bases
- 5. Inverse Trigonometric Functions
- 6. Hyperbolic Functions

## 10 Techniques of Integration

- 1. Powers of sine and cosine
- 2. Trigonometric Substitutions
- 3. Integration by Parts
- 4. Rational Functions
- 5. Numerical Integration
- 6. Additional exercises

## 11 More Applications of Integration

- 1. Center of Mass
- 2. Kinetic energy; improper integrals
- 3. Probability
- 4. Arc Length
- 5. Surface Area

## 12 Polar Coordinates, Parametric Equations

- 1. Polar Coordinates
- 2. Slopes in polar coordinates
- 3. Areas in polar coordinates
- 4. Parametric Equations
- 5. Calculus with Parametric Equations

## 13 Sequences and Series

- 1. Sequences
- 3. The Integral Test
- 4. Alternating Series
- 5. Comparison Tests
- 6. Absolute Convergence
- 7. The Ratio and Root Tests
- 8. Power Series
- 9. Calculus with Power Series
- 10. Taylor Series
- 11. Taylor's Theorem
- 12. Additional exercises

## 14 Three Dimensions

- 1. The Coordinate System
- 3. The Dot Product
- 4. The Cross Product
- 5. Lines and Planes
- 6. Other Coordinate Systems

## 15 Vector Functions

- 1. Space Curves
- 2. Calculus with vector functions
- 3. Arc length and curvature
- 4. Motion along a curve

## 16 Partial Differentiation

- 1. Functions of Several Variables
- 2. Limits and Continuity
- 3. Partial Differentiation
- 4. The Chain Rule
- 5. Directional Derivatives
- 6. Higher order derivatives
- 7. Maxima and minima
- 8. Lagrange Multipliers

## 17 Multiple Integration

- 1. Volume and Average Height
- 2. Double Integrals in Cylindrical Coordinates
- 3. Moment and Center of Mass
- 4. Surface Area
- 5. Triple Integrals
- 6. Cylindrical and Spherical Coordinates
- 7. Change of Variables

## 18 Vector Calculus

- 1. Vector Fields
- 2. Line Integrals
- 3. The Fundamental Theorem of Line Integrals
- 4. Green's Theorem
- 5. Divergence and Curl
- 6. Vector Functions for Surfaces
- 7. Surface Integrals
- 8. Stokes's Theorem
- 9. The Divergence Theorem

## 19 Differential Equations

- 1. First Order Differential Equations
- 2. First Order Homogeneous Linear Equations
- 3. First Order Linear Equations
- 4. Approximation
- 5. Second Order Homogeneous Equations
- 6. Second Order Linear Equations
- 7. Second Order Linear Equations, take two

## 20 Useful formulas

21 introduction to sage.

- 2. Differentiation
- 3. Integration

We have now seen some of the most generally useful methods for discovering antiderivatives, and there are others. Unfortunately, some functions have no simple antiderivatives; in such cases if the value of a definite integral is needed it will have to be approximated. We will see two methods that work reasonably well and yet are fairly simple; in some cases more sophisticated techniques will be needed.

Of course, we already know one way to approximate an integral: if we think of the integral as computing an area, we can add up the areas of some rectangles. While this is quite simple, it is usually the case that a large number of rectangles is needed to get acceptable accuracy. A similar approach is much better: we approximate the area under a curve over a small interval as the area of a trapezoid. In figure 10.5.1 we see an area under a curve approximated by rectangles and by trapezoids; it is apparent that the trapezoids give a substantially better approximation on each subinterval.

As with rectangles, we divide the interval into $n$ equal subintervals of length $\Delta x$. A typical trapezoid is pictured in figure 10.5.2 ; it has area $\ds{f(x_i)+f(x_{i+1})\over2}\Delta x$. If we add up the areas of all trapezoids we get $$ \eqalign{ {f(x_0)+f(x_1)\over2}\Delta x&+{f(x_1)+f(x_2)\over2}\Delta x+\cdots+ {f(x_{n-1})+f(x_n)\over2}\Delta x=\cr &\left({f(x_0)\over2}+f(x_1)+f(x_2)+\cdots+f(x_{n-1})+{f(x_n)\over2}\right) \Delta x.\cr} $$ This is usually known as the Trapezoid Rule . For a modest number of subintervals this is not too difficult to do with a calculator; a computer can easily do many subintervals.

In practice, an approximation is useful only if we know how accurate it is; for example, we might need a particular value accurate to three decimal places. When we compute a particular approximation to an integral, the error is the difference between the approximation and the true value of the integral. For any approximation technique, we need an error estimate , a value that is guaranteed to be larger than the actual error. If $A$ is an approximation and $E$ is the associated error estimate, then we know that the true value of the integral is between $A-E$ and $A+E$. In the case of our approximation of the integral, we want $E=E(\Delta x)$ to be a function of $\Delta x$ that gets small rapidly as $\Delta x$ gets small. Fortunately, for many functions, there is such an error estimate associated with the trapezoid approximation.

Theorem 10.5.1 Suppose $f$ has a second derivative $f''$ everywhere on the interval $[a,b]$, and $|f''(x)|\le M$ for all $x$ in the interval. With $\Delta x= (b-a)/n$, an error estimate for the trapezoid approximation is $$ E(\Delta x) = {b-a\over12}M(\Delta x)^2={(b-a)^3\over 12n^2}M. $$ $\qed$

Let's see how we can use this.

Example 10.5.2 Approximate $\ds\int_0^1 e^{-x^2}\,dx$ to two decimal places. The second derivative of $\ds f=e^{-x^2}$ is $\ds(4x^2-2)e^{-x^2}$, and it is not hard to see that on $[0,1]$, $\ds|(4x^2-2)e^{-x^2}|\le 2$. We begin by estimating the number of subintervals we are likely to need. To get two decimal places of accuracy, we will certainly need $E(\Delta x)< 0.005$ or $$ \eqalign{ {1\over12}(2){1\over n^2} &< 0.005\cr {1\over6}(200)&< n^2\cr 5.77\approx\sqrt{100\over3}&< n\cr} $$ With $n=6$, the error estimate is thus $\ds1/6^3< 0.0047$. We compute the trapezoid approximation for six intervals: $$ \left({f(0)\over2}+f(1/6)+f(2/6)+\cdots+f(5/6)+{f(1)\over2}\right){1\over6} \approx 0.74512. $$ So the true value of the integral is between $0.74512-0.0047=0.74042$ and $0.74512+0.0047=0.74982$. Unfortunately, the first rounds to $0.74$ and the second rounds to $0.75$, so we can't be sure of the correct value in the second decimal place; we need to pick a larger $n$. As it turns out, we need to go to $n=12$ to get two bounds that both round to the same value, which turns out to be $0.75$. For comparison, using $12$ rectangles to approximate the area gives $0.7727$, which is considerably less accurate than the approximation using six trapezoids.

In practice it generally pays to start by requiring better than the maximum possible error; for example, we might have initially required $E(\Delta x)< 0.001$, or $$ \eqalign{ {1\over12}(2){1\over n^2} &< 0.001\cr {1\over6}(1000)&< n^2\cr 12.91\approx\sqrt{500\over3}&< n\cr} $$ Had we immediately tried $n=13$ this would have given us the desired answer. $\square$

The trapezoid approximation works well, especially compared to rectangles, because the tops of the trapezoids form a reasonably good approximation to the curve when $\Delta x$ is fairly small. We can extend this idea: what if we try to approximate the curve more closely, by using something other than a straight line? The obvious candidate is a parabola: if we can approximate a short piece of the curve with a parabola with equation $\ds y=ax^2+bx+c$, we can easily compute the area under the parabola.

There are an infinite number of parabolas through any two given points, but only one through three given points. If we find a parabola through three consecutive points $(x_i,f(x_i))$, $(x_{i+1},f(x_{i+1}))$, $(x_{i+2},f(x_{i+2}))$ on the curve, it should be quite close to the curve over the whole interval $[x_i,x_{i+2}]$, as in figure 10.5.3 . If we divide the interval $[a,b]$ into an even number of subintervals, we can then approximate the curve by a sequence of parabolas, each covering two of the subintervals. For this to be practical, we would like a simple formula for the area under one parabola, namely, the parabola through $(x_i,f(x_i))$, $(x_{i+1},f(x_{i+1}))$, and $(x_{i+2},f(x_{i+2}))$. That is, we should attempt to write down the parabola $y=ax^2+bx+c$ through these points and then integrate it, and hope that the result is fairly simple. Although the algebra involved is messy, this turns out to be possible. The algebra is well within the capability of a good computer algebra system like Sage, so we will present the result without all of the algebra; you can see how to do it in this Sage worksheet.

To find the parabola, we solve these three equations for $a$, $b$, and $c$: $$ \eqalign{ f(x_i)&=a(x_{i+1}-\Delta x)^2+b(x_{i+1}-\Delta x)+c\cr f(x_{i+1})&=a(x_{i+1})^2+b(x_{i+1})+c\cr f(x_{i+2})&=a(x_{i+1}+\Delta x)^2+b(x_{i+1}+\Delta x)+c\cr} $$ Not surprisingly, the solutions turn out to be quite messy. Nevertheless, Sage can easily compute and simplify the integral to get $$ \int_{x_{i+1}-\Delta x}^{x_{i+1}+\Delta x} ax^2+bx+c\,dx= {\Delta x\over3}(f(x_i)+4f(x_{i+1})+f(x_{i+2})). $$ Now the sum of the areas under all parabolas is $$ \displaylines{ {\Delta x\over3}(f(x_0)+4f(x_{1})+f(x_{2})+f(x_2)+4f(x_{3})+f(x_{4})+\cdots +f(x_{n-2})+4f(x_{n-1})+f(x_{n}))=\cr {\Delta x\over3}(f(x_0)+4f(x_{1})+2f(x_{2})+4f(x_{3})+2f(x_{4})+\cdots +2f(x_{n-2})+4f(x_{n-1})+f(x_{n})).\cr} $$ This is just slightly more complicated than the formula for trapezoids; we need to remember the alternating 2 and 4 coefficients; note that $n$ must be even for this to make sense. This approximation technique is referred to as Simpson's Rule .

As with the trapezoid method, this is useful only with an error estimate:

Theorem 10.5.3 Suppose $f$ has a fourth derivative $f^{(4)}$ everywhere on the interval $[a,b]$, and $|f^{(4)}(x)|\le M$ for all $x$ in the interval. With $\Delta x= (b-a)/n$, an error estimate for Simpson's approximation is $$ E(\Delta x) = {b-a\over180}M(\Delta x)^4={(b-a)^5\over 180n^4}M. $$ $\qed$

Example 10.5.4 Let us again approximate $\ds\int_0^1 e^{-x^2}\,dx$ to two decimal places. The fourth derivative of $\ds f=e^{-x^2}$ is $\ds(16x^4-48x^2+12)e^{-x^2}$; on $[0,1]$ this is at most $12$ in absolute value. We begin by estimating the number of subintervals we are likely to need. To get two decimal places of accuracy, we will certainly need $E(\Delta x)< 0.005$, but taking a cue from our earlier example, let's require $E(\Delta x)< 0.001$: $$ \eqalign{ {1\over180}(12){1\over n^4} &< 0.001\cr {200\over3}&< n^4\cr 2.86\approx\root 4 \of {200\over3}&< n\cr} $$ So we try $n=4$, since we need an even number of subintervals. Then the error estimate is $\ds12/180/4^4< 0.0003$ and the approximation is $$ (f(0)+4f(1/4)+2f(1/2)+4f(3/4)+f(1)){1\over3\cdot4} \approx 0.746855. $$ So the true value of the integral is between $0.746855-0.0003=0.746555$ and $0.746855+0.0003=0.7471555$, both of which round to $0.75$. $\square$

The figure below compares the three methods we have discussed, computing the area under $y=\sin x$, $0\le x\le \pi/2$. Of course, it's easy to compute this exactly: the area is $1$.

## Exercises 10.5

In the following problems, compute the trapezoid and Simpson approximations using 4 subintervals, and compute the error estimate for each. (Finding the maximum values of the second and fourth derivatives can be challenging for some of these; you may use a graphing calculator or computer software to estimate the maximum values.) If you have access to Sage or similar software, approximate each integral to two decimal places. You can use this Sage worksheet to get started.

Ex 10.5.1 $\ds\int_1^3 x\,dx$ ( answer )

Ex 10.5.2 $\ds\int_0^3 x^2\,dx$ ( answer )

Ex 10.5.3 $\ds\int_2^4 x^3\,dx$ ( answer )

Ex 10.5.4 $\ds\int_1^3 {1\over x}\,dx$ ( answer )

Ex 10.5.5 $\ds\int_1^2 {1\over 1+x^2}\,dx$ ( answer )

Ex 10.5.6 $\ds\int_0^1 x\sqrt{1+x}\,dx$ ( answer )

Ex 10.5.7 $\ds\int_1^5 {x\over 1+x}\,dx$ ( answer )

Ex 10.5.8 $\ds\int_0^1 \sqrt{x^3+1}\,dx$ ( answer )

Ex 10.5.9 $\ds\int_0^1 \sqrt{x^4+1}\,dx$ ( answer )

Ex 10.5.10 $\ds\int_1^4 \sqrt{1+1/x}\,dx$ ( answer )

Ex 10.5.11 Using Simpson's rule on a parabola $f(x)$, even with just two subintervals, gives the exact value of the integral, because the parabolas used to approximate $f$ will be $f$ itself. Remarkably, Simpson's rule also computes the integral of a cubic function $f(x)=ax^3+bx^2+cx+d$ exactly. Show this is true by showing that $$ \int_{x_0}^{x_2} f(x)\,dx={x_2-x_0\over3\cdot2}(f(x_0)+4f((x_0+x_2)/2)+f(x_2)). $$ Note that the right hand side of this equation is exactly the Simpson approximation for the cubic. This does require a bit of messy algebra, so you may prefer to use this Sage worksheet.

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## Section 7.10 : Approximating Definite Integrals

For each of the following integrals use the given value of n to approximate the value of the definite integral using

- the Midpoint Rule,
- the Trapezoid Rule, and
- Simpson’s Rule.

Use at least 6 decimal places of accuracy for your work.

- \( \displaystyle \int_{1}^{7}{{\frac{1}{{{x^3} + 1}}\,dx}}\) using \(n = 6\) Solution
- \( \displaystyle \int_{{ - 1}}^{2}{{\sqrt {{{\bf{e}}^{ - \,{x^{\,2}}}} + 1} \,dx}}\) using \(n = 6\) Solution
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## Solved Problems on Numerical Integration. Integration/Integration Techniques/Solved Problems on Numerical Integration by M. Seppälä Review of the Subject.

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## Presentation on theme: "Solved Problems on Numerical Integration. Integration/Integration Techniques/Solved Problems on Numerical Integration by M. Seppälä Review of the Subject."— Presentation transcript:

Section 7.6 – Numerical Integration

1 When you see… Find the zeros You think…. 2 To find the zeros...

When you see… Find the zeros You think…. To find the zeros...

Numerical Integration

Adguary Calwile Laura Rogers Autrey~ 2nd Per. 3/14/11

6. 4 Integration with tables and computer algebra systems 6

§ 6.2 Areas and Riemann Sums. Area Under a Graph Riemann Sums to Approximate Areas (Midpoints) Riemann Sums to Approximate Areas (Left Endpoints) Applications.

When you see… Find the zeros You think….

Section 8.5 Riemann Sums and the Definite Integral.

Section 5.7 Numerical Integration. Approximations for integrals: Riemann Sums, Trapezoidal Rule, Simpson's Rule Riemann Sum: Trapezoidal Rule: Simpson’s.

Copyright © Cengage Learning. All rights reserved. 5 Integrals.

1 Example 2 Estimate by the six Rectangle Rules using the regular partition P of the interval [0, ] into 6 subintervals. Solution Observe that the function.

Index FAQ Numerical Approximations of Definite Integrals Riemann Sums Numerical Approximation of Definite Integrals Formulae for Approximations Properties.

5.2 Definite Integrals Quick Review Quick Review Solutions.

Riemann Sums and the Definite Integral Lesson 5.3.

Integrals 5.

1 Chapter 5 Numerical Integration. 2 A Review of the Definite Integral.

MAT 1235 Calculus II Section 7.7 Approximate (Numerical) Integration

1 Simpson’s 1/3 rd Rule of Integration. 2 What is Integration? Integration The process of measuring the area under a curve. Where: f(x) is the integrand.

## About project

© 2023 SlidePlayer.com Inc. All rights reserved.

- 3.6 Numerical Integration
- Introduction
- 1.1 Approximating Areas
- 1.2 The Definite Integral
- 1.3 The Fundamental Theorem of Calculus
- 1.4 Integration Formulas and the Net Change Theorem
- 1.5 Substitution
- 1.6 Integrals Involving Exponential and Logarithmic Functions
- 1.7 Integrals Resulting in Inverse Trigonometric Functions
- Key Equations
- Key Concepts
- Review Exercises
- 2.1 Areas between Curves
- 2.2 Determining Volumes by Slicing
- 2.3 Volumes of Revolution: Cylindrical Shells
- 2.4 Arc Length of a Curve and Surface Area
- 2.5 Physical Applications
- 2.6 Moments and Centers of Mass
- 2.7 Integrals, Exponential Functions, and Logarithms
- 2.8 Exponential Growth and Decay
- 2.9 Calculus of the Hyperbolic Functions
- 3.1 Integration by Parts
- 3.2 Trigonometric Integrals
- 3.3 Trigonometric Substitution
- 3.4 Partial Fractions
- 3.5 Other Strategies for Integration
- 3.7 Improper Integrals
- 4.1 Basics of Differential Equations
- 4.2 Direction Fields and Numerical Methods
- 4.3 Separable Equations
- 4.4 The Logistic Equation
- 4.5 First-order Linear Equations
- 5.1 Sequences
- 5.2 Infinite Series
- 5.3 The Divergence and Integral Tests
- 5.4 Comparison Tests
- 5.5 Alternating Series
- 5.6 Ratio and Root Tests
- 6.1 Power Series and Functions
- 6.2 Properties of Power Series
- 6.3 Taylor and Maclaurin Series
- 6.4 Working with Taylor Series
- 7.1 Parametric Equations
- 7.2 Calculus of Parametric Curves
- 7.3 Polar Coordinates
- 7.4 Area and Arc Length in Polar Coordinates
- 7.5 Conic Sections
- A | Table of Integrals
- B | Table of Derivatives
- C | Review of Pre-Calculus

## Learning Objectives

- 3.6.1 Approximate the value of a definite integral by using the midpoint and trapezoidal rules.
- 3.6.2 Determine the absolute and relative error in using a numerical integration technique.
- 3.6.3 Estimate the absolute and relative error using an error-bound formula.
- 3.6.4 Recognize when the midpoint and trapezoidal rules over- or underestimate the true value of an integral.
- 3.6.5 Use Simpson’s rule to approximate the value of a definite integral to a given accuracy.

The antiderivatives of many functions either cannot be expressed or cannot be expressed easily in closed form (that is, in terms of known functions). Consequently, rather than evaluate definite integrals of these functions directly, we resort to various techniques of numerical integration to approximate their values. In this section we explore several of these techniques. In addition, we examine the process of estimating the error in using these techniques.

## The Midpoint Rule

Earlier in this text we defined the definite integral of a function over an interval as the limit of Riemann sums . In general, any Riemann sum of a function f ( x ) f ( x ) over an interval [ a , b ] [ a , b ] may be viewed as an estimate of ∫ a b f ( x ) d x . ∫ a b f ( x ) d x . Recall that a Riemann sum of a function f ( x ) f ( x ) over an interval [ a , b ] [ a , b ] is obtained by selecting a partition

The Riemann sum corresponding to the partition P P and the set S S is given by ∑ i = 1 n f ( x i * ) Δ x i , ∑ i = 1 n f ( x i * ) Δ x i , where Δ x i = x i − x i − 1 , Δ x i = x i − x i − 1 , the length of the i th subinterval.

The midpoint rule for estimating a definite integral uses a Riemann sum with subintervals of equal width and the midpoints, m i , m i , of each subinterval in place of x i * . x i * . Formally, we state a theorem regarding the convergence of the midpoint rule as follows.

## Theorem 3.3

Assume that f ( x ) f ( x ) is continuous on [ a , b ] . [ a , b ] . Let n be a positive integer and Δ x = b − a n . Δ x = b − a n . If [ a , b ] [ a , b ] is divided into n n subintervals, each of length Δ x , Δ x , and m i m i is the midpoint of the i th subinterval, set

Then lim n → ∞ M n = ∫ a b f ( x ) d x . lim n → ∞ M n = ∫ a b f ( x ) d x .

As we can see in Figure 3.13 , if f ( x ) ≥ 0 f ( x ) ≥ 0 over [ a , b ] , [ a , b ] , then ∑ i = 1 n f ( m i ) Δ x ∑ i = 1 n f ( m i ) Δ x corresponds to the sum of the areas of rectangles approximating the area between the graph of f ( x ) f ( x ) and the x -axis over [ a , b ] . [ a , b ] . The graph shows the rectangles corresponding to M 4 M 4 for a nonnegative function over a closed interval [ a , b ] . [ a , b ] .

## Example 3.39

Using the midpoint rule with m 4 m 4.

Use the midpoint rule to estimate ∫ 0 1 x 2 d x ∫ 0 1 x 2 d x using four subintervals. Compare the result with the actual value of this integral.

Each subinterval has length Δ x = 1 − 0 4 = 1 4 . Δ x = 1 − 0 4 = 1 4 . Therefore, the subintervals consist of

The midpoints of these subintervals are { 1 8 , 3 8 , 5 8 , 7 8 } . { 1 8 , 3 8 , 5 8 , 7 8 } . Thus,

we see that the midpoint rule produces an estimate that is somewhat close to the actual value of the definite integral.

## Example 3.40

Using the midpoint rule with m 6 m 6.

Use M 6 M 6 to estimate the length of the curve y = 1 2 x 2 y = 1 2 x 2 on [ 1 , 4 ] . [ 1 , 4 ] .

The length of y = 1 2 x 2 y = 1 2 x 2 on [ 1 , 4 ] [ 1 , 4 ] is

Since d y d x = x , d y d x = x , this integral becomes ∫ 1 4 1 + x 2 d x . ∫ 1 4 1 + x 2 d x .

If [ 1 , 4 ] [ 1 , 4 ] is divided into six subintervals, then each subinterval has length Δ x = 4 − 1 6 = 1 2 Δ x = 4 − 1 6 = 1 2 and the midpoints of the subintervals are { 5 4 , 7 4 , 9 4 , 11 4 , 13 4 , 15 4 } . { 5 4 , 7 4 , 9 4 , 11 4 , 13 4 , 15 4 } . If we set f ( x ) = 1 + x 2 , f ( x ) = 1 + x 2 ,

## Checkpoint 3.22

Use the midpoint rule with n = 2 n = 2 to estimate ∫ 1 2 1 x d x . ∫ 1 2 1 x d x .

## The Trapezoidal Rule

We can also approximate the value of a definite integral by using trapezoids rather than rectangles. In Figure 3.14 , the area beneath the curve is approximated by trapezoids rather than by rectangles.

The trapezoidal rule for estimating definite integrals uses trapezoids rather than rectangles to approximate the area under a curve. To gain insight into the final form of the rule, consider the trapezoids shown in Figure 3.14 . We assume that the length of each subinterval is given by Δ x . Δ x . First, recall that the area of a trapezoid with a height of h and bases of length b 1 b 1 and b 2 b 2 is given by Area = 1 2 h ( b 1 + b 2 ) . Area = 1 2 h ( b 1 + b 2 ) . We see that the first trapezoid has a height Δ x Δ x and parallel bases of length f ( x 0 ) f ( x 0 ) and f ( x 1 ) . f ( x 1 ) . Thus, the area of the first trapezoid in Figure 3.14 is

The areas of the remaining three trapezoids are

Consequently,

After taking out a common factor of 1 2 Δ x 1 2 Δ x and combining like terms, we have

Generalizing, we formally state the following rule.

## Theorem 3.4

Assume that f ( x ) f ( x ) is continuous over [ a , b ] . [ a , b ] . Let n be a positive integer and Δ x = b − a n . Δ x = b − a n . Let [ a , b ] [ a , b ] be divided into n n subintervals, each of length Δ x , Δ x , with endpoints at P = { x 0 , x 1 , x 2 … , x n } . P = { x 0 , x 1 , x 2 … , x n } . Set

Then, lim n → + ∞ T n = ∫ a b f ( x ) d x . lim n → + ∞ T n = ∫ a b f ( x ) d x .

Before continuing, let’s make a few observations about the trapezoidal rule. First of all, it is useful to note that

That is, L n L n and R n R n approximate the integral using the left-hand and right-hand endpoints of each subinterval, respectively. In addition, a careful examination of Figure 3.15 leads us to make the following observations about using the trapezoidal rules and midpoint rules to estimate the definite integral of a nonnegative function. The trapezoidal rule tends to overestimate the value of a definite integral systematically over intervals where the function is concave up and to underestimate the value of a definite integral systematically over intervals where the function is concave down. On the other hand, the midpoint rule tends to average out these errors somewhat by partially overestimating and partially underestimating the value of the definite integral over these same types of intervals. This leads us to hypothesize that, in general, the midpoint rule tends to be more accurate than the trapezoidal rule.

## Example 3.41

Using the trapezoidal rule.

Use the trapezoidal rule to estimate ∫ 0 1 x 2 d x ∫ 0 1 x 2 d x using four subintervals.

The endpoints of the subintervals consist of elements of the set P = { 0 , 1 4 , 1 2 , 3 4 , 1 } P = { 0 , 1 4 , 1 2 , 3 4 , 1 } and Δ x = 1 − 0 4 = 1 4 . Δ x = 1 − 0 4 = 1 4 . Thus,

## Checkpoint 3.23

Use the trapezoidal rule with n = 2 n = 2 to estimate ∫ 1 2 1 x d x . ∫ 1 2 1 x d x .

## Absolute and Relative Error

An important aspect of using these numerical approximation rules consists of calculating the error in using them for estimating the value of a definite integral. We first need to define absolute error and relative error .

If B B is our estimate of some quantity having an actual value of A , A , then the absolute error is given by | A − B | . | A − B | . The relative error is the error as a percentage of the absolute value and is given by | A − B A | = | A − B A | · 100 % . | A − B A | = | A − B A | · 100 % .

## Example 3.42

Calculating error in the midpoint rule.

Calculate the absolute and relative error in the estimate of ∫ 0 1 x 2 d x ∫ 0 1 x 2 d x using the midpoint rule, found in Example 3.39 .

The calculated value is ∫ 0 1 x 2 d x = 1 3 ∫ 0 1 x 2 d x = 1 3 and our estimate from the example is M 4 = 21 64 . M 4 = 21 64 . Thus, the absolute error is given by | ( 1 3 ) − ( 21 64 ) | = 1 192 ≈ 0.0052 . | ( 1 3 ) − ( 21 64 ) | = 1 192 ≈ 0.0052 . The relative error is

## Example 3.43

Calculating error in the trapezoidal rule.

Calculate the absolute and relative error in the estimate of ∫ 0 1 x 2 d x ∫ 0 1 x 2 d x using the trapezoidal rule, found in Example 3.41 .

The calculated value is ∫ 0 1 x 2 d x = 1 3 ∫ 0 1 x 2 d x = 1 3 and our estimate from the example is T 4 = 11 32 . T 4 = 11 32 . Thus, the absolute error is given by | 1 3 − 11 32 | = 1 96 ≈ 0.0104 . | 1 3 − 11 32 | = 1 96 ≈ 0.0104 . The relative error is given by

## Checkpoint 3.24

In an earlier checkpoint, we estimated ∫ 1 2 1 x d x ∫ 1 2 1 x d x to be 24 35 24 35 using T 2 . T 2 . The actual value of this integral is ln 2 . ln 2 . Using 24 35 ≈ 0.6857 24 35 ≈ 0.6857 and ln 2 ≈ 0.6931 , ln 2 ≈ 0.6931 , calculate the absolute error and the relative error.

In the two previous examples, we were able to compare our estimate of an integral with the actual value of the integral; however, we do not typically have this luxury. In general, if we are approximating an integral, we are doing so because we cannot compute the exact value of the integral itself easily. Therefore, it is often helpful to be able to determine an upper bound for the error in an approximation of an integral. The following theorem provides error bounds for the midpoint and trapezoidal rules. The theorem is stated without proof.

## Theorem 3.5

Error bounds for the midpoint and trapezoidal rules.

Let f ( x ) f ( x ) be a continuous function over [ a , b ] , [ a , b ] , having a second derivative f ″ ( x ) f ″ ( x ) over this interval. If M M is the maximum value of | f ″ ( x ) | | f ″ ( x ) | over [ a , b ] , [ a , b ] , then the upper bounds for the error in using M n M n and T n T n to estimate ∫ a b f ( x ) d x ∫ a b f ( x ) d x are

We can use these bounds to determine the value of n n necessary to guarantee that the error in an estimate is less than a specified value.

## Example 3.44

Determining the number of intervals to use.

What value of n n should be used to guarantee that an estimate of ∫ 0 1 e x 2 d x ∫ 0 1 e x 2 d x is accurate to within 0.01 if we use the midpoint rule?

We begin by determining the value of M , M , the maximum value of | f ″ ( x ) | | f ″ ( x ) | over [ 0 , 1 ] [ 0 , 1 ] for f ( x ) = e x 2 . f ( x ) = e x 2 . Since f ′ ( x ) = 2 x e x 2 , f ′ ( x ) = 2 x e x 2 , we have

From the error-bound Equation 3.12 , we have

Now we solve the following inequality for n : n :

Thus, n ≥ 600 e 24 ≈ 8.24 . n ≥ 600 e 24 ≈ 8.24 . Since n n must be an integer satisfying this inequality, a choice of n = 9 n = 9 would guarantee that | ∫ 0 1 e x 2 d x − M n | < 0.01 . | ∫ 0 1 e x 2 d x − M n | < 0.01 .

We might have been tempted to round 8.24 8.24 down and choose n = 8 , n = 8 , but this would be incorrect because we must have an integer greater than or equal to 8.24 . 8.24 . We need to keep in mind that the error estimates provide an upper bound only for the error. The actual estimate may, in fact, be a much better approximation than is indicated by the error bound.

## Checkpoint 3.25

Use Equation 3.13 to find an upper bound for the error in using M 4 M 4 to estimate ∫ 0 1 x 2 d x . ∫ 0 1 x 2 d x .

## Simpson’s Rule

With the midpoint rule, we estimated areas of regions under curves by using rectangles. In a sense, we approximated the curve with piecewise constant functions. With the trapezoidal rule, we approximated the curve by using piecewise linear functions. What if we were, instead, to approximate a curve using piecewise quadratic functions? With Simpson’s rule , we do just this. We partition the interval into an even number of subintervals, each of equal width. Over the first pair of subintervals we approximate ∫ x 0 x 2 f ( x ) d x ∫ x 0 x 2 f ( x ) d x with ∫ x 0 x 2 p ( x ) d x , ∫ x 0 x 2 p ( x ) d x , where p ( x ) = A x 2 + B x + C p ( x ) = A x 2 + B x + C is the quadratic function passing through ( x 0 , f ( x 0 ) ) , ( x 0 , f ( x 0 ) ) , ( x 1 , f ( x 1 ) ) , ( x 1 , f ( x 1 ) ) , and ( x 2 , f ( x 2 ) ) ( x 2 , f ( x 2 ) ) ( Figure 3.16 ). Over the next pair of subintervals we approximate ∫ x 2 x 4 f ( x ) d x ∫ x 2 x 4 f ( x ) d x with the integral of another quadratic function passing through ( x 2 , f ( x 2 ) ) , ( x 2 , f ( x 2 ) ) , ( x 3 , f ( x 3 ) ) , ( x 3 , f ( x 3 ) ) , and ( x 4 , f ( x 4 ) ) . ( x 4 , f ( x 4 ) ) . This process is continued with each successive pair of subintervals.

To understand the formula that we obtain for Simpson’s rule, we begin by deriving a formula for this approximation over the first two subintervals. As we go through the derivation, we need to keep in mind the following relationships:

x 2 − x 0 = 2 Δ x , x 2 − x 0 = 2 Δ x , where Δ x Δ x is the length of a subinterval.

If we approximate ∫ x 2 x 4 f ( x ) d x ∫ x 2 x 4 f ( x ) d x using the same method, we see that we have

Combining these two approximations, we get

The pattern continues as we add pairs of subintervals to our approximation. The general rule may be stated as follows.

## Theorem 3.6

Assume that f ( x ) f ( x ) is continuous over [ a , b ] . [ a , b ] . Let n be a positive even integer and Δ x = b − a n . Δ x = b − a n . Let [ a , b ] [ a , b ] be divided into n n subintervals, each of length Δ x , Δ x , with endpoints at P = { x 0 , x 1 , x 2 ,… , x n } . P = { x 0 , x 1 , x 2 ,… , x n } . Set

Just as the trapezoidal rule is the average of the left-hand and right-hand rules for estimating definite integrals, Simpson’s rule may be obtained from the midpoint and trapezoidal rules by using a weighted average. It can be shown that S 2 n = ( 2 3 ) M n + ( 1 3 ) T n . S 2 n = ( 2 3 ) M n + ( 1 3 ) T n .

It is also possible to put a bound on the error when using Simpson’s rule to approximate a definite integral. The bound in the error is given by the following rule:

## Rule: Error Bound for Simpson’s Rule

Let f ( x ) f ( x ) be a continuous function over [ a , b ] [ a , b ] having a fourth derivative, f ( 4 ) ( x ) , f ( 4 ) ( x ) , over this interval. If M M is the maximum value of | f ( 4 ) ( x ) | | f ( 4 ) ( x ) | over [ a , b ] , [ a , b ] , then the upper bound for the error in using S n S n to estimate ∫ a b f ( x ) d x ∫ a b f ( x ) d x is given by

## Example 3.45

Applying simpson’s rule 1.

Use S 2 S 2 to approximate ∫ 0 1 x 3 d x . ∫ 0 1 x 3 d x . Estimate a bound for the error in S 2 . S 2 .

Since [ 0 , 1 ] [ 0 , 1 ] is divided into two intervals, each subinterval has length Δ x = 1 − 0 2 = 1 2 . Δ x = 1 − 0 2 = 1 2 . The endpoints of these subintervals are { 0 , 1 2 , 1 } . { 0 , 1 2 , 1 } . If we set f ( x ) = x 3 , f ( x ) = x 3 , then

S 4 = 1 3 · 1 2 ( f ( 0 ) + 4 f ( 1 2 ) + f ( 1 ) ) = 1 6 ( 0 + 4 · 1 8 + 1 ) = 1 4 . S 4 = 1 3 · 1 2 ( f ( 0 ) + 4 f ( 1 2 ) + f ( 1 ) ) = 1 6 ( 0 + 4 · 1 8 + 1 ) = 1 4 . Since f ( 4 ) ( x ) = 0 f ( 4 ) ( x ) = 0 and consequently M = 0 , M = 0 , we see that

This bound indicates that the value obtained through Simpson’s rule is exact. A quick check will verify that, in fact, ∫ 0 1 x 3 d x = 1 4 . ∫ 0 1 x 3 d x = 1 4 .

## Example 3.46

Applying simpson’s rule 2.

Use S 6 S 6 to estimate the length of the curve y = 1 2 x 2 y = 1 2 x 2 over [ 1 , 4 ] . [ 1 , 4 ] .

The length of y = 1 2 x 2 y = 1 2 x 2 over [ 1 , 4 ] [ 1 , 4 ] is ∫ 1 4 1 + x 2 d x . ∫ 1 4 1 + x 2 d x . If we divide [ 1 , 4 ] [ 1 , 4 ] into six subintervals, then each subinterval has length Δ x = 4 − 1 6 = 1 2 , Δ x = 4 − 1 6 = 1 2 , and the endpoints of the subintervals are { 1 , 3 2 , 2 , 5 2 , 3 , 7 2 , 4 } . { 1 , 3 2 , 2 , 5 2 , 3 , 7 2 , 4 } . Setting f ( x ) = 1 + x 2 , f ( x ) = 1 + x 2 ,

After substituting, we have

## Checkpoint 3.26

Use S 2 S 2 to estimate ∫ 1 2 1 x d x . ∫ 1 2 1 x d x .

## Section 3.6 Exercises

Approximate the following integrals using either the midpoint rule, trapezoidal rule, or Simpson’s rule as indicated. (Round answers to three decimal places.)

∫ 1 2 d x x ; ∫ 1 2 d x x ; trapezoidal rule; n = 5 n = 5

∫ 0 3 4 + x 3 d x ; ∫ 0 3 4 + x 3 d x ; trapezoidal rule; n = 6 n = 6

∫ 0 3 4 + x 3 d x ; ∫ 0 3 4 + x 3 d x ; trapezoidal rule; n = 3 n = 3

∫ 0 12 x 2 d x ; ∫ 0 12 x 2 d x ; midpoint rule; n = 6 n = 6

∫ 0 1 sin 2 ( π x ) d x ; ∫ 0 1 sin 2 ( π x ) d x ; midpoint rule; n = 3 n = 3

Use the midpoint rule with eight subdivisions to estimate ∫ 2 4 x 2 d x . ∫ 2 4 x 2 d x .

Use the trapezoidal rule with four subdivisions to estimate ∫ 2 4 x 2 d x . ∫ 2 4 x 2 d x .

Find the exact value of ∫ 2 4 x 2 d x . ∫ 2 4 x 2 d x . Find the error of approximation between the exact value and the value calculated using the trapezoidal rule with four subdivisions. Draw a graph to illustrate.

Approximate the integral to three decimal places using the indicated rule.

∫ 0 1 sin 2 ( π x ) d x ; ∫ 0 1 sin 2 ( π x ) d x ; trapezoidal rule; n = 6 n = 6

∫ 0 3 1 1 + x 3 d x ; ∫ 0 3 1 1 + x 3 d x ; trapezoidal rule; n = 6 n = 6

∫ 0 3 1 1 + x 3 d x ; ∫ 0 3 1 1 + x 3 d x ; trapezoidal rule; n = 3 n = 3

∫ 0 0.8 e − x 2 d x ; ∫ 0 0.8 e − x 2 d x ; trapezoidal rule; n = 4 n = 4

∫ 0 0.8 e − x 2 d x ; ∫ 0 0.8 e − x 2 d x ; Simpson’s rule; n = 4 n = 4

∫ 0 0.4 sin ( x 2 ) d x ; ∫ 0 0.4 sin ( x 2 ) d x ; trapezoidal rule; n = 4 n = 4

∫ 0 0.4 sin ( x 2 ) d x ; ∫ 0 0.4 sin ( x 2 ) d x ; Simpson’s rule; n = 4 n = 4

∫ 0.1 0.5 cos x x d x ; ∫ 0.1 0.5 cos x x d x ; trapezoidal rule; n = 4 n = 4

∫ 0.1 0.5 cos x x d x ; ∫ 0.1 0.5 cos x x d x ; Simpson’s rule; n = 4 n = 4

Evaluate ∫ 0 1 d x 1 + x 2 ∫ 0 1 d x 1 + x 2 exactly and show that the result is π / 4 . π / 4 . Then, find the approximate value of the integral using the trapezoidal rule with n = 4 n = 4 subdivisions. Use the result to approximate the value of π . π .

Approximate ∫ 2 4 1 ln x d x ∫ 2 4 1 ln x d x using the midpoint rule with four subdivisions to four decimal places.

Approximate ∫ 2 4 1 ln x d x ∫ 2 4 1 ln x d x using the trapezoidal rule with eight subdivisions to four decimal places.

Use the trapezoidal rule with four subdivisions to estimate ∫ 0 0.8 x 3 d x ∫ 0 0.8 x 3 d x to four decimal places.

Use the trapezoidal rule with four subdivisions to estimate ∫ 0 0.8 x 3 d x . ∫ 0 0.8 x 3 d x . Compare this value with the exact value and find the error estimate.

Using Simpson’s rule with four subdivisions, find ∫ 0 π / 2 cos ( x ) d x . ∫ 0 π / 2 cos ( x ) d x .

Show that the exact value of ∫ 0 1 x e − x d x = 1 − 2 e . ∫ 0 1 x e − x d x = 1 − 2 e . Find the absolute error if you approximate the integral using the midpoint rule with 16 subdivisions.

Given ∫ 0 1 x e − x d x = 1 − 2 e , ∫ 0 1 x e − x d x = 1 − 2 e , use the trapezoidal rule with 16 subdivisions to approximate the integral and find the absolute error.

Find an upper bound for the error in estimating ∫ 0 3 ( 5 x + 4 ) d x ∫ 0 3 ( 5 x + 4 ) d x using the trapezoidal rule with six steps.

Find an upper bound for the error in estimating ∫ 4 5 1 ( x − 1 ) 2 d x ∫ 4 5 1 ( x − 1 ) 2 d x using the trapezoidal rule with seven subdivisions.

Find an upper bound for the error in estimating ∫ 0 3 ( 6 x 2 − 1 ) d x ∫ 0 3 ( 6 x 2 − 1 ) d x using Simpson’s rule with n = 10 n = 10 steps.

Find an upper bound for the error in estimating ∫ 2 5 1 x − 1 d x ∫ 2 5 1 x − 1 d x using Simpson’s rule with n = 10 n = 10 steps.

Find an upper bound for the error in estimating ∫ 0 π 2 x cos ( x ) d x ∫ 0 π 2 x cos ( x ) d x using Simpson’s rule with four steps.

Estimate the minimum number of subintervals needed to approximate the integral ∫ 1 4 ( 5 x 2 + 8 ) d x ∫ 1 4 ( 5 x 2 + 8 ) d x with an error magnitude of less than 0.0001 using the trapezoidal rule.

Determine a value of n such that the trapezoidal rule will approximate ∫ 0 1 1 + x 2 d x ∫ 0 1 1 + x 2 d x with an error of no more than 0.01.

Estimate the minimum number of subintervals needed to approximate the integral ∫ 2 3 ( 2 x 3 + 4 x ) d x ∫ 2 3 ( 2 x 3 + 4 x ) d x with an error of magnitude less than 0.0001 using the trapezoidal rule.

Estimate the minimum number of subintervals needed to approximate the integral ∫ 3 4 1 ( x − 1 ) 2 d x ∫ 3 4 1 ( x − 1 ) 2 d x with an error magnitude of less than 0.0001 using the trapezoidal rule.

Use Simpson’s rule with four subdivisions to approximate the area under the probability density function y = 1 2 π e − x 2 / 2 y = 1 2 π e − x 2 / 2 from x = 0 x = 0 to x = 0.4 . x = 0.4 .

Use Simpson’s rule with n = 14 n = 14 to approximate (to three decimal places) the area of the region bounded by the graphs of y = 0 , y = 0 , x = 0 , x = 0 , and x = π / 2 . x = π / 2 .

The length of one arch of the curve y = 3 sin ( 2 x ) y = 3 sin ( 2 x ) is given by L = ∫ 0 π / 2 1 + 36 cos 2 ( 2 x ) d x . L = ∫ 0 π / 2 1 + 36 cos 2 ( 2 x ) d x . Estimate L using the trapezoidal rule with n = 6 . n = 6 .

The length of the ellipse x = a cos ( t ) , y = b sin ( t ) , 0 ≤ t ≤ 2 π x = a cos ( t ) , y = b sin ( t ) , 0 ≤ t ≤ 2 π is given by L = 4 a ∫ 0 π / 2 1 − e 2 cos 2 ( t ) d t , L = 4 a ∫ 0 π / 2 1 − e 2 cos 2 ( t ) d t , where e is the eccentricity of the ellipse. Use Simpson’s rule with n = 6 n = 6 subdivisions to estimate the length of the ellipse when a = 2 a = 2 and e = 1 / 3 . e = 1 / 3 .

Estimate the area of the surface generated by revolving the curve y = cos ( 2 x ) , 0 ≤ x ≤ π 4 y = cos ( 2 x ) , 0 ≤ x ≤ π 4 about the x -axis. Use the trapezoidal rule with six subdivisions.

Estimate the area of the surface generated by revolving the curve y = 2 x 2 , y = 2 x 2 , 0 ≤ x ≤ 3 0 ≤ x ≤ 3 about the x- axis. Use Simpson’s rule with n = 6 . n = 6 .

The growth rate of a certain tree (in feet) is given by y = 2 t + 1 + e − t 2 / 2 , y = 2 t + 1 + e − t 2 / 2 , where t is time in years. Estimate the growth of the tree through the end of the second year by using Simpson’s rule, using two subintervals. (Round the answer to the nearest hundredth.)

[T] Use a calculator to approximate ∫ 0 1 sin ( π x ) d x ∫ 0 1 sin ( π x ) d x using the midpoint rule with 25 subdivisions. Compute the relative error of approximation.

[T] Given ∫ 1 5 ( 3 x 2 − 2 x ) d x = 100 , ∫ 1 5 ( 3 x 2 − 2 x ) d x = 100 , approximate the value of this integral using the trapezoidal rule with 16 subdivisions and determine the absolute error.

Given that we know the Fundamental Theorem of Calculus, why would we want to develop numerical methods for definite integrals?

The table represents the coordinates ( x , y ) ( x , y ) that give the boundary of a lot. The units of measurement are meters. Use the trapezoidal rule to estimate the number of square meters of land that is in this lot.

Choose the correct answer. When Simpson’s rule is used to approximate the definite integral, it is necessary that the number of partitions be____

- an even number
- either an even or an odd number
- a multiple of 4

The “Simpson” sum is based on the area under a ____.

The error formula for Simpson’s rule depends on___.

- f ( x ) f ( x )
- f ′ ( x ) f ′ ( x )
- f ( 4 ) ( x ) f ( 4 ) ( x )
- the number of steps

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- Book title: Calculus Volume 2
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- Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
- Section URL: https://openstax.org/books/calculus-volume-2/pages/3-6-numerical-integration

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Choose an approximation Rule and then use the slider to investigate different types of approximations of a definite integral. Choose Rule: Midpoint Rule