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7.02: Trapezoidal Rule of Integration

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Lesson 1: Single-Application Trapezoidal Rule

Learning Objectives

After successful completion of this lesson, you should be able to:

1) derive the trapezoidal rule of integration,

2) use the trapezoidal rule of integration to solve problems,

Introduction

The trapezoidal rule is based on the Newton-Cotes formula that if one approximates the integrand by an nth order polynomial, then the integral of the function is approximated by the integral of that nth order polynomial. Integrating polynomials is simple and is based on the calculus formula.

For a function that is above the x-axis in the region between x=a and x=b, the integral of the function between a and b is the area under the curve and above the x-axis.
Figure 7.2.1.1. Integration of a function.

baxn dx=(bn+1an+1n+1),n1(7.2.1.1)

So if we want to approximate the integral

I=baf(x) dx(7.2.1.2)

to find the value of the above integral, one assumes

f(x)fn(x)

where

fn(x)=a0+a1x++an1xn1+anxn.(7.2.1.3)

where fn(x) is a nth order polynomial. The trapezoidal rule assumes n=1, that is, approximating the integral by a linear polynomial (straight line),

baf(x) dxbaf1(x) dx(7.2.1.4)

Method 1: Derived from Calculus

The value of the integral of a function between points a and b is approximated as the area of a trapezoidal region, bounded on the bottom by the x-axis, on the left and right sides by vertical lines at x=a and b, and on the top by an approximation of the function between the boundary points as a first-order polynomial.
Figure 7.2.1.2. Approximating the function by a first-order polynomial to derive the trapezoidal rule.

Approximating the integrand f(x) by a first-order polynomial f1(x), that is, f1(x)=a0+a1x,

baf(x) dxbaf1(x) dx=ba(a0+a1x) dx=a0(ba)+a1(b2a22)(7.2.1.5)

But what are a0 and a1? Now if one chooses (a,f(a)) and (b,f(b)) as the two points to approximate f(x) by a straight line from a to b,

f(a)=f1(a)=a0+a1af(b)=f1(b)=a0+a1b(7.2.1.6a,b)

Solving Equations (7.2.1.6a) and (7.2.1.6b) for a0 and a1, we get

a1=f(b)f(a)ba

a0=f(a)bf(b)aba(7.2.1.7a,b)

Substituting values of a0 and a1 from Equations (7.2.1.7a) and (7.2.1.7b) in Equation (7.2.1.5) gives

baf(x) dxf(a)bf(b)aba(ba)+f(b)f(a)bab2a22=(ba)[f(a)+f(b)2](7.2.1.8)

Note that the trapezoidal rule can be interpreted as the width of the interval [a,b] multiplied to an approximate average value of the function. Recall the definition of the mean of the function ¯f of a function f in an interval [a,b] is given by ¯f=1babaf(x) dx(7.2.1.9)

Example 7.2.1.1

The following integral is given

1.30.15xe2x dx

a) Use the trapezoidal rule to estimate the value of the integral.

b) Find the true error, Et, for part (a).

c) Find the absolute relative true error, |εt|, for part (a).

Solution

a) From Equation (7.2.1.8, which is repeated here,

abf(x) dx(ba)2[f(a)+f(b)]

where

a=0.1

b=1.3

Then

1.30.1f(x)dx(1.30.1)2[f(0.1)+f(1.3)]=0.6[f(0.1)+f(1.3)]=0.6[5(0.1)e2(0.1)+5(1.3)e2(1.3)]=0.6(0.4094+0.4828)=0.5353

b) The true error is

Et= True ValueApproximate Value=0.89390.5353=0.3586

c) The absolute relative true error, |εt|, would then be

|εt|=|True ErrorTrue Value|×100=|0.35860.8939|×100=40.12%

Example 7.2.1.2

The vertical distance covered by a rocket from t=8 to t=30 seconds is given by

x=308(2000ln[1400001400002100t]9.8t) dt

a) Use the trapezoidal rule to find the distance covered for t=8 to t=30 seconds.

b) Find the true error, Et for part (a).

c) Find the absolute relative true error for part (a).

Solution

a) I(ba)[f(a)+f(b)2], where

a=8

b=30

f(t)=2000ln[1400001400002100t]9.8t

f(8)=2000ln[1400001400002100(8)]9.8(8)=177.27 m/s

f(30)=2000ln[1400001400002100(30)]9.8(30)=901.67 m/s

I(308)[177.27+901.672]=11868 m

b) The exact value of the above integral is

x=308(2000ln[1400001400002100t]9.8t) dt=11061 m

so the true error is

Et=True ValueApproximate Value=1106111868=807 m

c) The absolute relative true error, |ϵt|, would then be

|ϵt|=|True ErrorTrue Value|×100=|110611186811061|×100=7.2958%

Audiovisual Lecture

Title: Single Application Trapezoidal Example

Summary: Learn the trapezoidal rule of integration through a simple example.

Appendix 1

This appendix shows other methods of deriving the trapezoidal rule of integration. The method of undetermined coefficients is significant to learn as it forms the basis of the Gauss quadrature rule.

Method 2: Derived from Calculus Using Newton’s divided difference polynomial

f1(x) can also be approximated by using Newton’s divided difference polynomial as

f1(x)=f(a)+f(b)f(a)ba(xa)(7.2.A1.1)

Hence

baf(x) dxbaf1(x) dx=ba[f(a)+f(b)f(a)ba(xa)] dx=[f(a)x+f(b)f(a)ba(x22ax)]ba=f(a)bf(a)a+(f(b)f(a)ba)(b22aba22+a2)=f(a)bf(a)a+(f(b)f(a)ba)(b22ab+a22)=f(a)bf(a)a+(f(b)f(a)ba)12(ba)2=f(a)bf(a)a+12(f(b)f(a))(ba)=f(a)bf(a)a+12f(b)b12f(b)a12f(a)b+12f(a)a=12f(a)b12f(a)a+12f(b)b12f(b)a=(ba)[f(a)+f(b)2](7.2.A1.2)

This gives the same result as Equation (7.2.1.8) because f1(x) is just a different form of writing the same first-order polynomial.

Method 3: Derived from Geometry

The trapezoidal rule can also be derived from geometry. Look at Figure 7.2.1.2. The area under the curve f1(x) is the area of a trapezoid. The integral

baf(x)dxArea of trapezoid=12(Sum of the length of parallel sides)           ×(Perpendicular distance between the parallel sides)=12(f(b)+f(a))(ba)=(ba)[f(a)+f(b)2](7.2.A1.3)

Audiovisual Lecture

Title: Single-Application Trapezoidal Rule Derivation

Summary: Learn how to derive the trapezoidal rule of integration.

Method 4: Derived from Method of Coefficients

Choose the integral baf(x) dx approximated as follows.

baf(x) dxc1f(a)+c2f(b)(7.2.A1.4)

The coefficients c1 and c2 are undetermined. We will find these coefficients such that the right-hand side is exact for integrals of a straight line a0+a1x.

So from exact integration

ba(a0+a1x) dx=[a0x+a1x22]ba=a0(ba)+a1(b2a22)(7.2.A1.5)

But we want the right-hand side formula to give the same result as Equation (7.2.A1.5) for f(x)=a0+a1x which is

c1f(a)+c2f(b)=c1(a0+a1b)+c2(a0+a1b)=a0(c1+c2)+a1(c1a+c2b)(7.2.A1.6)

Hence from Equations (7.2.A1.5) and (7.2.A1.6),

a0(ba)+a1(b2a22)=a0(c1+c2)+a1(c1a+c2b)(7.2.A1.7)

Since a0 and a1 are arbitrary constants for the chosen general straight line, the coefficients of a0 and a1 need to be equal. That gives

c1+c2=ba(7.2.A1.8a)

c1a+c2b=b2a22(7.2.A1.8b)

Multiplying Equation (7.2.A1.8a) by a and subtracting from Equation (7.2.A1.8b) gives

c2=ba2(7.2.A1.9a)

Substituting the value of c2 from Equation (7.2.A1.9a) in Equation (7.2.A1.8a) gives

c1=ba2(7.2.A1.9b)

Therefore, from Equation (7.2.A1.4), (7.2.A1.9a), and (7.2.A1.9b),

baf(x) dxc1f(a)+c2f(b)=ba2f(a)+ba2f(b)(7.2.A1.10)

Approximating the area under a curve between points (a, f(a)) and (b, f(b)) as two rectangles, with the width of each equal to half the distance between a and b, the height of the left rectangle equal to f(a), and the height of the right rectangle equal to f(b).
Figure 7.2.1.3. Area by the method of coefficients.

The interpretation is that f(x) is evaluated at points a and b, and each function evaluation is given a weight of ba2. Geometrically, Equation (7.2.1.8) is looked at as the area of a trapezoid, while Equation (7.2.A1.10) is viewed as the sum of the area of two rectangles, as shown in Figure 7.2.1.3.

Lesson 2: Composite Trapezoidal Rule

Learning Objectives

After successful completion of this lesson, you should be able to:

1) derive the composite (also called multiple-segment) trapezoidal rule of integration,

2) use the composite (also called multiple-segment) trapezoidal rule of integration to solve problems

Introduction

A single segment trapezoidal rule seldom gives you acceptable results for an integral. Instead for higher accuracy and its control, we can use the composite (also called multiple-segment) trapezoidal rule where the integral is broken into segments, and the single-segment trapezoidal rule is applied over each segment.

Derivation

Divide (ba) into n equal segments, as shown in Figure 7.2.2.1. Then the width of each segment is

h=ban(7.2.2.1)

The integral I can be broken into n integrals as

I=baf(x) dx=a+haf(x) dx+a+2ha+hf(x) dx+...+a+(n1)ha+(n2)hf(x) dx+ba+(n1)hf(x) dx(7.2.2.2)

The area under a curve between the points (a, f(a)) and (b, f(b)) is approximated by the area of 4 trapezoids, each trapezoid having the same width along the x-axis and the topmost segment of the trapezoid connecting the two points on the curve intersected by its vertical left and right sides.
Figure 7.2.2.1. Composite (n=4) trapezoidal rule

Applying single-segment trapezoidal rule on Equation (7.2.2.2) on each segment gives

baf(x) dx[(a+h)a][f(a)+f(a+h)2]+[(a+2h)(a+h)][f(a+h)+f(a+2h)2] +...............+[(a+(n1)h)(a+(n2)h)][f(a+(n2)h)+f(a+(n1)h)2]+[b(a+(n1)h)][f(a+(n1)h)+f(b)2]=h[f(a)+f(a+h)2]+h[f(a+h)+f(a+2h)2] +...............+h[f(a+(n2)h)+f(a+(n1)h)2]+h[f(a+(n1)h)+f(b)2]=h[f(a)+2f(a+h)+2f(a+2h)+...+2f(a+(n1)h)+f(b)2]=h2[f(a)+2{n1i=1f(a+ih)}+f(b)]=ba2n[f(a)+2{n1i=1f(a+ih)}+f(b)](7.2.2.3)

Audiovisual Lecture

Title: Derivation of Composite Trapezoidal Rule of Integration

Summary: Learn the derivation of and motivation behind the composite trapezoidal rule of integration.

Example 7.2.2.1

The following integral is given:

1.30.15xe2x dx

a) Use the composite trapezoidal rule to estimate the value of this integral. Use three segments.

b) Find the true error Et for part (a).

c) Find the absolute relative true error |εt| for part (a).

Solution

a) The solution using the composite trapezoidal rule with 3 segments is applied as follows.

Iba2n[f(a)+2n1i=1f(a+ih)+f(b)]

n=3

a=0.1

b=1.3

h=ban=1.30.13=0.4

From Equation (7.2.2.3),

I1.30.16[f(0.1)+231i=1f(0.1+0.4i)+f(1.3)]I1.30.16[f(0.1)+22i=1f(0.1+0.4i)+f(1.3)]=0.2[f(0.1)+2f(0.5)+2f(0.9)+f(1.3)]=0.2[5×0.1×e2(0.1)+2(5×0.5×e2(0.5))+2(5×0.9×e2(0.9))+5×1.3×e2(1.3)]=0.84385

b) The exact value of the above integral can be found by integration by parts and is

1.30.15xe2x dx=0.89387

So the true error is

Et=True ValueApproximate Value=0.893870.84385=0.05002

c) The absolute relative true error is

|εt|=|True ErrorTrue Value|×100=|0.050020.89387|×100=5.5959%

Example 7.2.2.2

The vertical distance covered by a rocket from t=8 to t=30 seconds is given by

x=308(2000ln[1400001400002100t]9.8t) dt

a) Use the composite trapezoidal rule to find the distance covered from t=8 to t=30 seconds. Use two segments.

b) Find the true error, Et, for part (a).

c) Find the absolute relative true error for part (a).

Solution

a) The solution using the composite trapezoidal rule with 2-segments is

Iba2n[f(a)+2{n1i=1f(a+ih)}+f(b)]

n=2

a=8

b=30

h=ban=3082=11

I3082(2)[f(8)+2{21i=1f(8+11i)}+f(30)]=224[f(8)+2f(19)+f(30)]=224[177.27+2(484.75)+901.67]=11266 m

b) The exact value of the above integral is

x=308(2000ln[1400001400002100t]9.8t)dt=11061 m

So the true error is

Et=True ValueApproximate Value=1106111266=205 m

c) The absolute relative true error,|ϵt|, would then be

|ϵt|=|True ErrorTrue Value|×100=|110611126611061|×100=1.8537%

Table 7.2.2.1. Values obtained using composite trapezoidal rule for x=308(2000ln[1400001400002100t]9.8t) dt
n Approximate Value Et |ϵt|% |ϵa|%
1 11868 807 7.296
2 11266 205 1.853 5.343
3 11153 91.4 0.8265 1.019
4 11113 51.5 0.4655 0.3594
5 11094 33.0 0.2981 0.1669
6 11084 22.9 0.2070 0.09082
7 11078 16.8 0.1521 0.05482
8 11074 12.9 0.1165 0.03560

Audiovisual Lecture

Title: Example for Composite Trapezoidal Rule of Integration

Summary: Learn how composite trapezoidal rule of integration is used via example.

Error in Composite Trapezoidal Rule

Without proof, the true error for a single-segment Trapezoidal rule for baf(x) dx is given by

Et=(ba)312f(ζ), a<ζ<b(7.2.2.4)

where

ζ  is some point in [a,b].

What is the error then in the composite trapezoidal rule for baf(x) dx?

It is given by

Et(ba)312n2¯f(7.2.2.5)

where

¯f= some approximate average value of the second derivativeof the function in [a,b].

n=number of segments.

Recall that the width of each segment is

h=ban

Appendix 2

The appendix gives you two more examples for showing the application of the composite trapezoidal rule.

Example 7.2.A2.1

Use the composite trapezoidal rule to find the area under the curve

f(x)=300x1+ex

from x=0 to x=10. Use 2 segments.

Solution

Using two segments, we get

h=1002=5

f(0)=300(0)1+e0=0

f(5)=300(5)1+e5=10.039

f(10)=300(10)1+e10=0.136

From Equation (7.2.2.3)

Iba2n[f(a)+2{n1i=1f(a+ih)}+f(b)]=1002(2)[f(0)+2{21i=1f(0+5)}+f(10)]=104[f(0)+2f(5)+f(10)]=104[0+2(10.039)+0.136]=50.537

So what is the true value of this integral?

100300x1+ex dx=246.59

The absolute relative true error is

|ϵt|=|246.5950.535246.59|×100=79.506%

Why is the true value different from the approximate values? Just take a look at Figure 7.2.2.2. As you can see, the area under the “trapezoids” (yeah, they really look like triangles now) covers a small portion of the area under the curve. As we add more segments, the approximated value quickly approaches the true value.

Approximating the area under the given function curve between x-values of 0 and 10, using the composite trapezoidal approximation with two trapezoids of width 5 each.
Figure 7.2.2.2. Composite trapezoidal rule approximation.
Table 7.2.A2.1. Values obtained using the composite trapezoidal rule for 100300x1+ex dx.
n Approximate Value Et |ϵt|
1 0.681 245.91 99.724%
2 50.535 196.05 79.505%
4 170.61 75.978 30.812%
8 227.04 19.546 7.927%
16 241.70 4.887 1.982%
32 245.37 1.222 0.495%
64 246.28 0.305 0.124%
Example 7.2.A2.2

Use the composite trapezoidal rule to find

I=201x dx

Solution

We cannot use the trapezoidal rule for this integral, as the value of the integrand at x=0 is infinite. However, it is known that discontinuities in a curve do not change the area under it. We can assume any value for the function at x=0. The algorithm to define the function so that we can use the composite trapezoidal rule is given below.

Function f(x)

If x=0 Then f=0

If x0 Then f=x(0.5)

End Function

Basically, we are just assigning the function a value of zero at x=0. Everywhere else, the function is continuous. This assumption means the true value of our integral is just that—true. Let’s see what happens using the composite trapezoidal rule.

Using two segments, we get

h=202=1

f(0)=0

f(1)=11=1

f(2)=12=0.70711

Iba2n[f(a)+2{n1i=1f(a+ih)}+f(b)]=202(2)[f(0)+2{21i=1f(0+1)}+f(2)]=24[f(0)+2f(1)+f(2)]=24[0+2(1)+0.70711]=1.3536

So what is the true value of this integral?

201x dx=2.8284

thus making the absolute relative true error

|ϵt|=|2.82841.35362.8284|×100=52.145%

Table 7.2.A2.2 shows how the approximate value converges to the true value as the number of segments is increased.

Table 7.2.A2.2. Values obtained using the composite trapezoidal rule for 201x dx.
n Approximate Value Et |ϵt|
2 1.354 1.474 52.14%
4 1.792 1.036 36.64%
8 2.097 0.731 25.85%
16 2.312 0.516 18.26%
32 2.463 0.365 12.91%
64 2.570 0.258 9.128%
128 2.646 0.182 6.454%
256 2.699 0.129 4.564%
512 2.737 0.091 3.227%
1024 2.764 0.064 2.282%
2048 2.783 0.045 1.613%
4096 2.796 0.032 1.141%

Lesson 3: Error Analysis of Trapezoidal Rule

Learning Objective

After successful completion of this lesson, you should be able to:

1) derive the formula for the true error in the composite (also called multiple-segment) trapezoidal rule of integration.

Error in Composite Trapezoidal Rule

The true error for a single segment trapezoidal rule is given by

Et=(ba)312f(ζ), a<ζ<b(7.2.3.1)

where ζ is some point in [a,b].

What is the error then in the composite trapezoidal rule? It is the sum of the errors from each of the n segments. The error in each segment is that of the single segment trapezoidal rule. The error in each segment is

E1=[(a+h)a]312f(ζ1),a<ζ1<a+h=h312f(ζ1) E2=[(a+2h)(a+h)]312f(ζ2),a+h<ζ2<a+2h=h312f(ζ2) Ei=[(a+ih)(a+(i1)h)]312f(ζi),a+(i1)h<ζi<a+ih=h312f(ζi)

En1=[{a+(n1)h}{a+(n2)h}]312f(ζn1),a+(n2)h<ζn1<a+(n1)h=h312f(ζn1) En=[b{a+(n1)h}]312f(ζn),a+(n1)h<ζn<b=h312f(ζn)

Hence the total error in the composite trapezoidal rule is

Et=ni=1Ei=h312f(ζ1)h312f(ζ2)h312f(ζn)=h312ni=1f(ζi)

Since h=ban,

Et=(ba)312n3ni=1f(ζi)=(ba)312n2ni=1f(ζi)n

The term ni=1f(ζi)n is an approximate average value of the second derivative f(x),a<x<b.

Hence

Et(ba)312n2 ¯f(7.2.3.2)

In Table 7.2.3.1, the approximate value of the integral

308(2000ln[1400001400002100t]9.8t) dt

is given as a function of the number of segments in a composite trapezoidal rule. You can visualize that as the number of segments is doubled, the true error gets approximately quartered.

Table 7.2.3.1. Values obtained using the composite trapezoidal rule for x=308(2000ln[1400001400002100t]9.8t) dt.
n Approximate Value Et |ϵt|% |ϵa|%
2 11266 205 1.853 5.343
4 11113 52 0.4701 0.3594
8 11074 13 0.1175 0.03560
16 11065 4 0.03616 0.00401

For example, for the composite trapezoidal rule with 2 segments, the true error is 205, and a quarter of that error is 51.25. That is close to the true error of 52 for the composite trapezoidal rule with 4 segments.

Can you answer the question, why is the true error not exactly 51.25? How does this information help us in numerical integration? You will find out that this forms the basis of Richardson’s extrapolation formula for integration based on the trapezoidal rule, where we use the argument that true error gets approximately quartered when the number of segments is doubled. Richardson’s extrapolation formula based on the trapezoidal rule is computationally more efficient than using the composite trapezoidal rule by itself.

Audiovisual Lecture

Title: Composite Trapezoidal Rule Error: Derivation

Summary: This video discusses how to derive the error formula for the composite trapezoidal rule.

Audiovisual Lecture

Title: Composite Trapezoidal Rule Error: Example

Summary: After watching this video, via an example, you will be able to use the formula for the true error in the composite trapezoidal rule. You will see how the true error is approximately inversely proportional to the square of the number of segments.

Richardson’s Extrapolation Formula for Trapezoidal Rule

From Equation (7.2.3.2), the true error, Et, in the composite trapezoidal rule with n segments is estimated as

Etα1n2(7.2.3.3)

Assuming C to be the constant of proportionality,

EtCn2(7.2.3.4)

Since

Et=True ValueApproximate ValueEt=(TV)In(7.2.3.5)

where

(TV)=true value

In=approximate value using n-segments,

then from Equations (7.2.3.4) and (7.2.3.5),

Cn2(TV)In(7.2.3.6)

If the number of segments is doubled from n to 2n in the composite trapezoidal rule,

C(2n)2(TV)I2n(7.2.3.7)

Equations (7.2.3.6) and (7.2.3.7) can be solved simultaneously for (TV) to get

(TV)I2n+I2nIn3

Example 7.2.3.1

The vertical distance in meters covered by a rocket from t=8 to t=30 seconds is given by

x=308(2000ln[1400001400002100t]9.8t)dt

a) Use Richardson’s extrapolation formula to find the distance covered. Use the composite trapezoidal rule results with 2 and 4 segments given in Table 7.2.3.1.

b) Find the true error for part (a).

c) Find the absolute relative true error for part (a).

Solution

a)

I2=11266 m

I4=11113 m

Using Richardson’s extrapolation formula for the trapezoidal rule, the true value is approximated by

TVI2n+I2nIn3

and choosing n=2,

TVI4+I4I23=11113+11113112663=11062 m]

b) The exact value of the above integral is

x=308(2000ln[1400001400002100t]9.8t)dt=11061 m

So the true error

Et=True ValueApproximate Value=1106111062=1 m

c) The absolute relative true error, |ϵt|, would then be

|ϵt|=True ErrorTrue Value×100=|110611106211061|×100=0.00904%

Table 7.2.3.2 shows Richardson’s extrapolation results using 1, 2, 4, and 8 segments. Results are compared with those with the composite trapezoidal rule.

Table 7.2.3.2. Values obtained using Richardson’s extrapolation formula for the trapezoidal rule for x=308(2000ln[1400001400002100t]9.8t)dt.
n Trapezoidal rule Et% for trapezoidal rule Richardson’s Extrapolation Et% for Richardson’s extrapolation
1 11868 7.296
2 11266 1.854 11065 0.03616
4 11113 0.4655 11062 0.009041
8 11074 0.1165 11061 0.0000

Multiple Choice Test

(1). What is the highest order of polynomial for which the composite trapezoidal rule of integration with two segments is exact?

(A) first

(B) second

(C) third

(D) fourth

(2). The estimated value of 2.20.2xex dx by the using single application trapezoidal rule is most nearly

(A) 11.672

(B) 11.807

(C) 20.099

(D) 24.119

(3). The estimated value of 2.20.2xex dx by using the composite trapezoidal rule with three segments is most nearly

(A) 11.672

(B) 11.807

(C) 12.811

(D) 14.633

(4). The velocity of a body is given by

v(t)=2t, 1t5=5t2+3, 5<t14

where t is given in seconds, and v is given in m/s. Use the composite trapezoidal rule with two segments to find the distance covered by the body from t=2 to t=9 seconds.

(A) 935.0m

(B) 1039.7m

(C) 1260.9m

(D) 5048.9m

(5). The shaded area shows a plot of land available for sale. Your best estimate of the area of the land is most nearly

A plot of land on the first quadrant of a coordinate plane, bounded on the left by the y-axis and on the bottom by the x-axis. The curve forming the other boundaries of the plot starts on the y-axis at y=60, remains roughly horizontal until x=60 when it drops to y=45, and remains roughly horizontal until x=100 where it drops to y=25 and moves left until x=75 before dropping to y=0.

(A) 2500 m2

(B) 4775 m2

(C) 5250 m2

(D) 6000 m2

(6). The following data of the velocity of a body is given as a function of time.

Time, t (s) 0 15 18 22 24
Velocity,v (m/s) 22 24 37 25 123

Using the trapezoidal rule with unequal segments, the distance covered by the body from t=12 s to t=18 s is

(A) 162.90 m

(B) 166.00 m

(C) 181.70 m

(D) 436.50 m

For complete solution, go to

http://nm.mathforcollege.com/mcquizzes/07int/quiz_07int_trapcontinous_solution.pdf

Problem Set

(1). Find the value of 2.20.2xex dx by using composite trapezoidal rule with two segments.

Answer

14.033

(2). Find the value of 2.23.4x2ex dx by using composite trapezoidal rule with four segments.

Answer

33.875

(3). The upward velocity of a rocket is given by

v(t)=200ln(t+1)10t,t>0

where t is given in s and v is given in m/s.

a) Use composite trapezoidal rule with two segments to calculate the distance covered by the rocket from t=0 s to t=5 s.

b) What is the true value of the distance covered by the rocket from t=0 s to t=5 s?

c) What is the true error in part (a)?

d) What is the relative true error in part (a)?

e) What is the absolute relative true error in percentage for part (a)?

f) Use composite trapezoidal rule with four segments to calculate the distance covered by the rocket from t=0 s to t=5 s.

g) What is the absolute relative approximate error in percentage for part (e), assuming the previous approximation is from part (a)?

h) Based on the answer from part (g), how many significant digits are correct in the answer in part (g).

Answer

a) 949.32 m
b) 1025.1 m
c) 75.790 m
d) 0.073933
e) 7.3933%
f) 1004.4 m
g) 5.4865%
h) 0

(4). Find the value of the integral 513e2t dt using composite trapezoidal rule with two segments.

Answer

0.42101

(5). For the integral 41xe2x dx, find the following

a) Exact integral using your calculus knowledge (Hint: u dv=uvv du+C)

b) Value of integral using composite trapezoidal rule with two segments.

c) Value of integral using composite trapezoidal rule with four segments.

d) Calculate the true error for part (b) and part (c). Is the true error for part (c) approximately a quarter of the true error for part (b)? Explain.

Answer

a) 0.10075
b) 0.12778
c) 0.10719
d) 0.027028, 0.00644

(6). In composite trapezoidal rule, if En is the true error using n (n1) segments, then the true error using composite trapezoidal rule with 2n sgements approximately is ________ of En.

Answer

1/4

(7). The upward velocity of a body is given by

v(t)=ln(1601609t),t>0

where t is given in seconds, and v in m/s.

What is the distance covered by the body from t=5 to t=9 seconds? Use composite trapezoidal rule with four segments.

Answer

2.0280 m

(8). The true error for the single application trapezoidal rule used to calculate the approximate value of the integral baf(x) dx is given by

Et=(ba)312f(ζ),aζb

For the integral633e1.1x dx, what is the value of ζ?

Answer

4.7386

(9). The approximate value of the integral of 51f(x) dx is found by using single application trapezoidal rule as 0.35958. Given f(2)=0.7447, f(3)=0.10454 and f(4)=0.50730, what is the approximate value of the integral of 51 f(x)dx using the composite trapezoidal rule with two segments.

Answer

0.38887

(10). The true error for trapezoidal rule used to calculate the approximate value of the integral baf(x) dx is given by

Et=(ba)312f(ζ),aζb

For the integral 52(3x2+dx+e) dx, where constants d and e are not given to you, the single application trapezoidal rule gives the value of the integral as 201.0. Find the exact value of the integral.

Et=(ba)312f(ζ),aζb

Answer

187.5

(11). A quadrature rule is developed by a scientist:

baf(x) dxc1f(a).

The value of c1 is found by assuming that the formula is exact for integrals of f(x)=a0x2, where a0 is an arbitrary constant. What is the value of c1 in terms of a and b?

Answer

b3a33a2


This page titled 7.02: Trapezoidal Rule of Integration is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Autar Kaw via source content that was edited to the style and standards of the LibreTexts platform.

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7.01: Prerequisites to Numerical Integration
7.05: Gauss Quadrature Rule of Integration