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AP Calculus AB : Trapezoidal sums

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Example Questions

Example Question #305 : Integrals

Using the Trapezoidal Rule with four subintervals, approximate to four decimal places:

$\int_{2}^{3} \log_{3}( x^{2} ) \mathrm{d} x$

Possible Answers:

1.6542

1.5620

2.0620

1.6558

1.7465

Correct answer:

1.6542

Explanation:

The Trapezoidal Rule states that a definite integral

$\int_{a}^{b} f(x) \textup{d}x$

can be approximated by computing the expression

$T_{N}=\frac{b-a}{2N} \left [f (x_{0} ) +2 f (x_{1}) +2 f ( x_{2} ) + ... 2f ( x_{N-1} )+ f (x_{N} ) \right ]$

where $x_{k} = a + \frac{b-a}{N} \cdot k$ for k = 0, 1, 2, 3, ...,N

Since we are dividing $\left [ 2,3 \right ]$ into four subintervals, set

a = 2, b =3, N = 4 ; the expression becomes

$T_{4}=\frac{1}{8} \left [ f(2 ) +2 f \left (2.25 \right ) +2 f \left (2.5 \right ) + 2f \left ( 2.75 \right ) + f (3 ) \right ]$

where

$f(x)= \log_{3}( x^{2} )$

$f(2)= \log_{3}( 2^{2} ) = \log_{3}4= \frac{\ln 4}{\ln 3} \approx 1.2619$

$f(2.25)= \log_{3}( 2.25^{2} ) = \log_{3}5.0625= \frac{\ln 5.0625}{\ln 3} \approx 1.4763$

$f(2.5)= \log_{3}( 2.5^{2} ) = \log_{3}6.25= \frac{\ln 6.25}{\ln 3} \approx 1.6681$

$f(2.75)= \log_{3}( 2.75^{2} ) = \log_{3}7.5625= \frac{\ln 7.5625}{\ln 3} \approx 1.8416$

$f(3)= \log_{3}( 3^{2} ) = \log_{3}9= 2$

$T_{4}=\frac{1}{8} \left [ 1.2619 +2 (1.4763 )+2 ( 1.6681)+ 2 ( 1.8416) +2 \right]$

$\approx 1.6542$

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Example Question #305 : Integrals

Solve in the integral by method of trapezoidal sums

$\int_{3}^{5}\cos(e^{3x})dx$

Possible Answers:

$\cos(e^{15})+\cos(e^{9})$

$\cos(e^{5})-\cos(e^{9})$

$\cos(e^{15})-\cos(e^{9})$

$\cos(e^{15})-\sin(e^{9})$

Correct answer:

$\cos(e^{15})-\cos(e^{9})$

Explanation:

To use the trapezoidal rule, we apply the following formula:

$\int_{a}^{b}f(x)dx\approx(b-a)\left [ \frac{f(a)-f(b)}{2}\right ]$

Using the integral from our problem statement, we get

$(5-3)\frac{(\cos(e^{15})-\cos(e^{9}))}{2}=\cos(e^{15})-\cos(e^{9})$

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Example Question #311 : Integrals

Evaluate the integral using the trapezoidal sums method:

$\int_{\frac{\pi}{2}}^{2\pi} e^{sin(x)} dx$

Possible Answers:

$\frac{3\pi}{2}(\frac{1+e}{2})$

$\frac{3\pi}{2}(\frac{2+e}{2})$

$\frac{\pi}{2}(\frac{1+e}{2})$

$\frac{3\pi}{2}(1+e)$

Correct answer:

$\frac{3\pi}{2}(\frac{1+e}{2})$

Explanation:

To solve the integral, we will use the formula for the trapezoidal sums method:

$\int_{a}^{b}f(x)dx=(b-a)\left [ \frac{f(a)+f(b)}{2}\right ]$

Using the integral from the problem statement, we get

$(2\pi-\frac{\pi}{2})\left [ \frac{e^{\sin(2\pi)}+e^{\sin(\frac{\pi}{2})}}{2}\right ]$

Simplifying, we end up with

$\frac{3\pi}{2}(\frac{1+e}{2})$

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Example Question #41 : Trapezoidal Sums

Evaluate using the trapezoidal approximation:

$\int_{0}^{3}x^2 dx$

Possible Answers:

$\frac{3}{2}$

$\frac{27}{2}$

$\frac{9}{2}$

Correct answer:

$\frac{27}{2}$

Explanation:

The trapezoidal approximation of a definite integral is given by

$\int_{a}^{b}f(x)dx\approx (b-a)(\frac{f(b)+f(a)}{2})$

For our integral, we get

$\int_{0}^{3}x^2 dx \approx (3-0)(\frac{9+0}{2})=\frac{27}{2}$

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Example Question #42 : Trapezoidal Sums

Use the method of trapezoidal sums to approximate the integral

$\int_{0}^{1}\frac{4x-e^x}{2x}dx$

Possible Answers:

$\frac{3-e}{7}$

$\frac{5-e}{10}$

$\frac{3-e}{10}$

$\frac{3+e}{10}$

Correct answer:

$\frac{3-e}{10}$

Explanation:

To approximate the integral using trapezoidal sums, we use the following formula:

$\int_{a}^{b} f(x)dx=(b-a)\left [ \frac{f(b)+f(a)}{2}\right ]$

Using the integral from the problem statement, we get

$(1-0)\left [ \frac{(\frac{4-e}{5})+(\frac{-1}{5})}{2} \right ]=\frac{3-e}{10}$

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Example Question #43 : Trapezoidal Sums

Using the method of trapezoidal sums, evaluate the following integral

$\int_{0}^{\frac{\pi}{2}}e^x\sin{x}dx$

Possible Answers:

$\frac{\pi}{2}e^\frac{\pi}{2}$

$\frac{\pi}{4}e^\frac{\pi}{4}$

$\frac{\pi}{4}e^\frac{\pi}{2}$

$\frac{\pi}{4}$

Correct answer:

$\frac{\pi}{4}e^\frac{\pi}{2}$

Explanation:

To find the value of the integral, we use the following formula

$\int_{a}^{b}f(x)dx=\left [ b-a\right ](\frac{f(a)+f(b)}{2})$

$\left [ \frac{\pi}{2}-0 \right ](\frac{\sin(\frac{\pi}{2})e^\frac{\pi}{2}+\sin(0)e^0}{2})=\frac{\pi}{4}e^\frac{\pi}{2}$

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Example Question #44 : Trapezoidal Sums

Using the method of trapezoidal sums, evaluate the following integral

$\int_{3}^{5}(2xe^{2x+5})dx$

Possible Answers:

$10e^{30}-6e^{14}$

$10e^{30}+6e^{14}$

$5e^{30}-3e^{14}$

$10e^{10}-6e^{3}$

Correct answer:

$10e^{30}+6e^{14}$

Explanation:

To use the method of trapezoidal sums, we follow the definition

$\int_{a}^{b}f(x)dx=(b-a)\left [ \frac{f(a)+f(b)}{2}\right ]$

Using the information from the problem statement, we get

$(5-3)(\frac{10e^{30}+6e^{14}}{2})=10e^{30}+6e^{14}$

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Example Question #46 : Trapezoidal Sums

Use the trapezoidal approximation to solve the definite integral, and find the difference between it and the actual integral:

$\int_{0}^{5} x^4 dx$

Possible Answers:

$\frac{3075}{2}$

$\frac{3125}{2}$

$\frac{1875}{2}$

Correct answer:

$\frac{1875}{2}$

Explanation:

The trapezoidal approximation to definite integrals is given by

$\int_{a}^{b} f(x)dx\approx (b-a)(\frac{f(b)+f(a)}{2})$

Using this formula for our integral, we get

$\int_{0}^{5} x^4 dx \approx (5-0)(\frac{625-0}{2})=\frac{3125}{2}$

Actually integrating, we get

$\int_{0}^{5} x^4 dx = \frac{x^5}{5}\left.\begin{matrix} 5\\0 \end{matrix}\right| = \frac{5^5}{5}=5^4=625$

The rule used for integration is

$\int x^n dx=\frac{x^{n+1}}{n+1}+C$

The difference between the approximation and the actual answer is

$\frac{3125}{2}-625=\frac{1875}{2}$

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Example Question #311 : Integrals

Approximate the value of $\int_{-2}^{4} x^2 dx$ using a trapezoidal sum with step size h = 2 . How far away is this approximation from the actual value of the integral above?

Possible Answers:

Correct answer:

Explanation:

Trapezoidal sums are found by creating trapezoids whose left and right end points are on the specified function, and whose widths are the step size. We then sum up their areas by remembering that the area of a trapezoid is the base times the average of the heights.

Thus, the calculation of the trapezoidal sum for this example would be $2 \cdot [\frac{f(-2)+f(0)}{2}] + 2 \cdot [\frac{f(0)+f(2)}{2}] + 2 \cdot [\frac{f(2)+f(4)}{2}]$

A more simplified version would be given by f(-2) + 2(f(0) + f(2)) + f(4)

Which evaluates to

4 + 2(0 + 4) + 16 = 28

The actual answer is found by evaluating the definite integral given, which would just be given by $\int_{-2}^{4} x^2 dx = \left . \frac{x^3}{3} \right ]^4_{-2} = \frac{4^3 - (-2)^3}{3} = \frac{64 + 8}{3} = \frac{72}{3} = 24$

The difference between the approximation and the the true answer is thus 28-24 = 4

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AP Calculus AB : Trapezoidal sums

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #305 : Integrals

Example Question #305 : Integrals

Example Question #311 : Integrals

Example Question #41 : Trapezoidal Sums

Example Question #42 : Trapezoidal Sums

Example Question #43 : Trapezoidal Sums

Example Question #44 : Trapezoidal Sums

Example Question #46 : Trapezoidal Sums

Example Question #311 : Integrals

All AP Calculus AB Resources

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