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GRE Subject Test: Math : Simpson's Rule

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Example Questions

Example Question #1 : Simpson's Rule

Solve the integral

$\int_{1}^{5}{\frac{1}{x-7}}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

-0.500

-1.6667

-0.1667

-8.2400

-1.0987

Correct answer:

-1.0987

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{5-1}{10})=0.4$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.35.50 pm

The sum of all the approximation terms is -8.23995 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(-8.23995)=-1.0987$

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Example Question #2 : Simpson's Rule

Solve the integral

$\int_{8}^{15}{ln(x)}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

2.0794

10.2289

8.6532

16.9852

10.6410

Correct answer:

16.9852

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx=S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{15-8}{10})=0.7$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.35.58 pm

The sum of all the approximation terms is 72.79378 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.7)*(72.79378)=16.9852$

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Example Question #3 : Simpson's Rule

Solve the integral

$\int_{-1}^{1}{\frac{1}{2+cos(x)}}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

0.3948

10.9830

0.7056

10.5872

1.2340

Correct answer:

0.7056

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx =S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{1+1}{10})=0.2$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.36.10 pm

The sum of all the approximation terms is 10.5840 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.2)*(10.5840)=0.7056$

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Example Question #4 : Simpson's Rule

Solve the integral

$\int_{0}^{9}{\frac{1}{1+e^{x}}}dx$

using Simpson's rule with 2n=10 subintervals.

Possible Answers:

0.6924

0.3482

0.8829

0.6599

1.2369

Correct answer:

0.6924

Explanation:

Simpson's rule is solved using the formula

$\int_{a}^{b}f(x))dx\approx S_{2m}=\frac{1}{3}(\frac{b-a}{2n})\left [ f(0)+4f(1)+2f(2)+...+2f(m-2)+4f(m-1)+f(m) \right ]$

where is the number of subintervals and f(m) is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{9-0}{10})=0.9$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 9.36.20 pm

The sum of all the approximation terms is 2.3081 therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.9)*(2.3081)=0.6924$

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GRE Subject Test: Math : Simpson's Rule

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Simpson's Rule

Example Question #2 : Simpson's Rule

Example Question #3 : Simpson's Rule

Example Question #4 : Simpson's Rule

All GRE Subject Test: Math Resources

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