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GRE Subject Test: Math : Numerical Integration

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Example Questions

Example Question #14 : Midpoint Riemann Sums

Solve the integral

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

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

Possible Answers:

-1.6667

-0.1667

-1.0987

-8.2400

-0.500

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 #1 : Numerical Integration

Solve the integral

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

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

Possible Answers:

10.6410

2.0794

10.2289

8.6532

16.9852

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 : Numerical Approximation

Solve the integral

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

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

Possible Answers:

0.7056

1.2340

10.5872

0.3948

10.9830

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 #21 : Midpoint Riemann Sums

Solve the integral

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

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

Possible Answers:

0.8829

0.6924

0.6599

1.2369

0.3482

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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Example Question #12 : Midpoint Riemann Sums

Solve the integral

$\int_{2}^{7}{\frac{1}{5x+3}}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

0.5233

0.4301

0.8603

0.9548

0.7601

Correct answer:

0.4301

Explanation:

Trapezoidal approximations are solved using the formula

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

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

For this problem, $(\frac{b-a}{2n})=(\frac{7-2}{2*5})=0.5$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.19.15 pm

The sum of all the approximation terms is 0.86027754 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.5)*(0.86027754)=0.4301$

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Example Question #13 : Midpoint Riemann Sums

Solve the integral

$\int_{1}^{2}{x\sin (x)}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

28.7867

2.5052

0.8415

2.8787

Correct answer:

2.8787

Explanation:

Trapezoidal approximations are solved using the formula

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

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

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{2*5})=0.1$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.32.39 pm

The sum of all the approximation terms is 28.7867 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)*(28.786699)=2.8787$

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Example Question #5 : Numerical Integration

Solve the integral

$\int_{0}^{4}{\sqrt{5+x^{^{3}}}}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

83.8900

33.5582

5.0000

14.3744

2.3547

Correct answer:

33.5582

Explanation:

Trapezoidal approximations are solved using the formula

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

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

For this problem, $(\frac{b-a}{2n})=(\frac{4-0}{2*5})=0.4$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.55.34 pm

The sum of all the approximation terms is 83.89553 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.4)*(83.89553)=33.5582$

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Example Question #1 : Numerical Integration

Solve the integral

$\int_{1}^{2}{\sqrt{ln(x)}}dx$

using the trapezoidal approximation with n=5 subintervals.

Possible Answers:

1.1726

0.6174

11.7257

0.8326

Correct answer:

1.1726

Explanation:

Trapezoidal approximations are solved using the formula

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

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

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{2*5})=0.1$ .

The value of each approximation term is below.

Screen shot 2015 06 11 at 8.55.45 pm

The sum of all the approximation terms is 11.7257431 , therefore

$(\frac{b-a}{2n})\left [ f(0)+2f(1)+...+f(m-1)+f(m) \right ]=(0.1)*(11.7257431)=1.1726$

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Example Question #1 : Numerical Integration

Evaluate $\int_{0}^{2} x^{x^{2}} dx$ using the Trapezoidal Rule, with n = 2.

Possible Answers:

9.0

16.0

8.5

5.0

2.0

Correct answer:

9.0

Explanation:

1) n = 2 indicates 2 equal subdivisions. In this case, they are from 0 to 1, and from 1 to 2.

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

3) For n = 2: $\int_{0}^2 x^{x^{2}} dx \approx [1-0]\left [ \frac{f(0)+f(1)}{2} \right ]+ [2-1]\left [ \frac{f(1)+f(2)}{2} \right ]$

4) Simplifying: $\int_{0}^2 x^{x^{2}} dx \approx [1]\left [ \frac{0+1}{2} \right ]+ [1]\left [ \frac{1+16}{2} \right ] = \frac{1}{2}+\frac{17}{2} = \frac{18}{2} = 9.$

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GRE Subject Test: Math : Numerical Integration

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #14 : Midpoint Riemann Sums

Example Question #1 : Numerical Integration

Example Question #3 : Numerical Approximation

Example Question #21 : Midpoint Riemann Sums

Example Question #12 : Midpoint Riemann Sums

Example Question #13 : Midpoint Riemann Sums

Example Question #5 : Numerical Integration

Example Question #1 : Numerical Integration

Example Question #1 : Numerical Integration

All GRE Subject Test: Math Resources

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