Simpson's Rule - GRE Subject Test: Math

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Question

Solve the integral

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

using Simpson's rule with subintervals.

Answer

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 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.

The sum of all the approximation terms is 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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Question

Solve the integral

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

using Simpson's rule with subintervals.

Answer

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 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.

The sum of all the approximation terms is 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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Question

Solve the integral

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

using Simpson's rule with subintervals.

Answer

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 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.

The sum of all the approximation terms is 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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Question

Solve the integral

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

using Simpson's rule with subintervals.

Answer

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 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.

The sum of all the approximation terms is 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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