Numerical Approximation - GRE Quantitative Reasoning

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Question

For which values of p is

$\sum_{n=1}^\infty \frac{1}{n^p}$

convergent?

Answer

We can solve this problem quite simply with the integral test. We know that if

$\int_1^{\infty} \frac{1}{x^p} dx$

converges, then our series converges.

We can rewrite the integral as

$\int x^{-p} dx$

and then use our formula for the antiderivative of power functions to get that the integral equals

$\frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty$ .

We know that this only goes to zero if . Subtracting p from both sides, we get

.

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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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Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{7-2}{25})=0.5$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

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Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{25})=0.1$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

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Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{4-0}{25})=0.4$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

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Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{25})=0.1$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

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Question

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

Answer

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

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

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

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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Question

For which values of p is

$\sum_{n=1}^\infty \frac{1}{n^p}$

convergent?

Answer

We can solve this problem quite simply with the integral test. We know that if

$\int_1^{\infty} \frac{1}{x^p} dx$

converges, then our series converges.

We can rewrite the integral as

$\int x^{-p} dx$

and then use our formula for the antiderivative of power functions to get that the integral equals

$\frac{x^{1-p}}{1-p} \Big|_{x=1}^\infty$ .

We know that this only goes to zero if . Subtracting p from both sides, we get

.

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{7-2}{25})=0.5$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

Compare your answer with the correct one above

Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{25})=0.1$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

Compare your answer with the correct one above

Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{4-0}{25})=0.4$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

Compare your answer with the correct one above

Question

Solve the integral

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

using the trapezoidal approximation with subintervals.

Answer

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 is the function evaluated at the midpoint.

For this problem, $(\frac{b-a}{2n})=(\frac{2-1}{25})=0.1$ .

The value of each approximation term is below.

The sum of all the approximation terms is , therefore

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

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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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