Numerical Approximation

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GRE Quantitative Reasoning › Numerical Approximation

Questions 1 - 10

1

Solve the integral

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

using Simpson's rule with subintervals.

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 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 therefore

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

2

Solve the integral

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

using Simpson's rule with subintervals.

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 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 therefore

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

3

Solve the integral

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

using Simpson's rule with subintervals.

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 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 therefore

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

4

Solve the integral

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

using the trapezoidal approximation with subintervals.

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 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 , therefore

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

5

Solve the integral

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

using the trapezoidal approximation with subintervals.

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 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 , therefore

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

6

Solve the integral

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

using the trapezoidal approximation with subintervals.

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 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 , therefore

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

7

For which values of p is

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

convergent?

only

All positive values of

it doesn't converge for any values of

Explanation

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

.

8

For which values of p is

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

convergent?

only

All positive values of

it doesn't converge for any values of

Explanation

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

.

9

For which values of p is

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

convergent?

only

All positive values of

it doesn't converge for any values of

Explanation

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

.

10

Solve the integral

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

using the trapezoidal approximation with subintervals.

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 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 , 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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