Trapezoidal Rule - GRE Quantitative Reasoning

Card 0 of 20

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

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

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

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

Compare your answer with the correct one above

Tap the card to reveal the answer