Trapezoidal Rule

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GRE Quantitative Reasoning › Trapezoidal Rule

Questions 1 - 5

1

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

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

2

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

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

3

Solve the integral

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

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

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 , therefore

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

4

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

Explanation

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

5

Solve the integral

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

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

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 , 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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