MHB Kendra N's question at Yahoo Answers regarding the Midpoint Rule

[MarkFL]

Here is the question:

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal?

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places

∫ 2 cos^5 (x) dx from x = 0 to x = π/2, n = 4

Here is a link to the question:

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Kendra N,

The Midpoint Rule is the approximation $\displaystyle \int_a^b f(x)\,dx\approx M_n$ where:

$\displaystyle M_n=\frac{b-a}{n}\sum_{k=1}^n\left[f\left(\frac{x_{k-1}+x_k}{2} \right) \right]$

We are asked to use this rule to approximate:

$\displaystyle 2\int_0^{\frac{\pi}{2}}\cos^5(x)\,dx$

Identifying:

$\displaystyle a=0,\,b=\frac{\pi}{2},\,n=4,\,f(x)=\cos^5(x),\,x_k=k\cdot\frac{\pi}{8}$, we have:

$\displaystyle M_4=2\frac{\frac{\pi}{2}-0}{4}\sum_{k=1}^4\left[\cos^5\left(\frac{(k-1)\cdot\frac{\pi}{8}+k\cdot\frac{\pi}{8}}{2} \right) \right]$

This simplifies to:

$\displaystyle M_4=\frac{\pi}{4}\sum_{k=1}^n\left[\cos^5\left(\frac{\pi}{16}(2k-1) \right) \right]$

Using a calculator/computer and rounding to 4 decimal places, we find:

$\displaystyle M_4\approx1.0667$

For comparison, the true value of the integral is:

$\displaystyle 2\int_0^{\frac{\pi}{2}}\cos^5(x)\,dx=\frac{16}{15}=1.0\bar{6}$.