MHB Ahmed's question at Yahoo Answers regarding the midpoint rule

[MarkFL]

Here is the original question:

Use the Midpoint Rule with n=4 to approximate?

1.5
Integral sqrt (x^4 +19) dx
-2.5

Here is a link to the question:

Use the Midpoint Rule with n=4 to approximate? - Yahoo! Answers

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

 

[MarkFL]

Hello Ahmed,

We are given to approximate:

$\displaystyle \int_{-2.5}^{1.5}\sqrt{x^4+19}\,dx$

using the midpoint rule with $n=4$

The Midpoint Rule is the approximation:

$\displaystyle \int_a^b f(x)\,dx\approx M_n$ where:

$\displaystyle M_n=\frac{b-a}{n}\cdot\sum_{k=0}^{n-1}\left[f\left(\frac{x_k+x_{k+1}}{2} \right) \right]$

where $\displaystyle x_k=a+k\frac{b-a}{n}$.

With $a=-2.5,\,b=1.5,\,n=4$ this becomes:

$\displaystyle M_n=\frac{1.5+2.5}{4}\cdot\sum_{k=0}^{3}\left[f\left(\frac{-2.5+k\frac{1.5+2.5}{4}-2.5+(k+1)\frac{1.5+2.5}{4}}{2} \right) \right]$

$\displaystyle M_n=\sum_{k=0}^{3}\left[f\left(k-2 \right) \right]$

Using the given integrand, this becomes:

$\displaystyle M_n=\sum_{k=0}^{3}\left[\sqrt{(k-2)^4+19} \right]$

$\displaystyle M_n=\sqrt{(0-2)^4+19}+\sqrt{(1-2)^4+19}+\sqrt{(2-2)^4+19}+\sqrt{(3-2)^4+19}$

$\displaystyle M_n=\sqrt{35}+\sqrt{20}+\sqrt{19}+\sqrt{20}=4\sqrt{5}+\sqrt{19}+\sqrt{35}\approx19.219250636639448$