☺'s question at Yahoo! Answers: a definite integral whose integrand has an absolute value factor.

MarkFL

Pessimist Singularitarian
Staff member
's question at Yahoo! Answers: a definite integral whose integrand has an absolute value factor.

Here is the question:

☺ said:
How to solve ʃ (-18 to 1) [s│81 - s^2 │] ds?

The 81 - s^2 is inside absolute value signs. I don't know how to do this. Can you show me how?
I have posted a link there to this thread so the OP can view my work.

MarkFL

Pessimist Singularitarian
Staff member
Re: 's question at Yahoo! Answers: a definite integral whose integrand has an absolute value factor

Hello ,

We are given to evaluate:

$$\displaystyle I=\int_{-18}^{1} s\left|81-s^2\right|\,ds$$

First, we should observe that the expression within the absolute value is negative on:

$$\displaystyle (-\infty,-9)\,\cup\,(9,\infty)$$

And so, using the definition of an absolute value, namely:

$$\displaystyle |x|=\begin{cases}x & 0\le x\\ -x & x<0 \\ \end{cases}$$

we may then write:

$$\displaystyle I=\int_{-18}^{-9} s\left(s^2-81\right)\,ds-\int_{-9}^{1}s\left(s^2-81\right)\,ds$$

Using the substitution:

$$\displaystyle u=s^2-81\,\therefore\,du=2s\,ds$$

we obtain:

$$\displaystyle I=\frac{1}{2}\left(\int_{-80}^{0} u\,du-\int_{0}^{3^5}u\,du\right)$$

Applying the FTOC, we get:

$$\displaystyle I=\frac{1}{4}\left(\left[u^2\right]_{-80}^{0}-\left[u^2\right]_{0}^{3^5}\right)$$

$$\displaystyle I=\frac{1}{4}\left(-80^2-9^5\right)=-\frac{65449}{4}$$

Hence, we may conclude:

$$\displaystyle \bbox[10px,border:2px solid #207498]{\int_{-18}^{1} s\left|81-s^2\right|\,ds=-\frac{65449}{4}}$$