$\displaystyle \int_{-5}^5 (8-|x|) dx$
the answer is 55 but I am hving throuble getting to the solution.
Sorry! I can solve it using geometry, but I still need a little more help.
The interval length I had before was 10/n, would it now change to 5/n ??
$\displaystyle \frac{5}{n} [\sum\limits_{k=1}^{n} (8+x) + \sum\limits_{k=1}^{n} (8-x) ] $
Using reimann's right sum $\displaystyle (a+k* \Delta )$
does it matter which reimann sum I use?
According to Captain Black's post, your area is $\displaystyle 2\int_{0}^{5}(8-x)dx$.
Now, recall that $\displaystyle \int_{a}^{b}f(x)dx=\lim_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$ where $\displaystyle \Delta x=\frac{b-a}{n}$ and $\displaystyle x_i=a+i \Delta x$.
In this case
$\displaystyle a=0$
$\displaystyle b=5$
$\displaystyle \Delta x = \frac{5}{n}$
$\displaystyle x_i=0+i\Delta x =\frac{5i}{n}$
$\displaystyle f(x_i)=8-\frac{5i}{n}$
required area = $\displaystyle 2 \times \lim_{n \to \infty}\sum_{i=1}^{n}( 8-\frac{5i}{n})\frac{5}{n}$
Evaluating this, you will get the value of the definite integral.
I'm afraid most of CalBear12's posts partake of the nature of the one you are commenting upon. I'm hopping he will learn to read the original and the other posts in a thread more carefully before shooting-form-the-hip before he earns a weeks ban due to accumulated infractions.
CB