# Piecewise Integral, question about breaking integrals up

• Dec 9th 2012, 04:34 PM
AZach
Piecewise Integral, question about breaking integrals up
$\int_{-\pi}^{\pi} f(x)dx$ where $f(x) = \left\{ {\begin{array}{rl} {6x^3,} & {\text{if }x~-\pi=

I broke up the integral like this $\int_{-\pi}^{0} 6x^{3} dx + \int_{0}^{\pi} 7sin(x)$ . When I look at the graph of these two functions I notice that they overlap each other. Does that affect the area value in any way? I know the antiderivates will be even functions, so I think I only have to find the integral of the 1st and 4th quadrant and then just double it. If I do : $2 \int_{0}^{\pi} 7sin(x)$ I get 28, but that's not correct. I'm sure I'm just not understanding something about these graphs, that's what I ask if the overlap feature means I need to break up the endpoints of the integral differently?
• Dec 9th 2012, 04:58 PM
Plato
Re: Piecewise Integral, question about breaking integrals up
Quote:

Originally Posted by AZach
$\int_{-\pi}^{\pi} f(x)dx$ where $f(x) = \left\{ {\begin{array}{rl} {6x^3,} & {\text{if }x~-\pi=

I do not understand most of your statement. In particular the bit about "overlapping".

Look at this graphic.
• Dec 9th 2012, 05:20 PM
AZach
Re: Piecewise Integral, question about breaking integrals up
When I graphed the two functions 6x^3 and 7sin(x) on my TI83 I got a graph that looked like this Attachment 26162

Did I have the integrals broken up correctly?
• Dec 9th 2012, 05:48 PM
skeeter
Re: Piecewise Integral, question about breaking integrals up
Quote:

Originally Posted by AZach
When I graphed the two functions 6x^3 and 7sin(x) on my TI83 I got a graph that looked like this Attachment 26162

Did I have the integrals broken up correctly?

$6x^3$ is for $x < 0$ and $7\sin{x}$ is for $x \ge 0$ ... they do not "overlap".

how to graph this piece-wise function in a TI-83 ...

$Y1 = 6x^3/(x < 0)$

$Y2 = 7\sin(x)/(x \ge 0)$

set your x window from $-\pi$ to $\pi$ and y window from -10 to 10
• Dec 9th 2012, 06:16 PM
AZach
Re: Piecewise Integral, question about breaking integrals up
Thanks Skeeter, I never knew how to do that before.