Piecewise Integral, question about breaking integrals up

$\displaystyle \int_{-\pi}^{\pi} f(x)dx $ where $\displaystyle f(x) = \left\{ {\begin{array}{rl} {6x^3,} & {\text{if }x~-\pi=<x<0} \\ {7sin(x),} & {\text{if }x~0=<x<=\pi} \\\end{array} } \right.$

I broke up the integral like this $\displaystyle \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 : $\displaystyle 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?

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Re: Piecewise Integral, question about breaking integrals up

Quote:

Originally Posted by

**AZach** $\displaystyle \int_{-\pi}^{\pi} f(x)dx $ where $\displaystyle f(x) = \left\{ {\begin{array}{rl} {6x^3,} & {\text{if }x~-\pi=<x<0} \\ {7sin(x),} & {\text{if }x~0=<x<=\pi} \\\end{array} } \right.$

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

Look at this graphic.

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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?

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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?

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

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

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

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

set your x window from $\displaystyle -\pi$ to $\displaystyle \pi$ and y window from -10 to 10

Re: Piecewise Integral, question about breaking integrals up

Thanks Skeeter, I never knew how to do that before.