calculating definate intregals

**softballchick** · February 11th 2011, 01:47 PM

Let

.
Find

and

Let

.

Find

and

So I found the first one, which is just 0.

How do I find the seecond one? I don't get the logic of it...

**zzzoak** · February 11th 2011, 02:27 PM

$<br /> \int_{-1}^{-3.5}(10f(x)-3)dx=-\int_{-3.5}^{-1}(10f(x)-3)dx=-10\int_{-3.5}^{-1}f(x)dx+3\int_{-3.5}^{-1}dx<br />$

**softballchick** · February 11th 2011, 02:54 PM

wouldn't it just be -10 (0) + 3(0) which is just zero? but that answer is wrong...

**zzzoak** · February 11th 2011, 03:09 PM

$<br /> \int_{-3.5}^{-1}dx=x|_{-3.5}^{-1}=(-1)-(-3.5)=2.5<br />$

**softballchick** · February 11th 2011, 03:42 PM

Sorry I still don't get it. What is dx or x in this problem? If I wsa given that, I could figure it ou, I know how to be integration...

Thread: calculating definate intregals