# Thread: Sum of two integrals

1. ## Sum of two integrals

I'm confused here...I'm guessing it's just using one that's above the x axis and subtracting the one that's below the x axis but I'm not too sure and the function is throwing me off.

Evaluate (x+3)sqrt[4-x^2] from -2 to 2 by writing it as a sum of two integrals and interpreting one of those in terms of an area.

2. Originally Posted by sfgiants13
I'm confused here...I'm guessing it's just using one that's above the x axis and subtracting the one that's below the x axis but I'm not too sure and the function is throwing me off.

Evaluate (x+3)sqrt[4-x^2] from -2 to 2 by writing it as a sum of two integrals and interpreting one of those in terms of an area.
ok, to begin: just expand, right?

$(x + 3) \sqrt{4 - x^2} = x \sqrt{4 - x^2} + 3 \sqrt{4 - x^2}$

further note that $y = \sqrt{4 - x^2}$ is the equation for the upper half of a circle

3. Originally Posted by Jhevon
ok, to begin: just expand, right?

$(x + 3) \sqrt{4 - x^2} = x \sqrt{4 - x^2} + 3 \sqrt{4 - x^2}$

further note that $y = \sqrt{4 - x^2}$ is the equation for the upper half of a circle
So it would be the derivatives of the 2nd equation subtracted from the 1st correct?

4. Originally Posted by sfgiants13
So it would be the derivatives of the 2nd equation subtracted from the 1st correct?
no, we're adding. there is a + sign between them

to do the problem as they said, you will need to use integration by parts for the first integral. otherwise, a simple substitution of $u = 4 - x^2$ would have finished this problem nicely.

5. Ok thanks I figured it out...after getting a nice messy equation I got 6pi.

### evaluate by writing it as a sum of two integrals and interpretingone as an area.

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