Leigh J. answered • 11/15/12

Biology, Chemistry, Math, and K-12 Tutor

Hi Rhonda!

To deal with the 1 in this integral, you can break up the original integral as follows:

∫[1+√(9-x^2)]dx = ∫1dx + ∫√(9-x^2)dx

*Note: I have indicated these as indefinite integrals because there is no way to nicely format definite integrals on here.

From there, you take the definite integrals each separately as you normally would, and add them together to find the answer.

Hope this helps! Let me know if you have any further questions.

-Leigh

Michael B.

Wow this editor doesn't work well at all... It messed up a bunch of characters. That last line should read "integral of (1/2)3^{2} times d-theta"

Test: ? θ θ T

11/15/12

Michael B.

Good answer!

It may not be apparent to the OP, but ?v(9-x

^{2})dx isn't a simple one, as you can't use u-substitution. There's two ways to handle this. First is to realize that this is the equation of a circle with radius=3, and thus the integral is asking for the area in the first quadrant. So you can just calculate area of the circle and divide it by 4 - that's the "graphical" method.To actually integrate this, the simplest method is to convert it to polar form:

y = v(9-x

^{2})y

^{2}= 9 - x^{2}x

^{2}+ y^{2}= 9r

^{2}= 9r = 3

? (1/2)(3)

^{2}d? (from 0 -> pi/2)11/15/12