# Evaluate integral by area

1. Dec 4, 2009

### synergix

1. The problem statement, all variables and given/known data
Evaluate the integral by intrepreting it in terms of areas

from -3 to 0

(1+ sqrt(9-x2)dx

2. Relevant equations

integral = F(0)-F(-3)

3. The attempt at a solution

first find F

F = x - [(9-x2)3/2]/3

I solved using integral = F(0)-F(-3)

and I got the incorrect answer I think it is because I am finding the definite integral and not the area. If this was my problem how would I find the area? Graphing?

Last edited: Dec 5, 2009
2. Dec 5, 2009

### Staff: Mentor

Yes - graph it. The function defines a fairly simple geometric object.

BTW, your work finding an antiderivative, besides not being the direction you're supposed to go, is incorrect. If you take the derivative of the function you found, you don't get 1 + sqrt(9 -x^2)

3. Dec 5, 2009

### synergix

ok i changed it i think its right now and i will try to graph it

4. Dec 5, 2009

### Staff: Mentor

If you want to get credit for your work, the first thing you should do is graph your function. This problem has nothing to do with finding antiderivatives.

5. Dec 5, 2009

### HallsofIvy

Staff Emeritus
It would help to think of this as two separate integrals:
$$\int_{-3}^0 1+ \sqrt{9- x^2} dx= \int_{-3}^0 1 dx+ \int_{-3}^0 \sqrt{9- x^2} dx$$
Graph each, if necessary, to recognise that those are simple figures and use basic area formulas.