Remember the rule ∫ ( f(x) + g(x) )dx = ∫ f(x)dx + ∫ g(x)dx. In other words, integration is
**additive** (distributes over addition). That is how the 1 is pulled out (see Leigh's answer.)

If you really wanted to use symbolic integration and plug in the bounds then use trigonometric substitution. Some of you tried to show this but had formatting trouble.

If x = 3 sin θ then dx = 3 cos θ dθ

We get

∫ [1+√(9-x^{2})] dx

= ∫ [1+√(9-9 sin^{2} θ)] * 3 cos θ dθ

= ∫ [3 cos θ + 9 cos^{2} θ] dθ

= ∫ [3 cos θ + 9(1+cos 2θ)/2 ] dθ

= ∫ [3 cos θ + 9/2 + (9/2) cos 2θ] dθ

= 3 sin θ + 9θ/2 + (9/4) sin 2θ + C

= 3 sin θ + 9θ/2 + (9/2) sin θ cos θ+ C

= x + (9/2) sin^{-1} (x/3) + (9/2) (x/3) √(1 - x^{2}/9) + C

= x + (9/2) sin^{-1} (x/3) + (x/2)√(9 - x^{2}) + C

Substituting the bounds -3 and 3 gives

[3+(9/2) sin^{-1} (3/3)+(1/2) (3) √(9 - 3^{2})] - [0+(9/2) sin^{-1} (0/3)+(1/2) (0) √(9 - 0^{2})]

= [3+(9/2) (π/2) + 0] - [0 + (9/2) (0) + 0]

= 3+9π/4

Same as the answer realized with the recognition that the integral equals the sum of a rectangle and quarter-circle.

## Comments

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)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