# Evaluating an integral interpreted as area...

• Apr 29th 2013, 04:57 PM
ILovePizza
Evaluating an integral interpreted as area...
Finishing the last chapter of Calculus 1, so go easy on me you guys :|

I know this is like the easiest and most obvious thing. I just can't figure it out though; I need help.

I don't know why the answer is 6π.

Evaluate ∫ from -2 to 2 of (x + 3)(4 - x^2)1/2 dx by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.

Why? And how is this done?
• Apr 29th 2013, 09:13 PM
mathguy25
Re: Evaluating an integral interpreted as area...
$\int_{-2}^2 (x+3)(4 - x^2)^{\frac{1}{2}} dx$
$\int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} + 3*(4 - x^2)^{\frac{1}{2}} dx$
$\int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + \int_{-2}^2 3*(4 - x^2)^{\frac{1}{2}} dx$
$\int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + 3 * \int_{-2}^2 (4 - x^2)^{\frac{1}{2}} dx$

Now, note that $\int_{-2}^2 (4 - x^2)^{\frac{1}{2}}$ is the semi unit circle of radius 2. Thus, the area is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi 2^2 = 2 \pi$

Note, use direct subsitution $u = 4 - x^2$ to evaluate $\int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx$

Then $u(-2) = u(2) = 0$ and $du = -2x dx$

Then $\int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx$ = $\int_{0}^0 \frac{-1}{2} u du$ which equals zero.

Therefore,

$\int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + \int_{-2}^2 3*(4 - x^2)^{\frac{1}{2}} dx = 0 + 3 * 2 \pi = 6 \pi$
• Apr 29th 2013, 10:26 PM
ILovePizza
Re: Evaluating an integral interpreted as area...
Omg :o

I have an exam on this stuff soon. This helped a lot!

Thank you :D