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?

Re: Evaluating an integral interpreted as area...

$\displaystyle \int_{-2}^2 (x+3)(4 - x^2)^{\frac{1}{2}} dx $

$\displaystyle \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} + 3*(4 - x^2)^{\frac{1}{2}} dx $

$\displaystyle \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + \int_{-2}^2 3*(4 - x^2)^{\frac{1}{2}} dx$

$\displaystyle \int_{-2}^2 x * (4 - x^2)^{\frac{1}{2}} dx + 3 * \int_{-2}^2 (4 - x^2)^{\frac{1}{2}} dx$

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

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

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

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

Therefore,

$\displaystyle \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 $

Re: Evaluating an integral interpreted as area...

Omg :o

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

Thank you :D