# Evaluating a definite integral

• Jun 8th 2012, 07:06 AM
Ragnarok
Evaluating a definite integral
Hello, I am having trouble with a calculus problem which requires us to evaluate this sum of two integrals:

$\displaystyle \int^0_{-5} x + 2 + \sqrt{4-x}\,dx + \int^4_0 2\sqrt{4-x}\,dx$

In the answer key the next step is written as:

$\displaystyle \left[ \frac{x^2}{2} + 2x - \frac{(4-x)^{3/2}}{3/2} \right]_{-5}^0 + 2\left[ \frac{(4-x)^{3/2}}{-3/2} \right]^4_0$

However, I don't understand why the third term in the first integral is being subtracted, and why the denominator in the second integral is negative. If it helps, the original problem was to find the area of the region bounded by $\displaystyle x = 4-y^2$ and $\displaystyle x = y-2$ by integrating with respect to x. I set up the integral properly but I don't understand why it is being evaluted like this.
• Jun 8th 2012, 07:15 AM
Plato
Re: Evaluating a definite integral
Quote:

Originally Posted by Ragnarok
Hello, I am having trouble with a calculus problem which requires us to evaluate this sum of two integrals:

$\displaystyle \int^0_{-5} x + 2 + \sqrt{4-x}\,dx + \int^4_0 2\sqrt{4-x}\,dx$

In the answer key the next step is written as:

$\displaystyle \left[ \frac{x^2}{2} + 2x - \frac{(4-x)^{3/2}}{3/2} \right]_{-5}^0 + 2\left[ \frac{(4-x)^{3/2}}{-3/2} \right]^4_0$

However, I don't understand why the third term in the first integral is being subtracted, and why the denominator in the second integral is negative. If it helps, the original problem was to find the area of the region bounded by $\displaystyle x = 4-y^2$ and $\displaystyle x = y-2$ by integrating with respect to x. I set up the integral properly but I don't understand why it is being evaluted like this.

Take the derivative of each. That should show you why.
• Jun 8th 2012, 07:19 AM
Ragnarok
Re: Evaluating a definite integral
Ah, okay. I forgot to use u-substitution. Thank you!