Evaluating a definite integral

**Ragnarok** · June 8th 2012, 07:06 AM

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

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

In the answer key the next step is written as:

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

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 $x = 4-y^2$ and $x = y-2$ by integrating with respect to x. I set up the integral properly but I don't understand why it is being evaluted like this.

**Plato** · June 8th 2012, 07:15 AM

Originally Posted by Ragnarok

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

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

In the answer key the next step is written as:

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

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 $x = 4-y^2$ and $x = y-2$ 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.

**Ragnarok** · June 8th 2012, 07:19 AM

Ah, okay. I forgot to use u-substitution. Thank you!

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