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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

again i just want to check if my answers were correct. i didn't include the complete solution because i'm in a hurry our finals is just around the corner. if my answers are not correct just say so and i will try to solve it again until i get it. thanks!

1. find the area bounded by $y^2=4x$ & $y=2x-4$.

points of intersection are x=4,-2

$\displaystyle\int_{-2}^4 (\frac{1}{2}x+2-\frac{1}{4}x^2)dx$ = $\left[\frac{x^2}{4}+2x-\frac{x^3}{12}\right]_{-2}^4$

it turns out that the area is 16 sq. units

2. find the area bounded by $y=x+3$ and $y=x^2+x-13$

points of intersection are x=4,-4

$\displaystyle\int_{-4}^4 (16-x^2)dx = \left[16x-\frac{x^3}{3}\right]_{-4}^4$

the area is 256/3 sq. units.

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