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I need to find the area of the region that is bounded by $y=x^2-4$ and $y=2x-1$

I think I solved it, but I don't know what the right answer is so I'm not sure!

I got:

$$da = wl$$ $$=(x^2-4)-(2x-1)dy$$ $$=(x^2-2x-3)dy$$ $$\int{}da = \int{(x^2-2x-3)dy}$$ $$a = \frac{x^3}{3}-x^2-3x+c$$

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Is your first function supposed to be $y=x^2-4$? And note that area is a number, not an expression here. You need limits on the integral. I would suggest drawing the graphs of your functions first. –  David Mitra Jul 26 '12 at 20:45
    
yes it was! How did you know haha? And sorry about that. And I did graph it first, I just didn't know you could get that cool graph thing on here –  user69 Jul 26 '12 at 20:47

1 Answer 1

up vote 3 down vote accepted

I take it you mean $y=x^2-4$ and $y=2x-1$.

Draw a picture. We get a familiar parabola, and a straight line. The straight line $y=2x-1$ meets the parabola where $x^2-4=2x-1$. This can be rearranged to $x^2-2x-3=0$. The quadratic factors as $(x-3)(x+1)$, so the meeting points are at $x=-1$ and $x=3$.

Note that the finite region caught between the two has the line above the curve. Thus our area is $$\int_{-1}^3\left((2x-1)-(x^2-4)\right)\,dx.$$ Before integrating, simplify the integrand a bit.

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