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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$\begingroup$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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