Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3

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- Aug 10th 2008, 08:45 PMMathGeek06Area of the Region Between Two Curves
Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3

- Aug 10th 2008, 08:52 PMChris L T521
Did you graph the bounded region? This will help you set up the integral.

http://img.photobucket.com/albums/v4...tuff/parab.jpg

Recall that the area between two curves is $\displaystyle A=\int_a^b \left(f(x)-g(x)\right) \,dx$

I hope this helps!

--Chris - Aug 10th 2008, 09:00 PMMathGeek06
Yes, I graphed the bounded region but when I set up my integrals, I got a negative answer, which clearly was incorrect.

I had an integral of ((4-3x) - x^3)) from x = -1 to x= 1 + the integral ((x^2) - (4-3x)) from x = 1 to x = 4/3. - Aug 10th 2008, 09:24 PMChris L T521
$\displaystyle \int_{-1}^1 (4-3x-x^2)\,dx+\int_1^{\frac{4}{3}}(x^2+3x-4)\,dx$

$\displaystyle \implies\left.\bigg[4x-\tfrac{3}{2}x^2-\tfrac{1}{3}x^3\bigg]\right|_{-1}^1+\left.\bigg[\tfrac{1}{3}x^3+\tfrac{3}{2}x^2-4x\bigg]\right|_1^{\frac{4}{3}}$

$\displaystyle \implies\bigg[\left(4-\tfrac{3}{2}-\tfrac{1}{3}\right)-\left(-4-\tfrac{3}{2}+\tfrac{1}{3}\right)\bigg]+\bigg[\left(\tfrac{64}{81}+\tfrac{8}{3}-\tfrac{16}{3}\right)-\left(\tfrac{1}{3}+\tfrac{3}{2}-4\right)\bigg]=\color{red}\boxed{\frac{1235}{162}}$

Does this make sense?

--Chris