# Thread: What am I doing wrong in this problem (area enclosed between curves)?

1. ## What am I doing wrong in this problem (area enclosed between curves)?

Find the area of the region enclosed by $y=x^2-5x$ and $y=x+7$.

What I did was:

$\displaystyle\int^7_{-1} (x+7)-(x^2-5x) dx$

$\frac{-x^3}{3}+\frac{6x^2}{2}+7$ from -1 to 7 (sorry, not sure how to type that)

$\left(-\frac{343}{3}+147+7\right) - \left(\frac{1}{3}+3-7\right)$

Area = $\frac{130}{3}$

This is what I got, but our computer program tells me that I am wrong. Is there a mistake I'm making?

2. Originally Posted by Dana_Scully
Find the area of the region enclosed by $y=x^2-5x$ and $y=x+7$.

What I did was:

$\displaystyle\int^7_{-1} (x+7)-(x^2-5x) dx$

$\frac{-x^3}{3}+\frac{6x^2}{2}+7$ from -1 to 7 (sorry, not sure how to type that)

$\left(-\frac{343}{3}+147+7\right) - \left(\frac{1}{3}+3-7\right)$

Area = $\frac{130}{3}$

This is what I got, but our computer program tells me that I am wrong. Is there a mistake I'm making?
a little hickup... the integral of 7 is not 7

by the way, you can type \Bigg|_{-1}^7 to get $\Bigg|_{-1}^7$. otherwise, see our LaTeX tutorial and learn how to use \right and \left tags

3. Just a minor mistake:

\begin{aligned} \int_{-1}^7 \left(x + 7 - (x^2 - 5x)\right) dx & = \int_{-1}^7 \left( - x^2 + 6x + 7\right) dx \\ & = \left(- \frac{x^3}{3} + \frac{6x^2}{2} + 7{\color{red}x} \right)\bigg|_{-1}^7 \\ & \ \ \vdots \end{aligned}

4. Thanks, I had actually written 7x but somehow confused it in my calculations...and thanks for telling me how to write the, uh, integral thing.