Area bounded by parametric equations

The curve $\displaystyle L$ is defined by the parametric equations

$\displaystyle x=e^t$ , $\displaystyle y=t-2$

Find the area of the region enclosed by the curve $\displaystyle L$, the x-axis and the lines $\displaystyle x=e$ and $\displaystyle x=e^3$

$\displaystyle Area=\int^{e^3}_et-2dx$

$\displaystyle \frac{dx}{dt}=e^t$

$\displaystyle Area=\int^{3}_1(t-2)e^tdt$

$\displaystyle =((3-2)e^3)-((1-2)e)$

$\displaystyle =e^3+e$

$\displaystyle =22.8 unit^2$

But answer is 9.34unit^2

Re: Area bounded by parametric equations

If you draw a picture of the region in question, you'll find that y is negative for t = 1 to 2 then goes positive. You'll need to split your integral up into two pieces to account for that.