# How do you find the area under the graph of f(x)=e^(-2lnx) on the interval [1, 2]?

May 16, 2018

$\frac{1}{2}$

#### Explanation:

$f \left(x\right) = {e}^{- 2 \ln x} = {x}^{-} 2$

Thus the area under the curve is

${\int}_{1}^{2} f \left(x\right) \mathrm{dx} = {\int}_{1}^{2} {x}^{-} 2 \mathrm{dx}$
$q \quad = {\left(- {x}^{-} 1\right)}_{1}^{2}$
$q \quad = - \frac{1}{2} - \left(- \frac{1}{1}\right) = 1 - \frac{1}{2} = \frac{1}{2}$