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

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Answer 1 · 2018-05-16T00:47:00

$\frac{1}{2}$ $1/2$

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

Thus the area under the curve is

$\int_{1}^{2} f (x) d x = \int_{1}^{2} x^{- 2} d x$ $int_1^2 f(x) dx = int_1^2 x^-2 dx$
$qquad = (-x^-1)_1^2$
$qquad = -1/2-(-1/1) = 1-1/2=1/2$

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