What is the integral from 1 to 2 of e^{-ln(x^2)} ?

The integral from 1 to 2 of e^{-ln(x^2)} is 1/2

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integral from 1 to 2 of e^{-ln(x^2)}

\int_{1}^{2} e^{- l n (x^{2})} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{1}^{2} e^{- l n (x^{2})} d x

Simplify

e^{- l n (x^{2})} : \frac{1}{x ^{2}}

= \int_{1}^{2} \frac{1}{x ^{2}} d x

Apply the Power Rule

= [- \frac{1}{x}]_{1}^{2}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

What is the integral from 1 to 2 of e^{-ln(x^2)} ?
The integral from 1 to 2 of e^{-ln(x^2)} is 1/2