What is the integral from 0 to ln(4) of pi(e^{-x})^2 ?

The integral from 0 to ln(4) of pi(e^{-x})^2 is (15pi)/(32)

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integral from 0 to ln(4) of pi(e^{-x})^2

\int_{0}^{l n (4)} π (e^{- x})^{2} d x

\frac{15 π}{32}

Solution steps

\int_{0}^{l n (4)} π (e^{- x})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{l n (4)} (e^{- x})^{2} d x

Apply u-substitution

: \int_{1}^{\frac{1}{4}} - u d u

= π \cdot \int_{1}^{\frac{1}{4}} - u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= π (- \int_{\frac{1}{4}}^{1} - u d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π (- (- \int_{\frac{1}{4}}^{1} u d u))

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{\frac{1}{4}}^{1}

= π (- (- [\frac{u ^{2}}{2}]_{\frac{1}{4}}^{1}))

Simplify

= π [\frac{u ^{2}}{2}]_{\frac{1}{4}}^{1}

Compute the boundaries:

\frac{15}{32}

= π \frac{15}{32}

Simplify

= \frac{15 π}{32}

\frac{\partial}{\partial x} ((1 + x^{2})^{z s i n (y)})

\frac{\partial}{\partial y} (y^{2} ln (1 + x^{2} + y^{2}))

\frac{d}{d x} (ln (\frac{x ln ( x ) - 1}{x ln ( x ) + 1}))

f (x) = \frac{( x + 1 )}{( x - 1 )}

\frac{3 t - 1}{t - 2}

What is the integral from 0 to ln(4) of pi(e^{-x})^2 ?
The integral from 0 to ln(4) of pi(e^{-x})^2 is (15pi)/(32)