What is the integral from 0 to ln(8) of pi(e^{-x})^2 ?

The integral from 0 to ln(8) of pi(e^{-x})^2 is (63pi)/(128)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to ln(8) of pi(e^{-x})^2

\int_{0}^{l n (8)} π (e^{- x})^{2} d x

\frac{63 π}{128}

Decimal Notation

1.54625 \dots

Solution steps

\int_{0}^{l n (8)} 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 (8)} (e^{- x})^{2} d x

Apply u-substitution

: \int_{1}^{\frac{1}{8}} - u d u

= π \cdot \int_{1}^{\frac{1}{8}} - u d u

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

= π (- \int_{\frac{1}{8}}^{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}{8}}^{1} u d u))

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{\frac{1}{8}}^{1}

= π (- (- [\frac{u ^{2}}{2}]_{\frac{1}{8}}^{1}))

Simplify

= π [\frac{u ^{2}}{2}]_{\frac{1}{8}}^{1}

Compute the boundaries:

\frac{63}{128}

= π \frac{63}{128}

Simplify

= \frac{63 π}{128}

\int_{- 1}^{3} 8 - 4 y^{2} d y

\int_{0}^{2 π} t^{2} 2 d t

\int_{0}^{2} 1 - c ∣ 1 - x ∣ d x

\int_{- 1}^{1} - x sin (nπ x) d x

\int_{2}^{5} \frac{3}{x} d x

What is the integral from 0 to ln(8) of pi(e^{-x})^2 ?
The integral from 0 to ln(8) of pi(e^{-x})^2 is (63pi)/(128)