What is the integral from 0 to 1 of 0.2e^{-0.2x} ?

The integral from 0 to 1 of 0.2e^{-0.2x} is 0.18126…

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integral from 0 to 1 of 0.2e^{-0.2x}

\int_{0}^{1} 0.2 e^{- 0.2 x} d x

0.18126 \dots

Solution steps

\int_{0}^{1} 0.2 e^{- 0.2 x} d x

Take the constant out:

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

= 0.2 \cdot \int_{0}^{1} e^{- 0.2 x} d x

Apply u-substitution

: \int_{0}^{- 0.2} - \frac{1}{0.2} e^{u} d u

= 0.2 \cdot \int_{0}^{- 0.2} - \frac{1}{0.2} e^{u} d u

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

= 0.2 (- \int_{- 0.2}^{0} - \frac{1}{0.2} e^{u} d u)

Take the constant out:

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

= 0.2 (- (- \frac{1}{0.2} \cdot \int_{- 0.2}^{0} e^{u} d u))

Simplify

= 0.2 (- (- 5 \cdot \int_{- 0.2}^{0} e^{u} d u))

Use the common integral:

\int e^{u} d u = e^{u}

= 0.2 (- (- 5 [e^{u}]_{- 0.2}^{0}))

Simplify

= [e^{u}]_{- 0.2}^{0}

Compute the boundaries:

0.18126 \dots

= 0.18126 \dots

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

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) (y + 9 y^{3}))

\frac{d y}{d x} = 2 - 3 y

\int_{0}^{1} \frac{2}{( x + 2 ) ^{2}} d x

\frac{d}{d x} (\frac{1 ( e ^{x} - e ^{- x} )}{2})

What is the integral from 0 to 1 of 0.2e^{-0.2x} ?
The integral from 0 to 1 of 0.2e^{-0.2x} is 0.18126…