What is the integral from 0 to 1 of (e^{-0.1x}) ?

The integral from 0 to 1 of (e^{-0.1x}) is 0.95162…

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integral from 0 to 1 of (e^{-0.1x})

\int_{0}^{1} (e^{- 0.1 x}) d x

0.95162 \dots

Solution steps

\int_{0}^{1} e^{- 0.1 x} d x

Apply u-substitution

: \int_{0}^{- 0.1} - \frac{1}{0.1} e^{u} d u

= \int_{0}^{- 0.1} - \frac{1}{0.1} e^{u} d u

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

= - \int_{- 0.1}^{0} - \frac{1}{0.1} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{0.1} \cdot \int_{- 0.1}^{0} e^{u} d u)

Simplify

= - (- 10 \cdot \int_{- 0.1}^{0} e^{u} d u)

Use the common integral:

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

= - (- 10 [e^{u}]_{- 0.1}^{0})

Simplify

= 10 [e^{u}]_{- 0.1}^{0}

Compute the boundaries:

0.09516 \dots

= 10 \cdot 0.09516 \dots

Simplify

= 0.95162 \dots

\frac{d}{d x} ((x + 2)^{3})

z \to 0 lim (z^{3} e^{\frac{1}{z}})

\int_{1}^{2} \frac{e ^{\frac{3}{x}}}{x ^{2}} d x

y^{^{''}} + 16 y^{^{'}} + 64 y = 0

\int_{- 2}^{2} x^{3} + 1 d x

What is the integral from 0 to 1 of (e^{-0.1x}) ?
The integral from 0 to 1 of (e^{-0.1x}) is 0.95162…