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

y = e^{2} x

\frac{d}{d z} (x^{2} + y^{2} + z^{2})

\int \frac{( x ^{3} + 1 ) ^{2}}{x ^{2}} d x

\frac{d}{d x} y = sec (2 x)

f (x) = 2 x x - 2, at x = 3

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…