What is the integral from 120 to 150 of e^{-x/(100)} ?

The integral from 120 to 150 of e^{-x/(100)} is 100(1/(e^{6/5)}-1/(e^{3/2)})

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integral from 120 to 150 of e^{-x/(100)}

\int_{120}^{150} e^{- \frac{x}{100}} d x

100 (\frac{1}{e ^{\frac{6}{5}}} - \frac{1}{e ^{\frac{3}{2}}})

Decimal Notation

7.80640 \dots

Solution steps

\int_{120}^{150} e^{- \frac{x}{100}} d x

Apply u-substitution

: \int_{- \frac{6}{5}}^{- \frac{3}{2}} - 100 e^{u} d u

= \int_{- \frac{6}{5}}^{- \frac{3}{2}} - 100 e^{u} d u

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

= - \int_{- \frac{3}{2}}^{- \frac{6}{5}} - 100 e^{u} d u

Take the constant out:

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

= - (- 100 \cdot \int_{- \frac{3}{2}}^{- \frac{6}{5}} e^{u} d u)

Use the common integral:

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

= - (- 100 [e^{u}]_{- \frac{3}{2}}^{- \frac{6}{5}})

Simplify

= 100 [e^{u}]_{- \frac{3}{2}}^{- \frac{6}{5}}

Compute the boundaries:

\frac{1}{e ^{\frac{6}{5}}} - \frac{1}{e ^{\frac{3}{2}}}

= 100 (\frac{1}{e ^{\frac{6}{5}}} - \frac{1}{e ^{\frac{3}{2}}})

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What is the integral from 120 to 150 of e^{-x/(100)} ?
The integral from 120 to 150 of e^{-x/(100)} is 100(1/(e^{6/5)}-1/(e^{3/2)})