What is the integral from 0 to 1 of e^{1-3t} ?

The integral from 0 to 1 of e^{1-3t} is (e^3-1)/(3e^2)

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integral from 0 to 1 of e^{1-3t}

\int_{0}^{1} e^{1 - 3 t} d t

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

Solution steps

\int_{0}^{1} e^{1 - 3 t} d t

Apply u-substitution

: \int_{1}^{- 2} - \frac{1}{3} e^{u} d u

= \int_{1}^{- 2} - \frac{1}{3} e^{u} d u

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

= - \int_{- 2}^{1} - \frac{1}{3} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 2}^{1} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 2}^{1})

Simplify

= \frac{1}{3} [e^{u}]_{- 2}^{1}

Compute the boundaries:

e - \frac{1}{e ^{2}}

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

Simplify

= \frac{e ^{3} - 1}{3 e ^{2}}

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What is the integral from 0 to 1 of e^{1-3t} ?
The integral from 0 to 1 of e^{1-3t} is (e^3-1)/(3e^2)