What is the integral from 0 to 3 of e^{1-6t} ?

The integral from 0 to 3 of e^{1-6t} is (e^{18}-1)/(6e^{17)}

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integral from 0 to 3 of e^{1-6t}

\int_{0}^{3} e^{1 - 6 t} d t

\frac{e ^{18} - 1}{6 e ^{17}}

Decimal

0.45304 \dots

Solution steps

\int_{0}^{3} e^{1 - 6 t} d t

Apply u-substitution

= \int_{1}^{- 17} - \frac{1}{6} e^{u} d u

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

= - \int_{- 17}^{1} - \frac{1}{6} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{6} \cdot \int_{- 17}^{1} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{6} [e^{u}]_{- 17}^{1})

Simplify

= \frac{1}{6} [e^{u}]_{- 17}^{1}

Compute the boundaries:

e - \frac{1}{e ^{17}}

= \frac{1}{6} (e - \frac{1}{e ^{17}})

Simplify

= \frac{e ^{18} - 1}{6 e ^{17}}

What is the integral from 0 to 3 of e^{1-6t} ?
The integral from 0 to 3 of e^{1-6t} is (e^{18}-1)/(6e^{17)}