What is the integral from 1 to 2 of e^{3t} ?

The integral from 1 to 2 of e^{3t} is (e^6-e^3)/3

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integral from 1 to 2 of e^{3t}

\int_{1}^{2} e^{3 t} d t

\frac{e ^{6} - e ^{3}}{3}

Decimal Notation

127.78108 \dots

Solution steps

\int_{1}^{2} e^{3 t} d t

Apply u-substitution

: \int_{3}^{6} e^{u} \frac{1}{3} d u

= \int_{3}^{6} e^{u} \frac{1}{3} 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_{3}^{6} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{3}^{6}

Compute the boundaries:

e^{6} - e^{3}

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

Simplify

= \frac{e ^{6} - e ^{3}}{3}

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What is the integral from 1 to 2 of e^{3t} ?
The integral from 1 to 2 of e^{3t} is (e^6-e^3)/3