What is the integral from 1 to 4 of e^{7t} ?

The integral from 1 to 4 of e^{7t} is (e^{28}-e^7)/7

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integral from 1 to 4 of e^{7t}

\int_{1}^{4} e^{7 t} d t

\frac{e ^{28} - e ^{7}}{7}

Decimal Notation

206608151884.9771

Solution steps

\int_{1}^{4} e^{7 t} d t

Apply u-substitution

: \int_{7}^{28} e^{u} \frac{1}{7} d u

= \int_{7}^{28} e^{u} \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int_{7}^{28} e^{u} d u

Use the common integral:

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

= \frac{1}{7} [e^{u}]_{7}^{28}

Compute the boundaries:

e^{28} - e^{7}

= \frac{1}{7} (e^{28} - e^{7})

Simplify

= \frac{e ^{28} - e ^{7}}{7}

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What is the integral from 1 to 4 of e^{7t} ?
The integral from 1 to 4 of e^{7t} is (e^{28}-e^7)/7