What is the integral from 3 to 5 of e^{6x} ?

The integral from 3 to 5 of e^{6x} is (e^{30}-e^{18})/6

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integral from 3 to 5 of e^{6x}

\int_{3}^{5} e^{6 x} d x

\frac{e ^{30} - e ^{18}}{6}

Decimal Notation

1781068153592.5515

Solution steps

\int_{3}^{5} e^{6 x} d x

Apply u-substitution

: \int_{18}^{30} e^{u} \frac{1}{6} d u

= \int_{18}^{30} e^{u} \frac{1}{6} 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_{18}^{30} e^{u} d u

Use the common integral:

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

= \frac{1}{6} [e^{u}]_{18}^{30}

Compute the boundaries:

e^{30} - e^{18}

= \frac{1}{6} (e^{30} - e^{18})

Simplify

= \frac{e ^{30} - e ^{18}}{6}

f (x) = 3 - 2 x

\int \frac{x ^{4} + 1}{x ^{2} + 1} d x

0.882 a^{0.842}

\int_{- 2}^{2} ∣ x ∣ - (x^{2} - 2) d x

(x^{3} + 1) e^{x}

What is the integral from 3 to 5 of e^{6x} ?
The integral from 3 to 5 of e^{6x} is (e^{30}-e^{18})/6