What is the integral from 0 to 5 of e^{3x} ?

The integral from 0 to 5 of e^{3x} is (e^{15}-1)/3

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integral from 0 to 5 of e^{3x}

\int_{0}^{5} e^{3 x} d x

\frac{e ^{15} - 1}{3}

Decimal Notation

1089672.12415 \dots

Solution steps

\int_{0}^{5} e^{3 x} d x

Apply u-substitution

: \int_{0}^{15} e^{u} \frac{1}{3} d u

= \int_{0}^{15} 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_{0}^{15} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{15}

Compute the boundaries:

e^{15} - 1

= \frac{1}{3} (e^{15} - 1)

Simplify

= \frac{e ^{15} - 1}{3}

4 y^{^{''}} - 16 y^{^{'}} + 19 y = 0

\frac{d}{d x} (2 x^{2} cos (x))

\frac{y ^{^{'}}}{y} = x

\frac{6 x}{2 + x ^{2}}

\frac{\partial}{\partial y} (e^{2 y} sin (- 2 x))

What is the integral from 0 to 5 of e^{3x} ?
The integral from 0 to 5 of e^{3x} is (e^{15}-1)/3