What is the integral from 0 to t of 18e^{5x} ?

The integral from 0 to t of 18e^{5x} is 18/5 (e^{5t}-1)

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integral from 0 to t of 18e^{5x}

\int_{0}^{t} 18 e^{5 x} d x

\frac{18}{5} (e^{5 t} - 1)

Solution steps

\int_{0}^{t} 18 e^{5 x} d x

Take the constant out:

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

= 18 \cdot \int_{0}^{t} e^{5 x} d x

Apply u-substitution

: \int_{0}^{5 t} e^{u} \frac{1}{5} d u

= 18 \cdot \int_{0}^{5 t} e^{u} \frac{1}{5} d u

Take the constant out:

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

= 18 \cdot \frac{1}{5} \cdot \int_{0}^{5 t} e^{u} d u

Use the common integral:

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

= 18 \cdot \frac{1}{5} [e^{u}]_{0}^{5 t}

Simplify

18 \cdot \frac{1}{5} [e^{u}]_{0}^{5 t} : \frac{18}{5} [e^{u}]_{0}^{5 t}

= \frac{18}{5} [e^{u}]_{0}^{5 t}

Compute the boundaries:

e^{5 t} - 1

= \frac{18}{5} (e^{5 t} - 1)

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What is the integral from 0 to t of 18e^{5x} ?
The integral from 0 to t of 18e^{5x} is 18/5 (e^{5t}-1)