What is the integral from 0 to t of e^{at} ?

The integral from 0 to t of e^{at} is (e^{at}-1)/a

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integral from 0 to t of e^{at}

\int_{0}^{t} e^{a t} d t

\frac{e ^{a t} - 1}{a}

Solution steps

\int_{0}^{t} e^{a t} d t

Apply u-substitution

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

= \int_{0}^{a t} \frac{e ^{u}}{a} d u

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{a} [e^{u}]_{0}^{a t}

Simplify

\frac{1}{a} [e^{u}]_{0}^{a t} : \frac{[ e ^{u} ] _{0}^{a t}}{a}

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

Compute the boundaries:

e^{a t} - 1

= \frac{e ^{a t} - 1}{a}

\int_{- 1}^{5} 2 x + 6 d x

\int_{0}^{1} x^{2} + \frac{x}{2} d x

\int_{- π}^{0} 6 x^{3} d x

\int_{0.25}^{0.6} 5 (1 - y)^{4} d y

\int_{0}^{x} \frac{( 1 - t )}{e ^{t}} d t

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