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

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

Popular >

integral from t to 0 of e^{ax}

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

1 - e^{a t}

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

\int_{0}^{4} π 4 x - x^{2} d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} x^{3} sin (x) d x

\int_{0}^{2} \frac{1}{4 - s ^{2}} d s

\int_{1}^{3} x^{2} - 4 d x

\int_{0}^{3} (5 x^{2} + 9 x) d x

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