What is the integral from a to b of e^{tx} ?

The integral from a to b of e^{tx} is (e^{bt}-e^{at})/t

Popular >

integral from a to b of e^{tx}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{b t} - e^{a t}

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

\int_{0}^{2} \frac{4 x}{2 + x ^{2}} d x

\int_{0}^{2} 4 x + 1 d x

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

\int_{2}^{4} (x^{2} - 2 x) d x

\int_{- 1}^{0} ln (x + 1) d x

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