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

The integral from 0 to x of e^{xy} is (e^{x^2}-1)/x

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integral from 0 to x of e^{xy}

\int_{0}^{x} e^{x y} d y

\frac{e ^{x^{2}} - 1}{x}

Solution steps

\int_{0}^{x} e^{x y} d y

Apply u-substitution

: \int_{0}^{x^{2}} \frac{e ^{u}}{x} d u

= \int_{0}^{x^{2}} \frac{e ^{u}}{x} d u

Take the constant out:

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

= \frac{1}{x} \cdot \int_{0}^{x^{2}} e^{u} d u

Use the common integral:

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

= \frac{1}{x} [e^{u}]_{0}^{x^{2}}

Simplify

\frac{1}{x} [e^{u}]_{0}^{x^{2}} : \frac{[ e ^{u} ] _{0}^{x^{2}}}{x}

= \frac{[ e ^{u} ] _{0}^{x^{2}}}{x}

Compute the boundaries:

e^{x^{2}} - 1

= \frac{e ^{x^{2}} - 1}{x}

\int_{2}^{8} (x^{3} - 2) d x

\int_{0}^{1} (6 x) d x

\int_{- 1}^{1} e^{- t} d t

\int_{0}^{1} \frac{88}{x ^{2} - 121} d x

\int_{1}^{2} \frac{e ^{3 x}}{x ^{2}} d x

What is the integral from 0 to x of e^{xy} ?
The integral from 0 to x of e^{xy} is (e^{x^2}-1)/x