What is the integral from 0 to 1 of e^{xy} ?

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

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integral from 0 to 1 of e^{xy}

\int_{0}^{1} e^{x y} d y

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

Solution steps

\int_{0}^{1} e^{x y} d y

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{x} - 1

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

\int_{- 3}^{3} (∣ x ∣ + ∣ x + 1 ∣) d x

\int_{0}^{3} 6 - 3 y^{2} d y

\int_{0}^{\frac{π}{2}} cos (sin (x)) d x

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

\int_{0}^{\frac{π}{2}} sin (x^{\frac{1}{2}}) d x

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