What is the integral from 0 to 1 of xye^{xy^2} ?

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

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integral from 0 to 1 of xye^{xy^2}

\int_{0}^{1} x y e^{x y^{2}} d y

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

Solution steps

\int_{0}^{1} x y e^{x y^{2}} d y

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{x} - 1

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

\int_{0}^{1} \frac{6}{7} (x^{2} + \frac{x y}{2}) d x

\int_{0}^{l n (4)} 4 e^{2 x} d x

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

\int_{- 5}^{1} x^{2} + 4 x - 5 d x

\int_{0}^{1} x 3 x^{2} + 1 d x

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