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

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

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integral from 0 to 1 of x^2e^{xy}

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

x (e^{x} - 1)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= x^{2} \cdot \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{x} - 1

= x (e^{x} - 1)

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

\int_{0}^{x^{3}} sin (x^{2}) d x

\int_{0}^{5} x e^{- \frac{x}{4}} d x

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

\int_{1}^{\infty} \frac{1}{4 x - 1} d x

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