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

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

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

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

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

Solution steps

\int_{\frac{1}{2} x}^{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_{\frac{1}{2} x}^{1} e^{x y} d y

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

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