What is the integral from 1 to e of yln(y) ?

The integral from 1 to e of yln(y) is (e^2+1)/4

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integral from 1 to e of yln(y)

\int_{1}^{e} y ln (y) d y

\frac{e ^{2} + 1}{4}

Decimal

2.09726 \dots

Solution steps

\int_{1}^{e} y ln (y) d y

Apply Integration By Parts

: [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \int \frac{y}{2} d y)]_{1}^{e}

= [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \int \frac{y}{2} d y)]_{1}^{e}

\int \frac{y}{2} d y = \frac{y ^{2}}{4}

= [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \frac{y ^{2}}{4})]_{1}^{e}

2 \cdot \frac{y ^{2}}{4} = \frac{y ^{2}}{2}

= [\frac{1}{2} (y^{2} ln (y) - \frac{y ^{2}}{2})]_{1}^{e}

Compute the boundaries:

\frac{e ^{2}}{4} + \frac{1}{4}

= \frac{e ^{2}}{4} + \frac{1}{4}

Simplify

= \frac{e ^{2} + 1}{4}

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What is the integral from 1 to e of yln(y) ?
The integral from 1 to e of yln(y) is (e^2+1)/4