What is the integral from 1 to e^2 of 2piyln(y) ?

The integral from 1 to e^2 of 2piyln(y) is (pi(3e^4+1))/2

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integral from 1 to e^2 of 2piyln(y)

\int_{1}^{e^{2}} 2 π y ln (y) d y

\frac{π ( 3 e ^{4} + 1 )}{2}

Decimal Notation

258.85851 \dots

Solution steps

\int_{1}^{e^{2}} 2 π y ln (y) d y

Take the constant out:

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

= 2 π \cdot \int_{1}^{e^{2}} 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^{2}}

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

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

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

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

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

Compute the boundaries:

\frac{3 e ^{4}}{4} + \frac{1}{4}

= 2 π (\frac{3 e ^{4}}{4} + \frac{1}{4})

Simplify

= \frac{π ( 3 e ^{4} + 1 )}{2}

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What is the integral from 1 to e^2 of 2piyln(y) ?
The integral from 1 to e^2 of 2piyln(y) is (pi(3e^4+1))/2