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

The integral from 1 to e of piln^2(y) is pi(e-2)

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

\int_{1}^{e} π ln^{2} (y) d y

π (e - 2)

Decimal

2.25654 \dots

Solution steps

\int_{1}^{e} π ln^{2} (y) d y

Take the constant out:

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

= π \cdot \int_{1}^{e} ln^{2} (y) d y

Apply Integration By Parts

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

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

\int 2 ln (y) d y = 2 (y ln (y) - y)

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

Compute the boundaries:

e - 2

= π (e - 2)

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What is the integral from 1 to e of piln^2(y) ?
The integral from 1 to e of piln^2(y) is pi(e-2)