What is the integral from 0 to 1 of 2pi(2-y)(2-2y^2) ?

The integral from 0 to 1 of 2pi(2-y)(2-2y^2) is (13pi)/3

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integral from 0 to 1 of 2pi(2-y)(2-2y^2)

\int_{0}^{1} 2 π (2 - y) (2 - 2 y^{2}) d y

\frac{13 π}{3}

Decimal

13.61356 \dots

Solution steps

\int_{0}^{1} 2 π (2 - y) (2 - 2 y^{2}) d y

Take the constant out:

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

= 2 π \cdot \int_{0}^{1} (2 - y) (2 - 2 y^{2}) d y

Apply Integration By Parts

= 2 π [(2 - y) (2 y - \frac{2 y ^{3}}{3}) - \int - 2 y + \frac{2 y ^{3}}{3} d y]_{0}^{1}

\int - 2 y + \frac{2 y ^{3}}{3} d y = - y^{2} + \frac{y ^{4}}{6}

= 2 π [(2 - y) (2 y - \frac{2 y ^{3}}{3}) - (- y^{2} + \frac{y ^{4}}{6})]_{0}^{1}

Simplify

2 π [(2 - y) (2 y - \frac{2 y ^{3}}{3}) - (- y^{2} + \frac{y ^{4}}{6})]_{0}^{1} : 2 π [\frac{1}{6} (3 y^{4} - 8 y^{3} - 6 y^{2} + 24 y)]_{0}^{1}

= 2 π [\frac{1}{6} (3 y^{4} - 8 y^{3} - 6 y^{2} + 24 y)]_{0}^{1}

Compute the boundaries:

\frac{13}{6}

= 2 π \frac{13}{6}

Simplify

= \frac{13 π}{3}

What is the integral from 0 to 1 of 2pi(2-y)(2-2y^2) ?
The integral from 0 to 1 of 2pi(2-y)(2-2y^2) is (13pi)/3