What is the integral from 0 to 8 of pi((24-3x)/4)^2 ?

The integral from 0 to 8 of pi((24-3x)/4)^2 is pi96

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integral from 0 to 8 of pi((24-3x)/4)^2

\int_{0}^{8} π (\frac{24 - 3 x}{4})^{2} d x

π 96

Decimal

301.59289 \dots

Solution steps

\int_{0}^{8} π (\frac{24 - 3 x}{4})^{2} d x

Take the constant out:

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

= π \cdot \int_{0}^{8} (\frac{24 - 3 x}{4})^{2} d x

Apply Integration By Parts

: [\frac{1}{16} x (24 - 3 x)^{2} - \int - \frac{3}{8} x (- 3 x + 24) d x]_{0}^{8}

= π [\frac{1}{16} x (24 - 3 x)^{2} - \int - \frac{3}{8} x (- 3 x + 24) d x]_{0}^{8}

\int - \frac{3}{8} x (- 3 x + 24) d x = - \frac{3}{8} (- x^{3} + 12 x^{2})

= π [\frac{1}{16} x (24 - 3 x)^{2} - (- \frac{3}{8} (- x^{3} + 12 x^{2}))]_{0}^{8}

Simplify

π [\frac{1}{16} x (24 - 3 x)^{2} - (- \frac{3}{8} (- x^{3} + 12 x^{2}))]_{0}^{8} : π [\frac{1}{16} (3 x^{3} - 72 x^{2} + 576 x)]_{0}^{8}

= π [\frac{1}{16} (3 x^{3} - 72 x^{2} + 576 x)]_{0}^{8}

Compute the boundaries:

96

= π 96

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What is the integral from 0 to 8 of pi((24-3x)/4)^2 ?
The integral from 0 to 8 of pi((24-3x)/4)^2 is pi96