What is the integral from 0 to 4 of pi((12-3x)/4)^2 ?

The integral from 0 to 4 of pi((12-3x)/4)^2 is pi12

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integral from 0 to 4 of pi((12-3x)/4)^2

\int_{0}^{4} π (\frac{12 - 3 x}{4})^{2} d x

π 12

Decimal

37.69911 \dots

Solution steps

\int_{0}^{4} π (\frac{12 - 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}^{4} (\frac{12 - 3 x}{4})^{2} d x

Apply Integration By Parts

: [\frac{1}{16} x (12 - 3 x)^{2} - \int - \frac{3}{8} x (- 3 x + 12) d x]_{0}^{4}

= π [\frac{1}{16} x (12 - 3 x)^{2} - \int - \frac{3}{8} x (- 3 x + 12) d x]_{0}^{4}

\int - \frac{3}{8} x (- 3 x + 12) d x = - \frac{3}{8} (- x^{3} + 6 x^{2})

= π [\frac{1}{16} x (12 - 3 x)^{2} - (- \frac{3}{8} (- x^{3} + 6 x^{2}))]_{0}^{4}

Simplify

π [\frac{1}{16} x (12 - 3 x)^{2} - (- \frac{3}{8} (- x^{3} + 6 x^{2}))]_{0}^{4} : π [\frac{1}{16} (3 x^{3} - 36 x^{2} + 144 x)]_{0}^{4}

= π [\frac{1}{16} (3 x^{3} - 36 x^{2} + 144 x)]_{0}^{4}

Compute the boundaries:

12

= π 12

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What is the integral from 0 to 4 of pi((12-3x)/4)^2 ?
The integral from 0 to 4 of pi((12-3x)/4)^2 is pi12