What is the integral from 0 to 4 of pi((20-5x)/4)^2 ?

The integral from 0 to 4 of pi((20-5x)/4)^2 is (100pi)/3

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integral from 0 to 4 of pi((20-5x)/4)^2

\int_{0}^{4} π (\frac{20 - 5 x}{4})^{2} d x

\frac{100 π}{3}

Decimal Notation

104.71975 \dots

Solution steps

\int_{0}^{4} π (\frac{20 - 5 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{20 - 5 x}{4})^{2} d x

Apply Integration By Parts

: [\frac{1}{16} x (20 - 5 x)^{2} - \int - \frac{5}{8} x (- 5 x + 20) d x]_{0}^{4}

= π [\frac{1}{16} x (20 - 5 x)^{2} - \int - \frac{5}{8} x (- 5 x + 20) d x]_{0}^{4}

\int - \frac{5}{8} x (- 5 x + 20) d x = - \frac{5}{8} (- \frac{5 x ^{3}}{3} + 10 x^{2})

= π [\frac{1}{16} x (20 - 5 x)^{2} - (- \frac{5}{8} (- \frac{5 x ^{3}}{3} + 10 x^{2}))]_{0}^{4}

Simplify

π [\frac{1}{16} x (20 - 5 x)^{2} - (- \frac{5}{8} (- \frac{5 x ^{3}}{3} + 10 x^{2}))]_{0}^{4} : π [\frac{1}{48} (25 x^{3} - 300 x^{2} + 1200 x)]_{0}^{4}

= π [\frac{1}{48} (25 x^{3} - 300 x^{2} + 1200 x)]_{0}^{4}

Compute the boundaries:

\frac{100}{3}

= π \frac{100}{3}

Simplify

= \frac{100 π}{3}

y = 4 x^{2} + 1

\int 2 x^{3} e^{x^{2}} d x

\frac{d}{d x} (x + y - 1)

\int (\frac{1}{2 x}) d x

y = 3 x, x = 0, x = 4

What is the integral from 0 to 4 of pi((20-5x)/4)^2 ?
The integral from 0 to 4 of pi((20-5x)/4)^2 is (100pi)/3