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

The integral from 0 to 5 of pi((20-4x)/5)^2 is (80pi)/3

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

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

\frac{80 π}{3}

Decimal Notation

83.77580 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{25} x (20 - 4 x)^{2} - \int - \frac{8}{25} x (20 - 4 x) d x]_{0}^{5}

= π [\frac{1}{25} x (20 - 4 x)^{2} - \int - \frac{8}{25} x (20 - 4 x) d x]_{0}^{5}

\int - \frac{8}{25} x (20 - 4 x) d x = - \frac{8}{25} (10 x^{2} - \frac{4 x ^{3}}{3})

= π [\frac{1}{25} x (20 - 4 x)^{2} - (- \frac{8}{25} (10 x^{2} - \frac{4 x ^{3}}{3}))]_{0}^{5}

Simplify

π [\frac{1}{25} x (20 - 4 x)^{2} - (- \frac{8}{25} (10 x^{2} - \frac{4 x ^{3}}{3}))]_{0}^{5} : π [\frac{1}{75} (16 x^{3} - 240 x^{2} + 1200 x)]_{0}^{5}

= π [\frac{1}{75} (16 x^{3} - 240 x^{2} + 1200 x)]_{0}^{5}

Compute the boundaries:

\frac{80}{3}

= π \frac{80}{3}

Simplify

= \frac{80 π}{3}

\int 3 u d u

(8.1) (24.9)

x^{2} y^{^{'}} = y^{2} + x y

(3, 3), (- 3, - 1)

\int \frac{3}{t ^{5}} d t

What is the integral from 0 to 5 of pi((20-4x)/5)^2 ?
The integral from 0 to 5 of pi((20-4x)/5)^2 is (80pi)/3