What is the integral from 0 to 8 of pi(1-y/8)^2 ?

The integral from 0 to 8 of pi(1-y/8)^2 is (8pi)/3

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integral from 0 to 8 of pi(1-y/8)^2

\int_{0}^{8} π (1 - \frac{y}{8})^{2} d y

\frac{8 π}{3}

Decimal Notation

8.37758 \dots

Solution steps

\int_{0}^{8} π (1 - \frac{y}{8})^{2} d y

Take the constant out:

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

= π \cdot \int_{0}^{8} (1 - \frac{y}{8})^{2} d y

Apply u-substitution

: \int_{1}^{0} - 8 u^{2} d u

= π \cdot \int_{1}^{0} - 8 u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= π (- \int_{0}^{1} - 8 u^{2} d u)

Take the constant out:

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

= π (- (- 8 \cdot \int_{0}^{1} u^{2} d u))

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{1}

= π (- (- 8 [\frac{u ^{3}}{3}]_{0}^{1}))

Simplify

= 8 π [\frac{u ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= 8 π \frac{1}{3}

Simplify

= \frac{8 π}{3}

\int \frac{1}{2 x ^{2} - x - 3} d x

\int sin^{7} (\frac{3}{5} x) d x

\int \frac{x}{x + 9} d x

\int (3^{(2 - x)} + (x + 1)^{2} e^{(1 - x)}) d x

y = 3 x^{2} - 6, (- 1, - 3)

What is the integral from 0 to 8 of pi(1-y/8)^2 ?
The integral from 0 to 8 of pi(1-y/8)^2 is (8pi)/3