What is the integral from 0 to 1 of pi(6-3/2 x)^2 ?

The integral from 0 to 1 of pi(6-3/2 x)^2 is pi 111/4

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integral from 0 to 1 of pi(6-3/2 x)^2

\int_{0}^{1} π (6 - \frac{3}{2} x)^{2} d x

π \frac{111}{4}

Decimal Notation

87.17919 \dots

Solution steps

\int_{0}^{1} π (6 - \frac{3}{2} x)^{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}^{1} (6 - \frac{3}{2} x)^{2} d x

Simplify

(6 - \frac{3}{2} x)^{2} : 36 - 18 x + \frac{9}{4} x^{2}

= π \cdot \int_{0}^{1} 36 - 18 x + \frac{9}{4} x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{0}^{1} 36 d x - \int_{0}^{1} 18 x d x + \int_{0}^{1} \frac{9}{4} x^{2} d x)

\int_{0}^{1} 36 d x = 36

\int_{0}^{1} 18 x d x = 9

\int_{0}^{1} \frac{9}{4} x^{2} d x = \frac{3}{4}

= π (36 - 9 + \frac{3}{4})

Simplify

π (36 - 9 + \frac{3}{4}) : π \frac{111}{4}

= π \frac{111}{4}

\frac{2}{s ^{2} + s}

y^{^{''}} + 9 y = 3 cos (2 x) + 3

\int 9 - x^{2} (- 2 x) d x

\frac{d}{d x} ((x - 2)^{2} (x + 1))

\frac{\partial}{\partial y} (x^{2} + y^{2} - 9)

What is the integral from 0 to 1 of pi(6-3/2 x)^2 ?
The integral from 0 to 1 of pi(6-3/2 x)^2 is pi 111/4