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

The integral from 0 to 1 of 2pi(8-x)(x^2) is (29pi)/6

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

\int_{0}^{1} 2 π (8 - x) (x^{2}) d x

\frac{29 π}{6}

Decimal Notation

15.18436 \dots

Solution steps

\int_{0}^{1} 2 π (8 - x) x^{2} d x

Take the constant out:

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

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

Expand

(8 - x) x^{2} : 8 x^{2} - x^{3}

= 2 π \cdot \int_{0}^{1} 8 x^{2} - x^{3} 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

= 2 π (\int_{0}^{1} 8 x^{2} d x - \int_{0}^{1} x^{3} d x)

\int_{0}^{1} 8 x^{2} d x = \frac{8}{3}

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

= 2 π (\frac{8}{3} - \frac{1}{4})

Simplify

2 π (\frac{8}{3} - \frac{1}{4}) : \frac{29 π}{6}

= \frac{29 π}{6}

\int_{- 2}^{2} 2 - x^{2} d x

(x^{2} + 1)^{x + 1}

\frac{d}{d x} (x^{3} + 4 x - 15)

f (x) = (3 x - 2)^{5} (x + 1)^{3}

What is the integral from 0 to 1 of 2pi(8-x)(x^2) ?
The integral from 0 to 1 of 2pi(8-x)(x^2) is (29pi)/6