What is the integral from 0 to 5 of 2pi(7-x)(25-x^2) ?

The integral from 0 to 5 of 2pi(7-x)(25-x^2) is (5125pi)/6

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integral from 0 to 5 of 2pi(7-x)(25-x^2)

\int_{0}^{5} 2 π (7 - x) (25 - x^{2}) d x

\frac{5125 π}{6}

Decimal

2683.44372 \dots

Solution steps

\int_{0}^{5} 2 π (7 - x) (25 - 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}^{5} (7 - x) (25 - x^{2}) d x

Expand

(7 - x) (25 - x^{2}) : 175 - 7 x^{2} - 25 x + x^{3}

= 2 π \cdot \int_{0}^{5} 175 - 7 x^{2} - 25 x + 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}^{5} 175 d x - \int_{0}^{5} 7 x^{2} d x - \int_{0}^{5} 25 x d x + \int_{0}^{5} x^{3} d x)

\int_{0}^{5} 175 d x = 875

\int_{0}^{5} 7 x^{2} d x = \frac{875}{3}

\int_{0}^{5} 25 x d x = \frac{625}{2}

\int_{0}^{5} x^{3} d x = \frac{625}{4}

= 2 π (875 - \frac{875}{3} - \frac{625}{2} + \frac{625}{4})

Simplify

2 π (875 - \frac{875}{3} - \frac{625}{2} + \frac{625}{4}) : \frac{5125 π}{6}

= \frac{5125 π}{6}

What is the integral from 0 to 5 of 2pi(7-x)(25-x^2) ?
The integral from 0 to 5 of 2pi(7-x)(25-x^2) is (5125pi)/6