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

The integral from 0 to 5 of 2pi(x)(5x-x^2) is (625pi)/6

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

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

\frac{625 π}{6}

Decimal

327.24923 \dots

Solution steps

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

Expand

x (5 x - x^{2}) : 5 x^{2} - x^{3}

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

\int_{0}^{5} 5 x^{2} d x = \frac{625}{3}

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

= 2 π (\frac{625}{3} - \frac{625}{4})

Simplify

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

= \frac{625 π}{6}

f (x) = x^{4} - 25 x^{2} + 144, at x = 2

\int_{0}^{4} (4 - x)^{2} d x

y^{^{''}} - (\frac{2}{x}) y^{^{'}} - (\frac{3}{x ^{2}}) y = 0

n = 0 \sum \infty cos (2 n)

r \to 1 lim (\frac{8 r + 1}{r + 3})

What is the integral from 0 to 5 of 2pi(x)(5x-x^2) ?
The integral from 0 to 5 of 2pi(x)(5x-x^2) is (625pi)/6