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

The integral from 0 to 1 of 2pi(x+8)(x^2) is (35pi)/6

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

\int_{0}^{1} 2 π (x + 8) (x^{2}) d x

\frac{35 π}{6}

Decimal

18.32595 \dots

Solution steps

\int_{0}^{1} 2 π (x + 8) 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} (x + 8) x^{2} d x

Expand

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

= 2 π \cdot \int_{0}^{1} x^{3} + 8 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

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

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

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

= 2 π (\frac{1}{4} + \frac{8}{3})

Simplify

2 π (\frac{1}{4} + \frac{8}{3}) : \frac{35 π}{6}

= \frac{35 π}{6}

d er i v a t i v e \frac{y , ( x + y )}{( x ^{2} + y ^{2} + x y + 3 )}

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

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{\partial}{\partial x} (x^{x})

x \to \infty lim (9.35 e^{(- 0.4 x)} + 5.52)

What is the integral from 0 to 1 of 2pi(x+8)(x^2) ?
The integral from 0 to 1 of 2pi(x+8)(x^2) is (35pi)/6