What is the integral from 0 to 1 of 2pi(x+9)x^2 ?

The integral from 0 to 1 of 2pi(x+9)x^2 is (13pi)/2

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integral from 0 to 1 of 2pi(x+9)x^2

\int_{0}^{1} 2 π (x + 9) x^{2} d x

\frac{13 π}{2}

Decimal Notation

20.42035 \dots

Solution steps

\int_{0}^{1} 2 π (x + 9) 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 + 9) x^{2} d x

Expand

(x + 9) x^{2} : x^{3} + 9 x^{2}

= 2 π \cdot \int_{0}^{1} x^{3} + 9 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} 9 x^{2} d x)

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

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

= 2 π (\frac{1}{4} + 3)

Simplify

2 π (\frac{1}{4} + 3) : \frac{13 π}{2}

= \frac{13 π}{2}

(1 + y) d x - (1 - x) d y = 0

x \to 10 lim (\frac{x}{2})

y = e^{7 x + 1}

\int x^{2} \cdot (1 + 2 x^{3})^{5} d x

\frac{d}{d x} (9 x - \frac{1}{x})

What is the integral from 0 to 1 of 2pi(x+9)x^2 ?
The integral from 0 to 1 of 2pi(x+9)x^2 is (13pi)/2