What is the integral from 0 to 1 of 2pi(x+10)x^2 ?

The integral from 0 to 1 of 2pi(x+10)x^2 is (43pi)/6

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integral from 0 to 1 of 2pi(x+10)x^2

\int_{0}^{1} 2 π (x + 10) x^{2} d x

\frac{43 π}{6}

Decimal Notation

22.51474 \dots

Solution steps

\int_{0}^{1} 2 π (x + 10) 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 + 10) x^{2} d x

Expand

(x + 10) x^{2} : x^{3} + 10 x^{2}

= 2 π \cdot \int_{0}^{1} x^{3} + 10 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} 10 x^{2} d x)

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

\int_{0}^{1} 10 x^{2} d x = \frac{10}{3}

= 2 π (\frac{1}{4} + \frac{10}{3})

Simplify

2 π (\frac{1}{4} + \frac{10}{3}) : \frac{43 π}{6}

= \frac{43 π}{6}

f (x) = 12 x^{2} - 48 x

ln ∣ x ∣

\frac{d}{d x} (x^{3} - 6 x^{2})

\frac{d x}{d t} = x^{2} + \frac{1}{36}

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

What is the integral from 0 to 1 of 2pi(x+10)x^2 ?
The integral from 0 to 1 of 2pi(x+10)x^2 is (43pi)/6