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

The integral from 0 to 1 of 2pi(x+2)x^2 is (11pi)/6

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

\int_{0}^{1} 2 π (x + 2) x^{2} d x

\frac{11 π}{6}

Decimal Notation

5.75958 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{3} x^{3} (x + 2) - \int \frac{x ^{3}}{3} d x]_{0}^{1}

= 2 π [\frac{1}{3} x^{3} (x + 2) - \int \frac{x ^{3}}{3} d x]_{0}^{1}

\int \frac{x ^{3}}{3} d x = \frac{x ^{4}}{12}

= 2 π [\frac{1}{3} x^{3} (x + 2) - \frac{x ^{4}}{12}]_{0}^{1}

Simplify

2 π [\frac{1}{3} x^{3} (x + 2) - \frac{x ^{4}}{12}]_{0}^{1} : 2 π [\frac{1}{12} (3 x^{4} + 8 x^{3})]_{0}^{1}

= 2 π [\frac{1}{12} (3 x^{4} + 8 x^{3})]_{0}^{1}

Compute the boundaries:

\frac{11}{12}

= 2 π \frac{11}{12}

Simplify

= \frac{11 π}{6}

\int_{- 1}^{y - 1} \frac{3}{2} x^{2} d x

\int_{- 1}^{2} (1 - ∣ x ∣) d x

\int_{2}^{3} x^{2} + 2 d x

\int_{1}^{t} \frac{1}{x} d x

\int_{0}^{1} 3 x sin (π) x d x

What is the integral from 0 to 1 of 2pi(x+2)x^2 ?
The integral from 0 to 1 of 2pi(x+2)x^2 is (11pi)/6