What is the integral from 0 to 1 of 6x^2(3-5x^3)^3 ?

The integral from 0 to 1 of 6x^2(3-5x^3)^3 is 13/2

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integral from 0 to 1 of 6x^2(3-5x^3)^3

\int_{0}^{1} 6 x^{2} (3 - 5 x^{3})^{3} d x

\frac{13}{2}

Decimal

6.5

Solution steps

\int_{0}^{1} 6 x^{2} (3 - 5 x^{3})^{3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{0}^{1} x^{2} (3 - 5 x^{3})^{3} d x

Expand

x^{2} (3 - 5 x^{3})^{3} : 27 x^{2} - 135 x^{5} + 225 x^{8} - 125 x^{11}

= 6 \cdot \int_{0}^{1} 27 x^{2} - 135 x^{5} + 225 x^{8} - 125 x^{11} 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

= 6 (\int_{0}^{1} 27 x^{2} d x - \int_{0}^{1} 135 x^{5} d x + \int_{0}^{1} 225 x^{8} d x - \int_{0}^{1} 125 x^{11} d x)

\int_{0}^{1} 27 x^{2} d x = 9

\int_{0}^{1} 135 x^{5} d x = \frac{45}{2}

\int_{0}^{1} 225 x^{8} d x = 25

\int_{0}^{1} 125 x^{11} d x = \frac{125}{12}

= 6 (9 - \frac{45}{2} + 25 - \frac{125}{12})

Simplify

6 (9 - \frac{45}{2} + 25 - \frac{125}{12}) : \frac{13}{2}

= \frac{13}{2}

- \frac{1}{2} x^{2} - \frac{5}{2} x - 2, [- 2, 7]

\int \frac{1}{x I n x} d x

\frac{d}{d x} (x^{2} - 6 x + 11)

n = 0 \sum \infty \frac{1}{n !}

\int (x^{\frac{1}{4}}) d x

What is the integral from 0 to 1 of 6x^2(3-5x^3)^3 ?
The integral from 0 to 1 of 6x^2(3-5x^3)^3 is 13/2