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

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

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

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

- \frac{1}{6}

Decimal Notation

- 0.16666 \dots

Solution steps

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

Take the constant out:

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

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

Expand

x^{2} (x^{3} - 1)^{3} : x^{11} - 3 x^{8} + 3 x^{5} - x^{2}

= 2 \cdot \int_{0}^{1} x^{11} - 3 x^{8} + 3 x^{5} - 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 (\int_{0}^{1} x^{11} d x - \int_{0}^{1} 3 x^{8} d x + \int_{0}^{1} 3 x^{5} d x - \int_{0}^{1} x^{2} d x)

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

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

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

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

= 2 (\frac{1}{12} - \frac{1}{3} + \frac{1}{2} - \frac{1}{3})

Simplify

2 (\frac{1}{12} - \frac{1}{3} + \frac{1}{2} - \frac{1}{3}) : - \frac{1}{6}

= - \frac{1}{6}

\int x^{2} - 16 d x

\frac{d}{d x} (2 ln (2 x))

\frac{\partial}{\partial r} (r^{2 + s^{2}})

\frac{t}{t ^{2} + 1}

y^{^{''}} + 15 y^{^{'}} + 56 y = 0

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