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

The integral from 0 to 1 of x^2(x^3-2)^3 is -5/4

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

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

- \frac{5}{4}

Solution steps

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

Apply u-substitution

: \int_{- 2}^{- 1} \frac{u ^{3}}{3} d u

= \int_{- 2}^{- 1} \frac{u ^{3}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{- 2}^{- 1} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 2}^{- 1}

= \frac{1}{3} [\frac{u ^{4}}{4}]_{- 2}^{- 1}

Compute the boundaries:

- \frac{15}{4}

= \frac{1}{3} (- \frac{15}{4})

Simplify

= - \frac{5}{4}

\frac{\partial}{\partial x} (\frac{t}{x})

\frac{d y}{d x} + (\frac{3}{x}) y = - 4 x - 10

\int \frac{1}{x ^{2} - 4} d x

\frac{d y}{d x} = \frac{2}{x}

\frac{d}{d x} (\frac{a}{1 + \frac{b}{x + c}})

What is the integral from 0 to 1 of x^2(x^3-2)^3 ?
The integral from 0 to 1 of x^2(x^3-2)^3 is -5/4