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

The integral from-1 to 2 of (3x^2-2x)^3 is 4836/35

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

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

\frac{4836}{35}

Decimal Notation

138.17142 \dots

Solution steps

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

Multiply by the conjugate of

3 x^{2} - 2 x : \frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x}

= \int_{- 1}^{2} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x

Find undefined (singularity) points:

x = 0, x = - \frac{2}{3}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 1}^{- \frac{2}{3}} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x + \int_{- \frac{2}{3}}^{0} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x + \int_{0}^{2} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x

\int_{- 1}^{- \frac{2}{3}} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x = \frac{55844}{2835}

\int_{- \frac{2}{3}}^{0} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x = \frac{6688}{2835}

\int_{0}^{2} (\frac{9 x ^{4} - 4 x ^{2}}{3 x ^{2} + 2 x})^{3} d x = \frac{4064}{35}

= \frac{55844}{2835} + \frac{6688}{2835} + \frac{4064}{35}

Simplify

\frac{55844}{2835} + \frac{6688}{2835} + \frac{4064}{35} : \frac{4836}{35}

= \frac{4836}{35}

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What is the integral from-1 to 2 of (3x^2-2x)^3 ?
The integral from-1 to 2 of (3x^2-2x)^3 is 4836/35