What is the integral from-1 to 2 of (3x-x^2)^3(6-4x) ?

The integral from-1 to 2 of (3x-x^2)^3(6-4x) is -120

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integral from-1 to 2 of (3x-x^2)^3(6-4x)

\int_{- 1}^{2} (3 x - x^{2})^{3} (6 - 4 x) d x

- 120

Solution steps

\int_{- 1}^{2} (3 x - x^{2})^{3} (6 - 4 x) d x

Multiply by the conjugate of

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

= \int_{- 1}^{2} (\frac{9 x ^{2} - x ^{4}}{3 x + x ^{2}})^{3} (6 - 4 x) d x

Find undefined (singularity) points:

x = 0

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}^{0} (\frac{9 x ^{2} - x ^{4}}{3 x + x ^{2}})^{3} (6 - 4 x) d x + \int_{0}^{2} (\frac{9 x ^{2} - x ^{4}}{3 x + x ^{2}})^{3} (6 - 4 x) d x

\int_{- 1}^{0} (\frac{9 x ^{2} - x ^{4}}{3 x + x ^{2}})^{3} (6 - 4 x) d x = - 128

\int_{0}^{2} (\frac{9 x ^{2} - x ^{4}}{3 x + x ^{2}})^{3} (6 - 4 x) d x = 8

= - 128 + 8

Simplify

= - 120

\int 8 sin (ln (x)) d x

2 x^{2} + x y - y^{2}

2 1 - y

\int x 3 81 + x^{2} d x

\frac{\partial}{\partial x} (7 - 2 x^{2} - y^{2})

What is the integral from-1 to 2 of (3x-x^2)^3(6-4x) ?
The integral from-1 to 2 of (3x-x^2)^3(6-4x) is -120