What is the integral from 0 to 1 of-12x^2(4x^3-1)^3 ?

The integral from 0 to 1 of-12x^2(4x^3-1)^3 is -20

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integral from 0 to 1 of-12x^2(4x^3-1)^3

\int_{0}^{1} - 12 x^{2} (4 x^{3} - 1)^{3} d x

- 20

Solution steps

\int_{0}^{1} - 12 x^{2} (4 x^{3} - 1)^{3} d x

Take the constant out:

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

= - 12 \cdot \int_{0}^{1} x^{2} (4 x^{3} - 1)^{3} d x

Apply u-substitution

: \int_{- 1}^{3} \frac{u ^{3}}{12} d u

= - 12 \cdot \int_{- 1}^{3} \frac{u ^{3}}{12} d u

Take the constant out:

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

= - 12 \cdot \frac{1}{12} \cdot \int_{- 1}^{3} u^{3} d u

Apply the Power Rule

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

= - 12 \cdot \frac{1}{12} [\frac{u ^{4}}{4}]_{- 1}^{3}

Simplify

- 12 \cdot \frac{1}{12} [\frac{u ^{4}}{4}]_{- 1}^{3} : - [\frac{u ^{4}}{4}]_{- 1}^{3}

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

Compute the boundaries:

20

= - 20

\frac{1}{1 + e ^{x^{2}}}

\int \frac{1}{25 - 36 x ^{2}} d x

\int 13 tan^{2} (x) sec^{3} (x) d x

\int - 2 x^{2} + 8 x d x

y^{^{''}} - 3 y^{^{'}} + y = 0

What is the integral from 0 to 1 of-12x^2(4x^3-1)^3 ?
The integral from 0 to 1 of-12x^2(4x^3-1)^3 is -20