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

The integral from-1 to 0 of 2x^2(4x^3-1)^3 is -26

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

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

- 26

Solution steps

\int_{- 1}^{0} 2 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

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

Apply u-substitution

: \int_{- 5}^{- 1} \frac{u ^{3}}{12} d u

= 2 \cdot \int_{- 5}^{- 1} \frac{u ^{3}}{12} d u

Take the constant out:

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

= 2 \cdot \frac{1}{12} \cdot \int_{- 5}^{- 1} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 5}^{- 1}

= 2 \cdot \frac{1}{12} [\frac{u ^{4}}{4}]_{- 5}^{- 1}

Simplify

2 \cdot \frac{1}{12} [\frac{u ^{4}}{4}]_{- 5}^{- 1} : \frac{1}{6} [\frac{u ^{4}}{4}]_{- 5}^{- 1}

= \frac{1}{6} [\frac{u ^{4}}{4}]_{- 5}^{- 1}

Compute the boundaries:

- 156

= \frac{1}{6} (- 156)

Simplify

= - 26

\int_{0}^{2} 2 x e^{x} - e^{x} + e^{- x} d x

\int_{0}^{4} x^{2} e^{x} d x

\int_{- 4}^{4} - e^{(\frac{1}{4}) x} d x

\int_{4}^{8} 10 - x d x

\int_{0}^{2} 4 x^{2} + 1 d x

What is the integral from-1 to 0 of 2x^2(4x^3-1)^3 ?
The integral from-1 to 0 of 2x^2(4x^3-1)^3 is -26