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

The integral from 0 to 1 of 3x^2(4x^3-5)^3 is -39

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

\int_{0}^{1} 3 x^{2} (4 x^{3} - 5)^{3} d x

- 39

Solution steps

\int_{0}^{1} 3 x^{2} (4 x^{3} - 5)^{3} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} x^{2} (4 x^{3} - 5)^{3} d x

Apply u-substitution

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

= 3 \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

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

Apply the Power Rule

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

= 3 \cdot \frac{1}{12} [\frac{u ^{4}}{4}]_{- 5}^{- 1}

Simplify

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

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

Compute the boundaries:

- 156

= \frac{1}{4} (- 156)

Simplify

= - 39

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

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

\int_{0}^{7} 4 d x

\int_{0}^{\infty} e^{- 1 x} d x

\int_{0}^{4} (x^{2} + 4 x) d x

What is the integral from 0 to 1 of 3x^2(4x^3-5)^3 ?
The integral from 0 to 1 of 3x^2(4x^3-5)^3 is -39