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

The integral from 0 to 1 of 3x^2(2x^3-3)^2 is 13/3

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

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

\frac{13}{3}

Decimal Notation

4.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{( 2 u - 3 ) ^{2}}{3} d u

= 3 \cdot \int_{0}^{1} \frac{( 2 u - 3 ) ^{2}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{0}^{1} (2 u - 3)^{2} d u

Expand

(2 u - 3)^{2} : 4 u^{2} - 12 u + 9

= 3 \cdot \frac{1}{3} \cdot \int_{0}^{1} 4 u^{2} - 12 u + 9 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{3} (\int_{0}^{1} 4 u^{2} d u - \int_{0}^{1} 12 u d u + \int_{0}^{1} 9 d u)

\int_{0}^{1} 4 u^{2} d u = \frac{4}{3}

\int_{0}^{1} 12 u d u = 6

\int_{0}^{1} 9 d u = 9

= 3 \cdot \frac{1}{3} (\frac{4}{3} - 6 + 9)

Simplify

3 \cdot \frac{1}{3} (\frac{4}{3} - 6 + 9) : \frac{13}{3}

= \frac{13}{3}

\frac{d}{d x} (4 (x + 2)^{3})

x \to - 3 lim (- (\frac{1}{x ^{2} - 9}))

\int \frac{3 x ^{3} + 6 x ^{2} + 3}{3 x + 3} d x

x \to \infty lim (0.6 6^{x})

What is the integral from 0 to 1 of 3x^2(2x^3-3)^2 ?
The integral from 0 to 1 of 3x^2(2x^3-3)^2 is 13/3