What is the integral from-1 to 0 of 6x^2(2x^3+3)^2 ?

The integral from-1 to 0 of 6x^2(2x^3+3)^2 is 26/3

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integral from-1 to 0 of 6x^2(2x^3+3)^2

\int_{- 1}^{0} 6 x^{2} (2 x^{3} + 3)^{2} d x

\frac{26}{3}

Decimal

8.66666 \dots

Solution steps

\int_{- 1}^{0} 6 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

= 6 \cdot \int_{- 1}^{0} x^{2} (2 x^{3} + 3)^{2} d x

Apply u-substitution

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

= 6 \cdot \int_{- 1}^{0} \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

= 6 \cdot \frac{1}{3} \cdot \int_{- 1}^{0} (2 u + 3)^{2} d u

Expand

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

= 6 \cdot \frac{1}{3} \cdot \int_{- 1}^{0} 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

= 6 \cdot \frac{1}{3} (\int_{- 1}^{0} 4 u^{2} d u + \int_{- 1}^{0} 12 u d u + \int_{- 1}^{0} 9 d u)

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

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

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

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

Simplify

6 \cdot \frac{1}{3} (\frac{4}{3} - 6 + 9) : \frac{26}{3}

= \frac{26}{3}

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What is the integral from-1 to 0 of 6x^2(2x^3+3)^2 ?
The integral from-1 to 0 of 6x^2(2x^3+3)^2 is 26/3