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

The integral from 0 to 1 of x^2(2x^3+3)^2 is 49/9

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

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

\frac{49}{9}

Decimal

5.44444 \dots

Solution steps

\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

= \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

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

Expand

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

= \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

= \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

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

Simplify

\frac{1}{3} (\frac{4}{3} + 6 + 9) : \frac{49}{9}

= \frac{49}{9}

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What is the integral from 0 to 1 of x^2(2x^3+3)^2 ?
The integral from 0 to 1 of x^2(2x^3+3)^2 is 49/9