What is the integral from 0 to 1 of-6x^3(3x^4-2)^5 ?

The integral from 0 to 1 of-6x^3(3x^4-2)^5 is 21/4

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integral from 0 to 1 of-6x^3(3x^4-2)^5

\int_{0}^{1} - 6 x^{3} (3 x^{4} - 2)^{5} d x

\frac{21}{4}

Solution steps

\int_{0}^{1} - 6 x^{3} (3 x^{4} - 2)^{5} d x

Take the constant out:

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

= - 6 \cdot \int_{0}^{1} x^{3} (3 x^{4} - 2)^{5} d x

Expand

x^{3} (3 x^{4} - 2)^{5} : 243 x^{23} - 810 x^{19} + 1080 x^{15} - 720 x^{11} + 240 x^{7} - 32 x^{3}

= - 6 \cdot \int_{0}^{1} 243 x^{23} - 810 x^{19} + 1080 x^{15} - 720 x^{11} + 240 x^{7} - 32 x^{3} d x

Apply the Sum Rule:

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

= - 6 (\int_{0}^{1} 243 x^{23} d x - \int_{0}^{1} 810 x^{19} d x + \int_{0}^{1} 1080 x^{15} d x - \int_{0}^{1} 720 x^{11} d x + \int_{0}^{1} 240 x^{7} d x - \int_{0}^{1} 32 x^{3} d x)

\int_{0}^{1} 243 x^{23} d x = \frac{81}{8}

\int_{0}^{1} 810 x^{19} d x = \frac{81}{2}

\int_{0}^{1} 1080 x^{15} d x = \frac{135}{2}

\int_{0}^{1} 720 x^{11} d x = 60

\int_{0}^{1} 240 x^{7} d x = 30

\int_{0}^{1} 32 x^{3} d x = 8

= - 6 (\frac{81}{8} - \frac{81}{2} + \frac{135}{2} - 60 + 30 - 8)

Simplify

- 6 (\frac{81}{8} - \frac{81}{2} + \frac{135}{2} - 60 + 30 - 8) : \frac{21}{4}

= \frac{21}{4}

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What is the integral from 0 to 1 of-6x^3(3x^4-2)^5 ?
The integral from 0 to 1 of-6x^3(3x^4-2)^5 is 21/4