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

The integral from 0 to 1 of 4x^3(3x^4-3)^5 is -81/2

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

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

- \frac{81}{2}

Decimal

- 40.5

Solution steps

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

Take the constant out:

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

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

Expand

x^{3} (3 x^{4} - 3)^{5} : 243 x^{23} - 1215 x^{19} + 2430 x^{15} - 2430 x^{11} + 1215 x^{7} - 243 x^{3}

= 4 \cdot \int_{0}^{1} 243 x^{23} - 1215 x^{19} + 2430 x^{15} - 2430 x^{11} + 1215 x^{7} - 243 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

= 4 (\int_{0}^{1} 243 x^{23} d x - \int_{0}^{1} 1215 x^{19} d x + \int_{0}^{1} 2430 x^{15} d x - \int_{0}^{1} 2430 x^{11} d x + \int_{0}^{1} 1215 x^{7} d x - \int_{0}^{1} 243 x^{3} d x)

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

\int_{0}^{1} 1215 x^{19} d x = \frac{243}{4}

\int_{0}^{1} 2430 x^{15} d x = \frac{1215}{8}

\int_{0}^{1} 2430 x^{11} d x = \frac{405}{2}

\int_{0}^{1} 1215 x^{7} d x = \frac{1215}{8}

\int_{0}^{1} 243 x^{3} d x = \frac{243}{4}

= 4 (\frac{81}{8} - \frac{243}{4} + \frac{1215}{8} - \frac{405}{2} + \frac{1215}{8} - \frac{243}{4})

Simplify

4 (\frac{81}{8} - \frac{243}{4} + \frac{1215}{8} - \frac{405}{2} + \frac{1215}{8} - \frac{243}{4}) : - \frac{81}{2}

= - \frac{81}{2}

What is the integral from 0 to 1 of 4x^3(3x^4-3)^5 ?
The integral from 0 to 1 of 4x^3(3x^4-3)^5 is -81/2