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

The integral from 0 to 2 of x^2(1+2x)^3 is 944/5

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

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

\frac{944}{5}

Decimal

188.8

Solution steps

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

Multiply by the conjugate of

1 + 2 x : \frac{1 - 4 x ^{2}}{1 - 2 x}

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

Find undefined (singularity) points:

x = \frac{1}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{1}{2}} x^{2} (\frac{1 - 4 x ^{2}}{1 - 2 x})^{3} d x + \int_{\frac{1}{2}}^{2} x^{2} (\frac{1 - 4 x ^{2}}{1 - 2 x})^{3} d x

\int_{0}^{\frac{1}{2}} x^{2} (\frac{1 - 4 x ^{2}}{1 - 2 x})^{3} d x = \frac{37}{160}

\int_{\frac{1}{2}}^{2} x^{2} (\frac{1 - 4 x ^{2}}{1 - 2 x})^{3} d x = \frac{30171}{160}

= \frac{37}{160} + \frac{30171}{160}

Simplify

\frac{37}{160} + \frac{30171}{160} : \frac{944}{5}

= \frac{944}{5}

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What is the integral from 0 to 2 of x^2(1+2x)^3 ?
The integral from 0 to 2 of x^2(1+2x)^3 is 944/5