What is the integral of (x^3*(1-x)^2)/2 ?

The integral of (x^3*(1-x)^2)/2 is 1/120 (10x^6-24x^5+15x^4)+C

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integral of (x^3*(1-x)^2)/2

\int \frac{x ^{3} \cdot ( 1 - x ) ^{2}}{2} d x

\frac{1}{120} (10 x^{6} - 24 x^{5} + 15 x^{4}) + C

Solution steps

\int \frac{x ^{3} ( 1 - x ) ^{2}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{3} (1 - x)^{2} d x

Apply Integration By Parts

: \frac{1}{4} x^{4} (1 - x)^{2} - \int - \frac{1}{2} x^{4} (- x + 1) d x

= \frac{1}{2} (\frac{1}{4} x^{4} (1 - x)^{2} - \int - \frac{1}{2} x^{4} (- x + 1) d x)

\int - \frac{1}{2} x^{4} (- x + 1) d x = - \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5})

= \frac{1}{2} (\frac{1}{4} x^{4} (1 - x)^{2} - (- \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5})))

Simplify

\frac{1}{2} (\frac{1}{4} x^{4} (1 - x)^{2} - (- \frac{1}{2} (- \frac{x ^{6}}{6} + \frac{x ^{5}}{5}))) : \frac{1}{120} (10 x^{6} - 24 x^{5} + 15 x^{4})

= \frac{1}{120} (10 x^{6} - 24 x^{5} + 15 x^{4})

Add a constant to the solution

= \frac{1}{120} (10 x^{6} - 24 x^{5} + 15 x^{4}) + C

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What is the integral of (x^3*(1-x)^2)/2 ?
The integral of (x^3*(1-x)^2)/2 is 1/120 (10x^6-24x^5+15x^4)+C