What is the integral of x(7-x)^2 ?

The integral of x(7-x)^2 is (49x^2)/2-(14x^3)/3+(x^4)/4+C

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integral of x(7-x)^2

\int x (7 - x)^{2} d x

\frac{49 x ^{2}}{2} - \frac{14 x ^{3}}{3} + \frac{x ^{4}}{4} + C

Solution steps

\int x (7 - x)^{2} d x

Expand

x (7 - x)^{2} : 49 x - 14 x^{2} + x^{3}

= \int 49 x - 14 x^{2} + 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

= \int 49 x d x - \int 14 x^{2} d x + \int x^{3} d x

\int 49 x d x = \frac{49 x ^{2}}{2}

\int 14 x^{2} d x = \frac{14 x ^{3}}{3}

\int x^{3} d x = \frac{x ^{4}}{4}

= \frac{49 x ^{2}}{2} - \frac{14 x ^{3}}{3} + \frac{x ^{4}}{4}

Add a constant to the solution

= \frac{49 x ^{2}}{2} - \frac{14 x ^{3}}{3} + \frac{x ^{4}}{4} + C

1 + x

\int 6 x - 5 d x

\int 5 x^{4} d x

\frac{d y}{d x} + \frac{y}{x} = y^{- 2}

\frac{( t ^{3} + t )}{( t ^{4} - 2 )}

What is the integral of x(7-x)^2 ?
The integral of x(7-x)^2 is (49x^2)/2-(14x^3)/3+(x^4)/4+C