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

The integral of-(x-3)^2 is -(x^3)/3+3x^2-9x+C

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integral of-(x-3)^2

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

- \frac{x ^{3}}{3} + 3 x^{2} - 9 x + C

Solution steps

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

Take the constant out:

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

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

Expand

(x - 3)^{2} : x^{2} - 6 x + 9

= - \int x^{2} - 6 x + 9 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 x^{2} d x - \int 6 x d x + \int 9 d x)

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

\int 6 x d x = 3 x^{2}

\int 9 d x = 9 x

= - (\frac{x ^{3}}{3} - 3 x^{2} + 9 x)

Simplify

= - \frac{x ^{3}}{3} + 3 x^{2} - 9 x

Add a constant to the solution

= - \frac{x ^{3}}{3} + 3 x^{2} - 9 x + C

\frac{d}{d x} (y^{2} - 2 y cos (x))

y = 10 x^{4} - 10 x^{2}, y = 9 x^{2}

\int 1 + (cos (x))^{2} d x

\int_{0}^{2} \frac{3}{4 x ^{2} + 5 x + 1} d x

\frac{\partial}{\partial x} (f (x))