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

The integral of 1-(x-3)^2 is -(x^3)/3+3x^2-8x+C

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integral of 1-(x-3)^2

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

- \frac{x ^{3}}{3} + 3 x^{2} - 8 x + C

Solution steps

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

Expand

1 - (x - 3)^{2} : - x^{2} + 6 x - 8

= \int - x^{2} + 6 x - 8 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 8 d x

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

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

\int 8 d x = 8 x

= - \frac{x ^{3}}{3} + 3 x^{2} - 8 x

Add a constant to the solution

= - \frac{x ^{3}}{3} + 3 x^{2} - 8 x + C

What is the integral of 1-(x-3)^2 ?
The integral of 1-(x-3)^2 is -(x^3)/3+3x^2-8x+C