What is the integral of 1x(4-2(x-3)) ?

The integral of 1x(4-2(x-3)) is -(2x^3)/3+5x^2+C

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integral of 1x(4-2(x-3))

\int 1 x (4 - 2 (x - 3)) d x

- \frac{2 x ^{3}}{3} + 5 x^{2} + C

Solution steps

\int 1 \cdot x (4 - 2 (x - 3)) d x

Simplify

1 \cdot x (4 - 2 (x - 3)) : x (- 2 x + 10)

= \int x (- 2 x + 10) d x

Expand

x (- 2 x + 10) : - 2 x^{2} + 10 x

= \int - 2 x^{2} + 10 x 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 2 x^{2} d x + \int 10 x d x

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

\int 10 x d x = 5 x^{2}

= - \frac{2 x ^{3}}{3} + 5 x^{2}

Add a constant to the solution

= - \frac{2 x ^{3}}{3} + 5 x^{2} + C

What is the integral of 1x(4-2(x-3)) ?
The integral of 1x(4-2(x-3)) is -(2x^3)/3+5x^2+C