What is the integral of (x+4)(x-7) ?

The integral of (x+4)(x-7) is (x^3)/3-(3x^2)/2-28x+C

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integral of (x+4)(x-7)

\int (x + 4) (x - 7) d x

\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 28 x + C

Solution steps

\int (x + 4) (x - 7) d x

Expand

(x + 4) (x - 7) : x^{2} - 3 x - 28

= \int x^{2} - 3 x - 28 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 3 x d x - \int 28 d x

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

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

\int 28 d x = 28 x

= \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 28 x

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 28 x + C

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What is the integral of (x+4)(x-7) ?
The integral of (x+4)(x-7) is (x^3)/3-(3x^2)/2-28x+C