Solve ∫ (from 0 to a) of ((0)-(x^2-9x)] wrt x

$$\int _ { 0 } ^ { a } ( ( 0 ) - ( x ^ { 2 } - 9 x ) ] d x$$

Evaluate

$$-\frac{a^{3}}{3}+\frac{9a^{2}}{2}$$

Solution Steps

To find the opposite of x^{2}-9x, find the opposite of each term.

$$\int _{0}^{a}0-x^{2}+9x\mathrm{d}x$$

Anything plus zero gives itself.

$$\int _{0}^{a}-x^{2}+9x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int -x^{2}+9x\mathrm{d}x$$

Integrate the sum term by term.

$$\int -x^{2}\mathrm{d}x+\int 9x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-\int x^{2}\mathrm{d}x+9\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -1 times \frac{x^{3}}{3}.

$$-\frac{x^{3}}{3}+9\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply 9 times \frac{x^{2}}{2}.

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

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$-\frac{a^{3}}{3}+\frac{9}{2}a^{2}-\left(-\frac{0^{3}}{3}+\frac{9}{2}\times 0^{2}\right)$$

Simplify.

$$\frac{9a^{2}}{2}-\frac{a^{3}}{3}$$

Differentiate w.r.t. a

$$a\left(9-a\right)$$