Solve ∫ (from 0 to x) of sqrt9-t^2dt

$$\int _ { 0 } ^ { x } \sqrt { 9 - t } ^ { 2 } d t$$

Evaluate

$$\frac{x\left(18-x\right)}{2}$$

Solution Steps

Calculate \sqrt{9-t} to the power of 2 and get 9-t.

$$\int _{0}^{x}9-t\mathrm{d}t$$

Evaluate the indefinite integral first.

$$\int 9-t\mathrm{d}t$$

Integrate the sum term by term.

$$\int 9\mathrm{d}t+\int -t\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 9\mathrm{d}t-\int t\mathrm{d}t$$

Find the integral of 9 using the table of common integrals rule \int a\mathrm{d}t=at.

$$9t-\int t\mathrm{d}t$$

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

$$9t-\frac{t^{2}}{2}$$

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.

$$9x-\frac{x^{2}}{2}-\left(9\times 0-\frac{0^{2}}{2}\right)$$

Simplify.

$$\frac{x\left(18-x\right)}{2}$$

Differentiate w.r.t. x

$$9-x$$