Solve ∫ (from 0 to 2) of (4-t^2)dt

$$\int _ { 0 } ^ { 2 } ( 4 - t ^ { 2 } ) d t$$

Evaluate

$$\frac{16}{3}\approx 5.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 4-t^{2}\mathrm{d}t$$

Integrate the sum term by term.

$$\int 4\mathrm{d}t+\int -t^{2}\mathrm{d}t$$

Factor out the constant in each of the terms.

$$\int 4\mathrm{d}t-\int t^{2}\mathrm{d}t$$

Find the integral of 4 using the table of common integrals rule \int a\mathrm{d}t=at.

$$4t-\int t^{2}\mathrm{d}t$$

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

$$4t-\frac{t^{3}}{3}$$

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.

$$4\times 2-\frac{2^{3}}{3}-\left(4\times 0-\frac{0^{3}}{3}\right)$$

Simplify.

$$\frac{16}{3}$$