Solve ∫ (from 0 to 2) of sqrt{4}-x^2 wrt x=

$$\int _ { 0 } ^ { 2 } \sqrt { 4 } - x ^ { 2 } d x =$$

Evaluate

$$\frac{4}{3}\approx 1.333333333$$

Solution Steps

Calculate the square root of 4 and get 2.

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

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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}.

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

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.

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

Simplify.

$$\frac{4}{3}$$