Solve ∫ (from 0 to frac { 1) of {2}}(1-4x^2) wrt x

$$\int _ { 0 } ^ { \frac { 1 } { 2 } } ( 1 - 4 x ^ { 2 } ) d x$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 1-4x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int -4x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x-4\int x^{2}\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -4 times \frac{x^{3}}{3}.

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

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.

$$\frac{1}{2}-\frac{4}{3}\times \left(\frac{1}{2}\right)^{3}-\left(0-\frac{4}{3}\times 0^{3}\right)$$

Simplify.

$$\frac{1}{3}$$