Solve ∫ (from 0 to 1) of 1/2v^2dv

$$\int _ { 0 } ^ { 1 } \frac { 1 } { 2 } v ^ { 2 } d v$$

Evaluate

$$\frac{1}{6}\approx 0.166666667$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{v^{2}}{2}\mathrm{d}v$$

Factor out the constant using \int af\left(v\right)\mathrm{d}v=a\int f\left(v\right)\mathrm{d}v.

$$\frac{\int v^{2}\mathrm{d}v}{2}$$

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

$$\frac{v^{3}}{6}$$

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.

$$\frac{1^{3}}{6}-\frac{0^{3}}{6}$$

Simplify.

$$\frac{1}{6}$$