Solve ∫ (from 0 to 4) of sqrt{2t}dt

$$\int _ { 0 } ^ { 4 } \sqrt { 2 t } d t$$

Evaluate

$$\frac{16\sqrt{2}}{3}\approx 7.542472333$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \sqrt{2t}\mathrm{d}t$$

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

$$\sqrt{2}\int \sqrt{t}\mathrm{d}t$$

Rewrite \sqrt{t} as t^{\frac{1}{2}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{\frac{1}{2}}\mathrm{d}t with \frac{t^{\frac{3}{2}}}{\frac{3}{2}}. Simplify.

$$\sqrt{2}\times \frac{2t^{\frac{3}{2}}}{3}$$

Simplify.

$$\frac{2\sqrt{2}t^{\frac{3}{2}}}{3}$$

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.

$$\frac{2}{3}\times 2^{\frac{1}{2}}\times 4^{\frac{3}{2}}-\frac{2}{3}\times 2^{\frac{1}{2}}\times 0^{\frac{3}{2}}$$

Simplify.

$$\frac{16\sqrt{2}}{3}$$