Solve ∫ (from 0 to 8) of (sqrt{2x}) wrt x

$$\int _ { 0 } ^ { 8 } ( \sqrt { 2 x } ) d x$$

Evaluate

$$\frac{64}{3}\approx 21.333333333$$

Solution Steps

Evaluate the indefinite integral first.

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

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

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

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

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

Simplify.

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

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.

$$\frac{2}{3}\times 2^{\frac{1}{2}}\times 8^{\frac{3}{2}}-\frac{2}{3}\times 2^{\frac{1}{2}}\times 0^{\frac{3}{2}}$$

Simplify.

$$\frac{64}{3}$$