Solve ∫ sqrt{2x-0*5 wrt x

$$\int{ \sqrt{ 2x-0 \cdot 5 } }d x$$

Evaluate

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

Solution Steps

Multiply 0 and 5 to get 0.

$$\int \sqrt{2x-0}\mathrm{d}x$$

Multiply -1 and 0 to get 0.

$$\int \sqrt{2x+0}\mathrm{d}x$$

Anything plus zero gives itself.

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

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.

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

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

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

Differentiate w.r.t. x

$$\sqrt{2x}$$