Solve ∫ sqrt{2y wrt y

$$\int \sqrt { 2 y } d y$$

Evaluate

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

Solution Steps

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

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

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

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

Simplify.

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

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

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

Differentiate w.r.t. y

$$\sqrt{2y}$$