Solve ∫ 4sqrt{u}du

$$\int 4 \sqrt { u } d u$$

Evaluate

$$\frac{8u^{\frac{3}{2}}}{3}+С$$

Solution Steps

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

$$4\int \sqrt{u}\mathrm{d}u$$

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

$$\frac{8u^{\frac{3}{2}}}{3}$$

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

$$\frac{8u^{\frac{3}{2}}}{3}+С$$

Differentiate w.r.t. u

$$4\sqrt{u}$$