Solve ∫ 4/3sqrt[3]{t}dt

$$\int \frac { 4 } { 3 } \sqrt[ 3 ] { t } d t$$

Evaluate

$$t^{\frac{4}{3}}+С$$

Solution Steps

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

$$\frac{4\int \sqrt[3]{t}\mathrm{d}t}{3}$$

Rewrite \sqrt[3]{t} as t^{\frac{1}{3}}. Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{\frac{1}{3}}\mathrm{d}t with \frac{t^{\frac{4}{3}}}{\frac{4}{3}}. Simplify.

$$t^{\frac{4}{3}}$$

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

$$t^{\frac{4}{3}}+С$$

Differentiate w.r.t. t

$$\frac{4\sqrt[3]{t}}{3}$$