Solve ∫ (from - 1 to 1) of (sqrt[3]{t}-2)dt

$$\int _ { - 1 } ^ { 1 } ( \sqrt[ 3 ] { t } - 2 ) d t$$

Evaluate

$$-4$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \sqrt[3]{t}-2\mathrm{d}t$$

Integrate the sum term by term.

$$\int \sqrt[3]{t}\mathrm{d}t+\int -2\mathrm{d}t$$

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.

$$\frac{3t^{\frac{4}{3}}}{4}+\int -2\mathrm{d}t$$

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}t=at.

$$\frac{3t^{\frac{4}{3}}}{4}-2t$$

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.

$$\frac{3}{4}\times 1^{\frac{4}{3}}-2-\left(\frac{3}{4}\left(-1\right)^{\frac{4}{3}}-2\left(-1\right)\right)$$

Simplify.

$$-4$$