Solve ∫ (from -1 to 1) of 2sqrt[3]{x wrt x

$$\int_{ -1 }^{ 1 } 2 \sqrt[3]{ x } d x$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

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

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

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

Rewrite \sqrt[3]{x} as x^{\frac{1}{3}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{3}}\mathrm{d}x with \frac{x^{\frac{4}{3}}}{\frac{4}{3}}. Simplify. Multiply 2 times \frac{3x^{\frac{4}{3}}}{4}.

$$\frac{3x^{\frac{4}{3}}}{2}$$

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.

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

Simplify.

$$0$$