Solve ∫ (from 0 to 1) of sqrt[3]{x}-sqrt{x wrt x

$$\int_{ 0 }^{ 1 } \sqrt[3]{ x } - \sqrt{ x } d x$$

Evaluate

$$\frac{1}{12}\approx 0.083333333$$

Solution Steps

Evaluate the indefinite integral first.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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.

$$\frac{3x^{\frac{4}{3}}}{4}-\int \sqrt{x}\mathrm{d}x$$

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

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

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.

$$\frac{3}{4}\times 1^{\frac{4}{3}}-\frac{2}{3}\times 1^{\frac{3}{2}}-\left(\frac{3}{4}\times 0^{\frac{4}{3}}-\frac{2}{3}\times 0^{\frac{3}{2}}\right)$$

Simplify.

$$\frac{1}{12}$$