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

$$\int _ { a } ^ { 1 } \sqrt[ 3 ] { x } - x + 1 d x$$

Evaluate

$$\frac{a^{2}}{2}-\frac{3a^{\frac{4}{3}}}{4}-a+\frac{5}{4}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \sqrt[3]{x}-x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int \sqrt[3]{x}\mathrm{d}x+\int -x\mathrm{d}x+\int 1\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int \sqrt[3]{x}\mathrm{d}x-\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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.

$$\frac{3x^{\frac{4}{3}}}{4}-\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{3x^{\frac{4}{3}}}{4}-\frac{x^{2}}{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Simplify.

$$\frac{5}{4}-\frac{3a^{\frac{4}{3}}}{4}+\frac{a^{2}}{2}-a$$

Differentiate w.r.t. a

$$a-\sqrt[3]{a}-1$$