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\int \sqrt[3]{x}-2\mathrm{d}x
Evaluate the indefinite integral first.
\int \sqrt[3]{x}\mathrm{d}x+\int -2\mathrm{d}x
Integrate the sum term by term.
\frac{3x^{\frac{4}{3}}}{4}+\int -2\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}-2x
Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.
\frac{3}{4}\times 1^{\frac{4}{3}}-2-\left(\frac{3}{4}\left(-1\right)^{\frac{4}{3}}-2\left(-1\right)\right)
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.
-4
Simplify.