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Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 40

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Find the integrals $\displaystyle \int^8_1 \frac{x-1}{\sqrt[3]{x^2}} dx$

$ \begin{equation} \begin{aligned} \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \int \left( \frac{x}{\sqrt[3]{x^2}} - \frac{1}{\sqrt[3]{x^2}} \right) dx\\ \\ \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \int \left( \frac{x}{x^{\frac{2}{3}}} - \frac{1}{x^{\frac{2}{3}}} \right) dx\\ \\ \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \int \left( x^{\frac{1}{3}} - x^{\frac{-2}{3}} \right) dx\\ \\ \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \int x^{\frac{1}{3}} dx - \int x^{\frac{-2}{3}} dx\\ \\ \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1} - \frac{x^{\frac{-2}{3}+1}}{\frac{-2}{3}+1} + C\\ \\ \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{x^{\frac{4}{3}}}{\frac{4}{3}} - \left( \frac{x^{\frac{1}{3}}}{\frac{1}{3}} \right) + C\\ \\ \int\frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{3x^{\frac{4}{3}}}{4} - 3x^{\frac{1}{3}} + C\\ \\ \int^8_1 \frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{3(8)^{\frac{4}{3}}}{4} - 3(8)^{\frac{1}{3}} + C - \left[ \frac{3(1)^{\frac{4}{3}}}{4} - 3(1)^{\frac{1}{3}} + C \right]\\ \\ \int^8_1 \frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{3\left[(8)^{\frac{1}{3}}\right]}{4} - 3 (2) + C - \frac{3}{4} + 3 - C\\ \\ \int^8_1 \frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{3(2)^4}{4} - 6 - \frac{3}{4} + 3 \\ \\ \int^8_1 \frac{x-1}{\sqrt[3]{x^2}} dx &= 12- \frac{3}{4} - 3\\ \\ \int^8_1 \frac{x-1}{\sqrt[3]{x^2}} dx &= \frac{33}{4} \end{aligned} \end{equation} $