What is the integral from 8/27 to 1 of ((28x^{4/3}-26x^{1/3}))/(x^2) ?

The integral from 8/27 to 1 of ((28x^{4/3}-26x^{1/3}))/(x^2) is -83/4

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integral from 8/27 to 1 of ((28x^{4/3}-26x^{1/3}))/(x^2)

\int_{\frac{8}{27}}^{1} \frac{( 28 x ^{\frac{4}{3}} - 26 x ^{\frac{1}{3}} )}{x ^{2}} d x

- \frac{83}{4}

Decimal

- 20.75

Solution steps

\int_{\frac{8}{27}}^{1} \frac{28 x ^{\frac{4}{3}} - 26 x ^{\frac{1}{3}}}{x ^{2}} d x

Expand

\frac{28 x ^{\frac{4}{3}} - 26 x ^{\frac{1}{3}}}{x ^{2}} : \frac{28}{x ^{\frac{2}{3}}} - \frac{26}{x ^{\frac{5}{3}}}

= \int_{\frac{8}{27}}^{1} \frac{28}{x ^{\frac{2}{3}}} - \frac{26}{x ^{\frac{5}{3}}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{\frac{8}{27}}^{1} \frac{28}{x ^{\frac{2}{3}}} d x - \int_{\frac{8}{27}}^{1} \frac{26}{x ^{\frac{5}{3}}} d x

\int_{\frac{8}{27}}^{1} \frac{28}{x ^{\frac{2}{3}}} d x = 28

\int_{\frac{8}{27}}^{1} \frac{26}{x ^{\frac{5}{3}}} d x = \frac{195}{4}

= 28 - \frac{195}{4}

Simplify

28 - \frac{195}{4} : - \frac{83}{4}

= - \frac{83}{4}

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What is the integral from 8/27 to 1 of ((28x^{4/3}-26x^{1/3}))/(x^2) ?
The integral from 8/27 to 1 of ((28x^{4/3}-26x^{1/3}))/(x^2) is -83/4