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Find a primitive and apply the second fundamental theorem of calculus

Let

    \[ f(x) = 2x^{\frac{1}{3}} - x^{-\frac{1}{3}} \qquad x> 0. \]

Find a function P(x) such that P'(x) = f(x) (i.e., a primitive of f). Use the second fundamental theorem of calculus to evaluate

    \[ \int_a^b f(x) \, dx. \]


The function

    \[ P(x) = \frac{3}{2} x^{\frac{4}{3}} - \frac{3}{2} x^{\frac{2}{3}} \]

is a primitive of f since

    \[ P'(x) = 2x^{\frac{1}{3}} - x^{-\frac{1}{3}} = f(x). \]

Then, by the second fundamental theorem of calculus we have

    \begin{align*}   \int_a^b f(x) \, dx &= P(b) - P(a) \\  &= \frac{3}{2} \left( b^{\frac{4}{3}} - a^{\frac{4}{3}} + b^{\frac{2}{3}} - a^{\frac{2}{3}} \right).  \end{align*}

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