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Find a primitive and apply the second fundamental theorem of calculus

by RoRi

Let

    \[ f(x) = \sqrt{2x} + \sqrt{\frac{1}{2} x}. \]

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

    \[ \int_a^b f(x) \, dx. \]

First, let’s rewrite f(x),

    \begin{align*} f(x) &= \sqrt{2x} + \sqrt{\frac{1}{2} x} \\ &= \left( \sqrt{2} + \frac{1}{\sqrt{2}} \right) x^{\frac{1}{2}} \\ &= \frac{3\sqrt{2}}{2} \sqrt{x}. \end{align*}

Then,

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

is a primitive of f since

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

Then, by the second fundamental theorem of calculus we have

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

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