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Find a primitive and apply the second fundamental theorem of calculus

by RoRi

Let

    \[ f(x) = \left( 1 + \sqrt{x} \right)^2 \qquad x> 0. \]

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

    \[ \int_a^b f(x) \, dx. \]

First, let’s expand f(x),

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

Then,

    \[ P(x) = \frac{1}{2}x^2 + \frac{4}{3} x^{\frac{3}{2}} + x \]

is a primitive of f since

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

Then, by the second fundamental theorem of calculus we have

    \begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{1}{2} (b^2-a^2) + \frac{4}{3} \left( b^{\frac{3}{2}} - a^{\frac{3}{2}} \right) + (b-a). \end{align*}

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