Home » Blog » Find a primitive and apply the second fundamental theorem of calculus

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

Let

    \[ f(x) = \frac{2x^2 - 6x + y}{2\sqrt{x}}, \qquad x > 0. \]

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 rewrite f(x),

    \begin{align*} f(x) &= \frac{2x^2 - 6x + y}{2 \sqrt{x}} \\ &= x^{\frac{3}{2}} - 3x^{\frac{1}{2}} + \frac{7}{2} x^{-\frac{1}{2}} \end{align*}

Then,

    \[ P(x) = \frac{2}{5} x^{\frac{5}{2}} - 2x^{\frac{3}{2}} + 7 x^{\frac{1}{2}} \]

is a primitive of f since

    \[ P'(x) = x^{\frac{3}{2}} - 3x^{\frac{1}{2}} + \frac{7}{2} x^{-\frac{1}{2}} = f(x). \]

Then, by the second fundamental theorem of calculus we have

    \begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{2}{5} \left( b^{\frac{5}{2}} - a^{\frac{5}{2}} \right) - 2 \left( b^{\frac{3}{2}} - a^{\frac{3}{2}} \right) + 7 \left( b^{\frac{1}{2}} - a^{\frac{1}{2}} \right). \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):