Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = \frac{x^4 + x - 3}{x^3}, \qquad x \neq 0.$

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 $f(x)$ ,

$f(x) = \frac{x^4 + x - 3}{x^3} = x + x^{-2} - 3x^{-3}.$

Then,

$P(x) = \frac{1}{2} x^2 - x^{-1} + \frac{3}{2} x^{-2}$

is a primitive of $f$ since

$P'(x) = x + x^{-2} - 3x^{-3} = \frac{x^4 + x - 3}{x^3} = f(x).$

Then, by the second fundamental theorem of calculus we have

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

Stumbling Robot