Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = (x+1)(x^3-2).$

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 $f(x)$ ,

$f(x) = (x+1)(x^3 -2) = x^4 + x^3 - 2x -2.$

Then,

$P(x) = \frac{1}{5} x^5 + \frac{1}{4} x^4 - x^2 - 2x$

is a primitive of $f$ since

$P'(x) = x^4 + x^3 - 2x - 2 = (x+1)(x^3-2) = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{1}{5}(b^5 - a^5) + \frac{1}{4} (b^4 - a^4) - (b^2 - a^2) - 2 (b-a). \end{align*}$

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