Calculus I - Indefinite Integrals - Pauls Online Math Notes

Let’s actually start by getting the derivative of this function to help us see how we’re going to have to approach this problem. The derivative of this function is,

\[f'\left( x \right) = 4{x^3} + 3\]

The point of this was to remind us of how differentiation works. When differentiating powers of \(x\) we multiply the term by the original exponent and then drop the exponent by one.

Now, let’s go back and work the problem. In fact, let’s just start with the first term. We got \({x^4}\) by differentiating a function and since we drop the exponent by one it looks like we must have differentiated \({x^5}\). However, if we had differentiated \({x^5}\) we would have \(5{x^4}\) and we don’t have a 5 in front our first term, so the 5 needs to cancel out after we’ve differentiated. It looks then like we would have to differentiate \(\frac{1}{5}{x^5}\) in order to get \({x^4}\).

Likewise, for the second term, in order to get 3x after differentiating we would have to differentiate \(\frac{3}{2}{x^2}\). Again, the fraction is there to cancel out the 2 we pick up in the differentiation.

The third term is just a constant and we know that if we differentiate \(x\) we get 1. So, it looks like we had to differentiate -9\(x\) to get the last term.

Putting all of this together gives the following function,

\[F\left( x \right) = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x\]

Our answer is easy enough to check. Simply differentiate \(F\left( x \right)\).

\[F'\left( x \right) = {x^4} + 3x - 9 = f\left( x \right)\]

So, it looks like we got the correct function. Or did we? We know that the derivative of a constant is zero and so any of the following will also give \(f\left( x \right)\) upon differentiating.

\[\begin{align*}F\left( x \right) & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + 10\\ F\left( x \right) & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x - 1954\\ F\left( x \right) & = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + \frac{{3469}}{{123}}\\ & etc.\end{align*}\]

In fact, any function of the form,

\[F\left( x \right) = \frac{1}{5}{x^5} + \frac{3}{2}{x^2} - 9x + c,\,\,\hspace{0.25in}c{\mbox{ is a constant}}\]

will give \(f\left( x \right)\) upon differentiating.