Integral calculus: integral variable substitution confusion

  • #1

Main Question or Discussion Point

Recently I started seeing integral calculus and right now we are covering the topic of the antiderivative. At first sign it was not very difficult, until we started seeing integral variable substitution. The problem starts right here:
Let's suppose that we have a function like this:

[itex]\int \sqrt{x+2}dx[/itex]

We can easily convert that [itex]\sqrt{x+2}[/itex] into a [itex](x+2)^{1/2}[/itex]. However we cannot expand this polynomial to a fractional power so instead we rename [itex]x+2[/itex] as [itex]u[/itex] so [itex]x+2=u[/itex] and [itex]dx[/itex] would become into [itex]du[/itex]. Having this we can redefine the integral to have something as this:

[itex]\int u^{1/2}du[/itex]

Now we can use the power rule to solve this integral:

[itex]\frac{2}{3}u^{3/2}+C[/itex]

The only thing we now have to do is to express the solution in terms of [itex]x[/itex] so we have that:

[itex]\frac{2}{3}(x+2)^{3/2}+C[/itex]

And this was just the beginning; later on we saw more difficult examples involving trigonometric functions. My main doubt focuses on what my math teacher did in class.

We started having this another integral:

[itex]\int 2x\sqrt{4x-3}dx[/itex]

It's a very similar problem like the previous one, now having a factor multiplying a square root. Following the similar algorithm for the last case I replaced the whole expression [itex]4x-3[/itex] into [itex]u[/itex] and to express the remaining [itex]x[/itex] in terms of [itex]u[/itex] I only isolated for [itex]x[/itex] in the [itex]u[/itex]. However what my teacher did to find [itex]du[/itex] was equalling it to [itex]4dx[/itex] and that's where I don't understand why instead of [itex]du=dx[/itex] he did that. I hope someone can answer me this to solve my homework problems. Thanks
 

Answers and Replies

  • #2
pwsnafu
Science Advisor
1,080
85
The substitution was ##u = 4x - 3## and hence ##\frac{du}{dx} = 4##. Your teacher is then rearranging to get ##du = 4 \, dx##
(I personally don't like this method, the derivative is not a fraction of du divided by dx).
 
  • #3
33,287
4,991
The substitution was ##u = 4x - 3## and hence ##\frac{du}{dx} = 4##. Your teacher is then rearranging to get ##du = 4 \, dx##
(I personally don't like this method, the derivative is not a fraction of du divided by dx).
I agree that technically du/dx is not a fraction, but most of the time it does no harm to treat it as if it were. This sort of arithmetic is done all the time in integration substitutions, including trig substitutions, as well as in integration by parts.
 

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