- #1

thegreengineer

- 54

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**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