Difficulty to find this integral

  • #1

Homework Statement


Hi! I'm not being able to solve this integral: [tex]\int \sqrt{2ax}\; dx[/tex] We have started with integral yesterday and I don't know much yet.



Homework Equations





The Attempt at a Solution


This is what I tried to solve the integral
[tex]\int \sqrt{2ax}\; dx = \int u^{\frac{1}{2}}\; du = \frac{2}{3}{}u^{\frac{3}{2}}=\frac{2}{3}\sqrt[3]{(2ax)^2}[/tex]
 

Answers and Replies

  • #2
DryRun
Gold Member
838
4
Assuming that a is a real constant,
[tex]\int \sqrt{2ax}\; dx = \int \sqrt{2a}\sqrt x\; dx= \sqrt{2a} \int \sqrt x\; dx[/tex]
Since [itex]\sqrt x[/itex] is the same as [itex]x^{\frac{1}{2}}[/itex], you can easily find its integral, and then multiply by [itex]\sqrt {2a}[/itex] to get the final answer.
 
  • #3
34,936
6,699

The Attempt at a Solution


This is what I tried to solve the integral
[tex]\int \sqrt{2ax}\; dx = \int u^{\frac{1}{2}}\; du = \frac{2}{3}{}u^{\frac{3}{2}}=\frac{2}{3}\sqrt[3]{(2ax)^2}[/tex]
When you use a substitution, as you obviously did above, you need to write down what the substitution is. IOW, you should have u = <...> somewhere close by.

Also, u3/2 = ##\sqrt{u^3}##, not ##\sqrt[3]{u^2}##, which is what you had.
 

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