Can't Find a Correct Method to Integrate \int (t - 2)^2\sqrt{t}\,dt?

  • #1
KungPeng Zhou
22
7
Homework Statement
\int x^{2}\sqrt{2+x}dx
Relevant Equations
The Substitution Rule,
Table of Indefinite Integrals
When I encountereD this kind of question before.For example
\int x\sqrt{2+x^{2}}dx
We make the Substitution t=x^{2}+2,because its differential is dt=2xdx,so we get \int x\sqrt{2+x^{2}}=1/2\int\sqrt{t}dt,then we can get the answer easily
But the question,it seems that I can't use the way to solve the question.I can't find a correct commutation method.
 
Physics news on Phys.org
  • #2
Try putting ##[tex]## at head and ##[/tex]## at tail of your math code to show it properly.
 
Last edited:
  • #3
Sorry,but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.
 
  • #5
KungPeng Zhou said:
Homework Statement: [itex]\int x^{2}\sqrt{2+x}\,dx[/itex]
Relevant Equations: The Substitution Rule,
Table of Indefinite Integrals

When I encountereD this kind of question before.For example
[tex]\int x\sqrt{2+x^{2}}\,dx[/tex]
We make the Substitution [itex]t=x^{2}+2[/itex],because its differential is [itex]dt=2xdx[/tex],so we get [tex]\int x\sqrt{2+x^{2}}\,dx =\frac12 \int\sqrt{t}\,dt,[/tex] then we can get the answer easily
But the question,it seems that I can't use the way to solve the question.I can't find a correct commutation method.

Did you consider [itex]t = x + 2[/itex], [itex]dx = dt[/itex]?
 
  • #6
KungPeng Zhou said:
Sorry,but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.
1693222343057.png


Example for \frac{\pi}{2}
##\frac{\pi}{2}##
[tex]\frac{\pi}{2}[/tex]

Why don't you try these two ways for Latex in your post ?
 
  • #7
KungPeng Zhou said:
Sorry,but I can't understand you.Could you please tell how to show my math code properly?Now I just can use these math code.
Look at https://www.physicsforums.com/help/latexhelp/.
KungPeng Zhou said:
Homework Statement: ##\int x^{2}\sqrt{2+x}dx##
Relevant Equations: The Substitution Rule,
Table of Indefinite Integrals

When I encountereD this kind of question before.For example
##\int x\sqrt{2+x^{2}}dx##
We make the Substitution ##t=x^{2}+2##,because its differential is ##dt=2xdx##,so we get ##\int x\sqrt{2+x^{2}}=1/2\int\sqrt{t}dt##,then we can get the answer easily
But the question,it seems that I can't use the way to solve the question.I can't find a correct commutation method.
I'm not quite sure what you actually want to know. The LaTeX issue is addressed above and in the link.

The integral works as follows and is explained here:
https://www.physicsforums.com/threads/substitution-in-a-definite-integral.1054611/#post-6919864
 
  • #8
pasmith said:
Did you consider [itex]t = x + 2[/itex], [itex]dx = dt[/itex]?
Ithe seems that we still can't solve it with this way...
 
  • #9
KungPeng Zhou said:
Ithe seems that we still can't solve it with this way...

You can't integrate [itex]\int (t - 2)^2\sqrt{t}\,dt = \int t^{5/2} - 4t^{3/2} + 4t^{1/2}\,dt[/itex]?
 
  • Like
Likes KungPeng Zhou
  • #10
pasmith said:
You can't integrate [itex]\int (t - 2)^2\sqrt{t}\,dt = \int t^{5/2} - 4t^{3/2} + 4t^{1/2}\,dt[/itex]?
Yes,you are right.
 
Last edited by a moderator:

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
863
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
905
  • Calculus and Beyond Homework Help
Replies
3
Views
689
  • Calculus and Beyond Homework Help
Replies
3
Views
689
  • Calculus and Beyond Homework Help
Replies
2
Views
421
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
670
  • Calculus and Beyond Homework Help
Replies
7
Views
883
  • Calculus and Beyond Homework Help
Replies
3
Views
942
Back
Top