Solve Integration by Parts: Help with First Steps

In summary, the student has tried a question multiple times without success and is requesting help with the first couple of lines. The question involves finding the integral of 2cte^{-ct^2}dt where c is a positive constant. The student attempted to use integration by parts, but was advised to use a simple u-substitution instead. The student expresses gratitude for the assistance.
  • #1
Millhawk
3
0

Homework Statement


I have tried this question a number of times, but to no avail. Could somebody please help with the first couple of lines, thank you.

[tex]\int_0^x2cte^{-ct^2}dt[/tex] where [tex]c[/tex] is a postive constant


Homework Equations





The Attempt at a Solution


Let u=t and let dv=[tex]2ce^{-ct^2}dt[/tex]
 
Last edited:
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  • #2
Millhawk said:

Homework Statement


I have tried this question a number of times, but to no avail. Could somebody please help with the first couple of lines, thank you.

[tex]\int_0^x2cte^{-ct^2}dt[/tex] where [tex]c[/tex] is a postive constant

Homework Equations


The Attempt at a Solution


Let u=t and let dv=[tex]2ce^{-ct^2}dt[/tex]

just change variables to [itex]x=ct^2[/itex]. you don't need "integration by parts".
 
  • #3

Homework Statement



The above was edited wrong.

I have tried this question a number of times, but to no avail. Could somebody please help with the first couple of lines, thank you.

[tex]\int_0^x2cte^{-ct^2}dt[/tex] where [tex]c[/tex] is a postive constant


Homework Equations





The Attempt at a Solution


Let u=t and let dv=[tex]2ce^{-ct^2}dt[/tex]
 
  • #4
ditto
 
  • #5
Take olgranpappy's advice, all you need is a simple u-sub.
 
  • #6
Thank you, it's amazing how a simple question can transform itself into Electromagnetic Field Theory with such ease when a tired human brain is applied. Thank you again.
 
  • #7
Happens to me all the time, no worries :-]
 

What is integration by parts?

Integration by parts is a method used to solve integrals that involve products of functions. It involves splitting the integral into two parts and using a formula to rewrite it in a different form.

When should I use integration by parts?

Integration by parts is useful when the integral involves a product of functions that cannot be easily integrated using other methods, such as substitution or trigonometric identities.

How do I choose which function to differentiate and which function to integrate?

A common method is to use the acronym "LIATE" which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. In general, it is best to choose the function that becomes simpler when differentiated.

What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du. This means that the integral of the product of two functions u and dv is equal to the product of the first function u and the integral of the second function dv, minus the integral of the product of the first function u and the derivative of the second function du.

Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for definite integrals. After applying the formula, you can use the limits of integration to evaluate the integral.

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