What is the Integral Solution for a Calculus Problem with Two Variables?

In summary, the conversation is discussing a problem with an expected answer that contains both n and m, but the solution provided only contains n. The expert summarizer suggests starting with the original equation and integrating it to find the correct solution.
  • #1
eas123
9
0

Homework Statement



See attached.

Homework Equations




The Attempt at a Solution



I integrated the equation with respect to x to obtain
∫[itex]\frac{d}{dx}[/itex](x[itex]e^{-x}[/itex][itex]\frac{df}{dx}[/itex])dx+∫n[itex]e^{-x}[/itex]fdx= constant
The first term on the left hand side goes to zero as x, df/dx are bounded at 0, infinity. This leaves the expression ∫n[itex]e^{-x}[/itex]fdx= constant which is not the one given.
 

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  • #2
hi eas123! :smile:

what happened to m ? :confused:
 
  • #3
Hi. :-)

What do you mean? I don't know how to derive the expression.
 
  • #4
your expected answer, ##\int_0^{\infty} e^{-x}f_n(x)f_m(x) dx = 0##, has an "m" in it

i don't see an "m" in your actual work
 
  • #5
So where have I gone wrong?
 
  • #6
eas123 said:
So where have I gone wrong?

i'm completely confused :redface:

you seem to be solving a different problem :confused:

start with ##e^{-x}f_n(x)f_m(x) dx = 0##, and integrate it :smile:
 
  • #7
eas123 said:

Homework Statement



See attached.

Homework Equations




The Attempt at a Solution



I integrated the equation with respect to x to obtain
∫[itex]\frac{d}{dx}[/itex](x[itex]e^{-x}[/itex][itex]\frac{df}{dx}[/itex])dx+∫n[itex]e^{-x}[/itex]fdx= constant
The first term on the left hand side goes to zero as x, df/dx are bounded at 0, infinity. This leaves the expression ∫n[itex]e^{-x}[/itex]fdx= constant which is not the one given.
The following is an image of the attachment. Please notice that the result you are to prove contains both n and m . That's what tiny-tim is telling you.

attachment.php?attachmentid=58464&d=1367661781.jpg
 

FAQ: What is the Integral Solution for a Calculus Problem with Two Variables?

What is calculus and why is it important?

Calculus is a branch of mathematics that deals with rates of change and accumulation. It is important because it provides a powerful tool for solving complex problems in fields such as physics, engineering, economics, and statistics.

What are the basic concepts in calculus?

The basic concepts in calculus are derivatives and integrals. Derivatives measure the rate of change of a function, while integrals measure the accumulation of a function over a given interval.

How do you solve a basic calculus problem?

To solve a basic calculus problem, you need to identify the function and the variable you are working with, apply the necessary calculus rules (such as the power rule or the chain rule), and simplify the expression to find the solution.

What are the common applications of calculus?

Calculus has many applications in real-world scenarios, such as calculating the speed and acceleration of moving objects, finding the optimal solution in optimization problems, and determining the growth and decay of populations or resources.

What are the common mistakes to avoid in basic calculus problems?

Some common mistakes to avoid in basic calculus problems include incorrect application of rules, forgetting to simplify the expression, and not checking for extraneous solutions. It is also important to carefully label units and pay attention to the domain and range of the function.

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