Should Integration by Parts Be Used on Functions Like \( x \cdot y(x) \)?

  • Thread starter Thread starter Fluidman117
  • Start date Start date
  • Tags Tags
    Integration
Fluidman117
Messages
34
Reaction score
0

Homework Statement



I want to take an antiderivative of a function with respect to x. But in addition the function includes a term y (x) that is a function of x itself. Do I have to apply the reverse power rule also to y(x) also? The integral can be seen as an indefinite.

Homework Equations



A=\int x*y(x)*dx

The Attempt at a Solution



I think I should just apply the reverse power rule to x. So:

A= \frac{x^{2}}{2}*y(x) + C
 
Physics news on Phys.org
Fluidman117 said:

Homework Statement



I want to take an antiderivative of a function with respect to x. But in addition the function includes a term y (x) that is a function of x itself. Do I have to apply the reverse power rule also to y(x) also? The integral can be seen as an indefinite.

Homework Equations



A=\int x*y(x)*dx

The Attempt at a Solution



I think I should just apply the reverse power rule to x. So:

A= \frac{x^{2}}{2}*y(x) + C

You can answer your own question---and that is the best way to learn! Just try it out on some examples. What do you get if you use your formula on the function ##y(x) = x^2##? What happens if you use your formula on ##y(x) = 1/x^2##? Are you getting correct results?
 
Last edited:
What you are looking for is something called integration by parts.
 
Okay, then I believe I was mistaken. It seems that I also need to apply the reverse power rule to ##y(x)##.
So an example:
##y(x)=2*x##

A=\int^{2}_{1} x*y(x)*dx=\int^{2}_{1} x*2*x*dx
A=\left|2\frac{x^{3}}{3}\right|^{2}_{1}=4.6667

Is this correct?

I also looked up the integration by parts and it seems that in the above example it was possible to do without the integration by parts. But can someone give a good example why and when is it necessary to turn to the integration by parts technique?
 
Fluidman117 said:
But can someone give a good example why and when is it necessary to turn to the integration by parts technique?

Try \displaystyle\int_0^\pi x \sin x\,dx or \displaystyle\int_0^1 x e^{-x}\,dx.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top