An integral by parts problem - please advise

In summary, the function is increasing and has a inverse f^-1, which can be shown by using integration by parts. If y = f(x), then the following is true: \int _{a ^b} f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy.
  • #1
insane0hflex
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Homework Statement



The function is increasing and has a inverse f^-1
Also assume f′is continuous and f'(x) > 0 over the state interval of integration [a,b]

PLEASE NOTE! a is lower limit, b is upper limit (same for alpha and beta symbol later on)

Used integration by parts to show that:
[itex]\int f(x)dx=bf(b)−af(a)−∫ba xf′(x)dx[/itex]

Then if y = f(x), then the following is true (used the substitution rule)

[itex] \int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy [/itex]

Homework Equations



The question: Now, I need to show that if α = f(a), and β = f(b), then

[itex]∫βαf^-1(x)dx=β f^-1(β)−α f^-1(α)−∫f^-1(β)f^-1(α)f(x)dx[/itex]

The Attempt at a Solution



trying to fix the problem's display. B should be the upper limit, a should be lower limit when next to an integral sign

Im lost. I'm sure its something relatively easy to do, maybe another substitution or relevant manipulation?

Help appreciated!
 
Last edited:
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  • #2
insane0hflex said:

Homework Statement



The function is increasing and has a inverse f^-1
Also assume f′is continuous and f'(x) > 0 over the state interval of integration [a,b]

PLEASE NOTE! a is lower limit, b is upper limit (same for alpha and beta symbol later on)

Used integration by parts to show that:
[itex]\int f(x)dx=bf(b)−af(a)−∫ba xf′(x)dx[/itex]

It isn't true. Try f(x) = x, a = 1, b = 2.
 
  • #3
insane0hflex said:

Homework Statement



The function is increasing and has a inverse f^-1
Also assume f′is continuous and f'(x) > 0 over the state interval of integration [a,b]

PLEASE NOTE! a is lower limit, b is upper limit (same for alpha and beta symbol later on)

Used integration by parts to show that:
[itex]\int f(x)dx=bf(b)−af(a)−\int_{\,b}^{\,a} xf'(x)dx[/itex]

Then if y = f(x), then the following is true (used the substitution rule)

[itex] \int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy [/itex]

Homework Equations



The question: Now, I need to show that if α = f(a), and β = f(b), then

[itex]\displaystyle \int_β^αf^{-1}(x)dx=β f^{-1}(β)−α f^-1(α)−\int_{f^{-1}(β)}^{f^{-1}(α)}f(x)dx[/itex]

The Attempt at a Solution



trying to fix the problem's display. B should be the upper limit, a should be lower limit when next to an integral sign

I'm lost. I'm sure its something relatively easy to do, maybe another substitution or relevant manipulation?

Help appreciated!
You have a few (apparent) typos, which I attempted to fix.

You have worked out most of the result.

(To view the LaTeX code, right click the desired expression and choose "Show Math As: TEX Commands".)
 

FAQ: An integral by parts problem - please advise

What is an integral by parts problem?

An integral by parts problem is a type of integration problem in calculus where the integral of a product of two functions is solved by using a specific formula known as the integration by parts formula.

How do I know when to use integration by parts?

You can use integration by parts when the integral of a product of two functions cannot be solved using basic integration techniques such as substitution or trigonometric identities.

What is the integration by parts formula?

The integration by parts formula is ∫udv = uv - ∫vdu, where u and v are the two functions being integrated and dv and du are their respective differentials. This formula helps to simplify the integral and make it easier to solve.

Can you provide an example of solving an integral by parts problem?

Sure, let's say we want to solve the integral ∫xsin(x)dx. We can use the integration by parts formula by letting u = x and dv = sin(x)dx. This gives us du = dx and v = -cos(x). Substituting these into the formula, we get ∫xsin(x)dx = -xcos(x) - ∫-cos(x)dx. We can then integrate the second term using basic integration techniques, giving us the final answer of -xcos(x) + sin(x) + C.

Are there any tips for solving integration by parts problems?

Yes, a helpful tip is to choose u and dv in a way that makes the integral simpler to solve. This can often be done by choosing u to be the more complicated function and dv to be the simpler function. It may also be helpful to use the acronym "LIATE" to remember the order in which to choose u and dv: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential.

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