Finding Integral of a Divided Function (Hyperbola?)

Hope that helps!In summary, the conversation was about how to integrate the given expression and the attempts at solving it using different methods. The experts provided a hint to divide the numerator by the denominator, and also suggested using rationalizing of the denominator. The conversation ended with a thank you message from the person seeking help.
  • #1
Mathpower
28
0

Homework Statement


Integrate between y=0 and y=-0.5:
∫((y+0.5)/(0.5-y)) dy

Homework Equations


Can you please show me how to integrate it...Then I will be able to take it from there and substitute in the appropriate values.


The Attempt at a Solution


Quotient rule in reverse?...wouldn't work
Using ln ...wouldn't work
Try to factorise...nope
Use rationalising of denominator (conjugate idea)...that just makes it even more complicated
Now I have run out of ideas ...this is out of my scope. (Scholarship paper NZ)

Thank you in advance,
Kindest Regards.
 
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  • #2
Mathpower said:

Homework Statement


Integrate between y=0 and y=-0.5:
∫((y+0.5)/(0.5-y)) dy

Homework Equations


Can you please show me how to integrate it...Then I will be able to take it from there and substitute in the appropriate values.


The Attempt at a Solution


Quotient rule in reverse?...wouldn't work
Using ln ...wouldn't work
Try to factorise...nope
Use rationalising of denominator (conjugate idea)...that just makes it even more complicated
Now I have run out of ideas ...this is out of my scope. (Scholarship paper NZ)

Thank you in advance,
Kindest Regards.

Hint: Divide the numerator by the denominator to make the degree of the numerator one less than the degree of the denominator.
 
  • #3
Hmmm...Thanks for the hint, but I don't get what you meant here...
Isn't the numerator already being divided by the denominator and aren't they to the same degree anyway?
 
  • #4
Write it as$$
-\frac{y + \frac 1 2}{y -\frac 1 2}$$
and do one step of a long division writing it as quotient + remainder/divisor.

Alternatively you could write it as$$
-\frac {(y-\frac 1 2)+1}{(y-\frac 1 2)}$$and break it into two terms. It's the same result either way.
 
  • #5
Mathpower said:
Hmmm...Thanks for the hint, but I don't get what you meant here...
Isn't the numerator already being divided by the denominator and aren't they to the same degree anyway?
Yes, they're the same degree, so use some sort of division algorithm to find a quotient & remainder.
 
  • #6
Nice! I like the alternative (because its the only thing I understood) from LCKurtz...rationalising the denominator.
The rest of it didn't make much sense...how can you integrate with a remainder! (I must be missing something)
Thank you so much for all your help.

P.S.: Sorry for not being able to understand what you guys said.

Kindest Regards
 
  • #7
Mathpower said:
Nice! I like the alternative (because its the only thing I understood) from LCKurtz...rationalising the denominator.
The rest of it didn't make much sense...how can you integrate with a remainder! (I must be missing something)
Thank you so much for all your help.

P.S.: Sorry for not being able to understand what you guys said.

Kindest Regards
Here's how that remainder works:

Write the remainder of 1 as a fraction.

[itex]\displaystyle \frac{1}{y-\frac{1}{2}}[/itex]

So that [itex]\displaystyle -\frac {(y+\frac 1 2)}{(y-\frac 1 2)}=-\left(1+\frac{1}{y-\frac{1}{2}}\right)[/itex]

Although expressions like the one here, I often use the same method as LCKurtz showed you. For more complicated situations, you should know how to express the remainder as a fraction.
 

1. What is the process for finding the integral of a divided function?

The process for finding the integral of a divided function, also known as the antiderivative, is to use the power rule and u-substitution. First, divide the function into its separate parts and then use the power rule to find the antiderivative of each part. Finally, use u-substitution to combine the parts back together and find the overall antiderivative.

2. Can you explain the power rule and how it is used in finding the antiderivative?

The power rule states that the integral of x^n is equal to (x^(n+1))/(n+1). This rule is used to find the antiderivative of each individual part of a divided function. For example, if the function is (x^2)/(x+1), the antiderivative of x^2 is (x^(2+1))/(2+1) or (x^3)/3.

3. What is u-substitution and when is it used in finding the antiderivative?

U-substitution is a method used to combine the individual antiderivatives of a divided function back together. It is used when the function contains a term that is a function of x, such as (x+1)^2. In this case, u-substitution involves substituting u for the term (x+1) and then using the chain rule to find the antiderivative.

4. Are there any other methods for finding the antiderivative of a divided function?

Yes, there are other methods such as integration by parts and partial fractions. However, these methods can be more complex and may not always be necessary for finding the antiderivative. The power rule and u-substitution are usually sufficient for solving most divided functions.

5. Can the antiderivative of a divided function be checked using differentiation?

Yes, the antiderivative of a divided function can be checked by differentiating the result. If the resulting derivative is equal to the original function, then the antiderivative is correct. This is a useful way to double-check your work and ensure that the antiderivative was calculated correctly.

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