Definite integral of rational function

In summary, The definite integral of (t^3 + t -1)/(sin(t)) from 2 to x^2 is f'(x), where f(x) = \int_2^{x^2} \frac{(t^3 + t - 1)dt}{sin(t)}. This problem involves the use of the Fundamental Theorem of Calculus and requires you to find f'(x) using the given function f(x).
  • #1
nlsherrill
323
1

Homework Statement



The definite integral of (t^3 + t -1)/(sin(t)) from 2 to x^2

Homework Equations





The Attempt at a Solution



First off, I don't have the solution anywhere, my teacher just gave this to us to work on for the final exam review.

I can think of a few things. I know for the definite integrals you can first simplify the expression as an indefinite integral, then use the fundamental theorem to solve. I know the integral of 1/sin(t)=ln(sint), and integral of (t^3 + t -1)=(t^4/4 + t^2/2 -t) so for the indefinite integral I have..

(t^4/4 + t^2/2 -t)ln(sint). do I then just plug in the upper and lower limits to simplify? I kind of feel like I'm on the wrong track because it looks sloppy and usually he gives us problems that have reasonable looking solutions.

Another attempt is that I recognized 1/sin(t)=csc(t). So then I am left with the integral of (t^3 + t -1)(csc(t))... but I'm not sure if substitution would work.
 
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  • #2
nlsherrill said:

Homework Statement



The definite integral of (t^3 + t -1)/(sin(t)) from 2 to x^2

Homework Equations





The Attempt at a Solution



First off, I don't have the solution anywhere, my teacher just gave this to us to work on for the final exam review.

I can think of a few things. I know for the definite integrals you can first simplify the expression as an indefinite integral, then use the fundamental theorem to solve. I know the integral of 1/sin(t)=ln(sint),
Then you know something that isn't true! You seem to be thinking that
nlsherrill said:
and integral of (t^3 + t -1)=(t^4/4 + t^2/2 -t) so for the indefinite integral I have..

(t^4/4 + t^2/2 -t)ln(sint).
No, that's not right, either.
nlsherrill said:
do I then just plug in the upper and lower limits to simplify? I kind of feel like I'm on the wrong track because it looks sloppy and usually he gives us problems that have reasonable looking solutions.

Another attempt is that I recognized 1/sin(t)=csc(t). So then I am left with the integral of (t^3 + t -1)(csc(t))... but I'm not sure if substitution would work.
 
  • #3
What is the complete problem statement? I have the feeling that you have omitted an important detail.
 
  • #4
Mark44 said:
What is the complete problem statement? I have the feeling that you have omitted an important detail.

I am sorry but all I have put down is all that's on the paper. Is there more information needed to solve it?
 
  • #5
The integral you gave looks to be pretty difficult. I have a suspicion that the problem is something like this:
[tex]f(x) = \int_2^{x^2} \frac{(t^3 + t - 1)dt}{sin(t)}[/tex]
Find f'(x).​
 
  • #6
Mark44 said:
The integral you gave looks to be pretty difficult. I have a suspicion that the problem is something like this:
[tex]f(x) = \int_2^{x^2} \frac{(t^3 + t - 1)dt}{sin(t)}[/tex]
Find f'(x).​

yes that is exactly it. Sorry I don't know how to do the Latex.

This problem is way harder than the others my prof gave us for the final exam review(calculus 1). I guess its possible I wrote it down incorrectly.
 
  • #7
Whether you know how to use LaTeX or not, you need to give the problem statement exactly as given, which you did not do. If you don't provide the given information we are likely to waste a lot of time trying 1) to solve a different problem than the one that was given, and 2) to solve a problem that is much more difficult than it needs to be.

You omitted very important information!

To do this problem you don't need to carry out the integration, but you do need to know about the Fundamental Theorem of Calculus.
 

What is a definite integral of a rational function?

A definite integral of a rational function is the result of integrating a rational function over a specific interval, which results in a single numerical value. It represents the area under the curve of the function within the given bounds.

How is the definite integral of a rational function calculated?

The definite integral of a rational function is calculated by using the fundamental theorem of calculus, which states that the integral of a function can be calculated by finding the antiderivative of that function and evaluating it at the upper and lower bounds of the interval.

What are the steps involved in finding the definite integral of a rational function?

The steps involved in finding the definite integral of a rational function include: 1) finding the antiderivative of the function, 2) plugging in the upper and lower bounds of the interval into the antiderivative, 3) subtracting the lower bound value from the upper bound value, and 4) simplifying the resulting expression.

What is the significance of the definite integral of a rational function?

The definite integral of a rational function has various applications in mathematics and science. It can be used to find the area under a curve, calculate the average value of a function, and solve problems involving accumulation or rates of change.

Are there any special cases or exceptions when finding the definite integral of a rational function?

Yes, there are some special cases and exceptions when finding the definite integral of a rational function. These include cases where the function is undefined at one or both of the bounds, or when the function has a vertical asymptote within the interval. In these cases, the integral may not exist or may require special techniques to evaluate.

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