Solving Improper Integrals: Evaluate 4/(x-3)^3 From 1 to 8

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In summary: This technique will work for evaluating the integral over the interval [0,1], but it will not work for evaluating the integral over the interval [0,8].You are right that 3 is in the interval [1,8] and that the function has a discontinuity at x=3. Do you know of a technique that can get the point 3 in the limits of integration?There is no surefire way to get the point 3 in the limits of integration, but you can try some methods. One option is to use the derivative to find points of discontinuity in the function, and then use a calculus method
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lha08
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Homework Statement


Determine whether the integral is divergent or convergent. If it is convergent, evaluate it.
Integral from (1 to 8) 4/(x-3)^3 dx


Homework Equations





The Attempt at a Solution


So firstly, i pulled out the 4 and then i set the lim as t goes to 8 from the left of the integral. From there, i applied the u sub with u=x-3 and du=dx. I then did the antiderivative of it which gives me: 4 lim (as t goes to 8 from left) integral from 1 to t (1/(-2u^2)) and then i changed the limits of integration...t=t-3 and 1=-2. but when i plugged them in and evaluated i got 0.42..which is wrong...i don't know where I'm going wrong. thanks.
 
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  • #2
Are you sure it's convergent? What happens at x=3?
 
  • #3
Dick said:
Are you sure it's convergent? What happens at x=3?

ohh the bottom heads the zero so at x=3 it does not exist, right? so it diverges..
 
  • #4
In this case, yes, it diverges. But just having a zero denominator doesn't prove that. You are dealing with improper integrals. 1/sqrt(x) has a finite integral on [0,1]. Why is your case different?
 
  • #5
Dick said:
In this case, yes, it diverges. But just having a zero denominator doesn't prove that. You are dealing with improper integrals. 1/sqrt(x) has a finite integral on [0,1]. Why is your case different?

Um I'm not really sure but the first thought that comes to mind is that 4/0 does not exist and that 3 is found in the limits of integration..but I'm really not sure...
 
  • #6
You are doing 'improper integrals', right? To figure out whether 1/(x-3)^3 has an integral on [0,3], don't you do the integral from 0 to 3-e where e is some small positive number and then take the limit as e tends to zero?
 
  • #7
lha08 said:
Um I'm not really sure but the first thought that comes to mind is that 4/0 does not exist and that 3 is found in the limits of integration..but I'm really not sure...

You are right that 3 is in the interval [1,8] and that the function has a discontinuity at x=3. Do you know of a technique that can get the point 3 in the limits of integration?

Note that in the case of [itex]1/\sqrt{x}[/itex], integrated over the interval [0,1], the discontinuity happens at an endpoint of the interval. Then the integral can be evaluated by:
[tex]\int_0^1 \frac{1}{\sqrt{x}} \,dx = \lim_{t\to 0^+} \int_t^1 \frac{1}{\sqrt{x}} \,dx[/tex]
 

FAQ: Solving Improper Integrals: Evaluate 4/(x-3)^3 From 1 to 8

What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite or the integrand has a vertical asymptote within the integration interval. This means that the integral cannot be evaluated using the basic rules of integration and requires a different approach to solve.

How do you solve an improper integral?

To solve an improper integral, you must first identify which type of improper integral you are dealing with: type 1, where the upper or lower limit of integration is infinite, or type 2, where the integrand has a vertical asymptote. Then, you can use appropriate techniques, such as substitution, partial fractions, or comparison to a known convergent integral, to evaluate the integral.

What is the strategy for solving this specific improper integral?

The strategy for solving this specific improper integral, 4/(x-3)^3 from 1 to 8, is to first rewrite the integrand as a power function, 4x^-3, and then evaluate the integral using the power rule. This will result in a new limit of integration at infinity, which can then be evaluated using the limit definition of the integral.

Can you explain the limit definition of an integral?

The limit definition of an integral is a mathematical expression that represents the limit of a Riemann sum as the number of subintervals approaches infinity. It is used to evaluate improper integrals where the upper or lower limit of integration is infinite. The limit definition involves taking the limit of a function as the upper or lower limit of integration approaches infinity, and then evaluating the integral using the basic rules of integration.

What is the significance of evaluating improper integrals?

Evaluating improper integrals is important in many areas of science and mathematics, such as physics, engineering, and statistics. It allows us to calculate important quantities, such as areas under curves, volumes of solids, and probabilities, that cannot be found using basic integration techniques. Improper integrals also play a crucial role in understanding and solving differential equations, which are used to model many natural phenomena.

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