What am I doing wrong with this simple integral?

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In summary, the conversation discusses integrating x^-2 between -1 to 1 and the resulting answer of -2. However, the area is actually infinite between the given bounds. The person is confused about the negative answer and questions the validity of the integration. The other person explains that an infinite jump in a function can correspond to a finite jump in its integral. They also mention that in certain cases, a function may "blow up" but still have a finite integral.
  • #1
Physics_wiz
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I want to integrat x^-2 between -1 to 1. The answer shouldn't exist (should probably be something divided by 0) because the area is infinite between -1 to 1. But I integrate, and I get -x^-1 from -1 to 1. I then plug the bounds in and I get: [-(1)^-1] - [-(-1)^-1] which is -1 -1 or -2. What am I doing wrong?
 
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  • #2
Physics_wiz said:
because the area is infinite between -1 to 1

Not true. You're right with your answer. One thing you will learn with integrals is that an infinite jump at any particular point in a function f(x) may correspond to a finite jump in its integral.
 
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  • #3
vsage said:
Not true. You're right with your answer. One thing you will learn with integrals is that an infinite jump at any particular point in a function f(x) may correspond to a finite jump in its integral.

Why is the answer negative? I attached a picture of the graph, it still looks like it should be infinite to me.
 

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  • #4
Ok my apologies. You were correct in your arithmetic, but your answer is complete garbage. As far as I know you cannnot integrate over that interval. There are however cases where a function "blows up" but has a finite integral, so you must be careful (the only case I know of however is the dirac delta function, whose integral is 1 over infinite limits)
 
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FAQ: What am I doing wrong with this simple integral?

How do I know if I am using the correct method for solving this integral?

When solving an integral, it is important to first identify the type of integral you are dealing with. This will determine which method you should use. Some common methods include substitution, integration by parts, and partial fractions. Make sure to carefully analyze the integral before choosing a method to ensure it is the most appropriate one.

Why does my answer to the integral not match the solution provided?

There are a few reasons why your answer may not match the provided solution. First, check for any mistakes in your calculations. It is easy to make a small error when dealing with complex integrals. Additionally, it is possible that the solution provided is simplified or written in a different form. Make sure to double check your work and consider alternate forms of the solution.

Can I use a calculator to solve this integral?

While calculators can be helpful in solving integrals, it is important to understand the steps involved in solving the integral by hand. Calculators may not always provide the most accurate or simplified answer, so it is important to use them as a tool in conjunction with your own understanding of the problem.

How do I know if I need to use a trigonometric substitution for this integral?

Trigonometric substitutions are necessary when the integrand contains expressions involving square roots of the form a2 - x2, a2 + x2, or x2 - a2. If the integral does not contain any of these expressions, a trigonometric substitution is not needed.

How can I check if my answer to the integral is correct?

If you are unsure if your answer to the integral is correct, you can always check by taking the derivative of your solution. If the derivative matches the original integrand, then your answer is correct. Additionally, you can use online integral calculators or ask a colleague to check your work. It is always important to double check your solutions to avoid any mistakes.

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