Solving improper integrals explained simply

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The differential and integral calculus is part of the mathematics lesson in the upper level at the grammar school. As a student, you will sooner or later come across so-called improper integrals that differ from the Differentiate between "ordinary" integrals, but not much more difficult to solve with the right tools are.

What are improper integrals?

Improper integrals are integrals which at first sight do not have to differ from ordinary integrals. The best way to visualize improper integrals is to make a sketch. If you integrate any function, the integral corresponds to the area under the curve. But what if the function tends to reach infinity at an integration limit?

  • The same difficulty arises when the function under consideration has a horizontal or vertical asymptote.
  • At first you might not notice the problem, but start doing the integral like used to solve, then you will notice at the latest when the limits set in that you are not get ahead.
  • For example, consider Euler's function f (x) = e x and try to integrate these from minus infinity to zero. If you do this and put in the boundaries, you get the term "e0-e-∞ ", but what does this expression mean to you?

Solving improper integrals

  1. You can solve improper integrals very easily if you replace the "problematic" integration limit with a variable that Solve integral and then carry out a limit value analysis in which you run the variable against the original "problem value" permit.
    2. Integral dx - this is how you solve the task

    Even clever math people can get confused: integral signs and ...

  2. In the example above you solve the integral ex dx with the integration limits u and 0. The antiderivative of f (x) = ex is F (x) = ex, because we have F '(x) = f (x).
  3. If you now insert the integration limits, you will get the term e0-eu = 1-eu.
  4. Now form the limit value for u -> -∞. You get limu 1-eu = 1.

Another example of improper integrals

  1. The function g (x) = 1 / x2 should be integrated on the interval 0 to 1. You know that the function g has a pole at the point x = 0.
  2. First you determine the antiderivative of the function g with G (x) = -1 / x.
  3. For the lower integration limit, first substitute v for 0, which gives you for the area A = -1 - (- 1 / v).
  4. Now consider the limit value (limv-> 0) for v against 0. For v towards 0, 1 / v tends towards + ∞ and since there are two minus signs in front of the expression, the area A consequently tends towards infinity.

You see, solving improper integrals is not that difficult at all. You just need to know where to start.

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