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Improper Integral Calculator
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Improper integral is definite integral which has its limits as infinity or the end point approaches the limits.
The Integral $\int^{a}_{b}$ f(x) dx is called an improper integral if either a = $\infty$, b = - $\infty$ or f(x) is not defined at point [a,b].

Improper Integral Calculator integrates the given improper integral, simplifies the value and finds whether the given is convergent or divergent.

A function with its limits are given below as a default input. Integrate the given function with the given limits and evaluate to get the value of improper integral, when on clicking "Integrate".
 

Steps for Improper Integral Calculator

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Step 1 :  

Read the given integral and find out whether the given integral is improper by seeing the limits, if so carry on with the below steps.



Step 2 :  

Integrate the given integral and apply the limits to get the answer.



Problems on Improper Integral Calculator

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  1. Examine the type of integral and evaluate when the function f(x)= $\frac{1}{x^{2}}$ is integrated in an interval [+3, $\infty$].


    Step 1 :  

    The given function is f(x) = $\frac{1}{x^{2}}$ and the given limits are [3, $\infty$].



    Step 2 :  

    $\int_{+3}^{\infty}$ $\frac{1}{x^{2}}$ dx =lim t-> $\infty$ $\int_{+3}^{\infty}$$\frac{1}{x^{2}}$ dt
                                         = lim t->$\infty$  - [$\frac{1}{x}$]t 3


                                         = lim t->$\infty$ - [ $\frac{1}{t}$ - $\frac{1}{3}$ ]


                                         = $\frac{1}{3}$



    Answer  :  

    $\int_{+3}^{\infty}$ $\frac{1}{x^{2}}$ dx = $\frac{1}{3}$.



  2. Examine the type of integral and evaluate when the function f(x)= $\frac{1}{x^{3}}$ in an interval [2,$\infty$]


    Step 1 :  

    The given integral is $\frac{1}{x^{3}}$ and limits are [2, $\infty$].



    Step 2 :  

    $\int_{+2}^{\infty}$ $\frac{1}{x^{3}}$ dx =lim 2 ->$\infty$ - $\frac{1}{2x^{2}}$

                                       = lim t -> $\infty$ - [$\frac{1}{2t^{2}}$ - $\frac{1}{8}$]


                                       = $\frac{1}{8}$



    Answer  :  

    $\int_{+2}^{\infty}$ $\frac{1}{x^{3}}$ dx = $\frac{1}{8}$ $\approx$ 0.125.



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