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# Improper Integral Calculator

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A definite integral where either the limit of integration is infinite i.e., $\infty$ or -$\infty$ or if there is a discontinuity in the limit. If $\int_{b}^{a}f(x)$ is an integral, then the improper integral is of the form
$\lim_{a \to \infty }$$\int_{a}^{b}$$f(x) dx$ or $\lim_{b \to -\infty }$$\int_{a}^{b}$$f(x) dx$

or

$\lim_{b \to c^{-} }$$\int_{a}^{b}$$f(x) dx$ or $\lim_{b \to a^{+} }$$\int_{b}^{c}$$f(x) dx$

Plug in the function with their limits into the improper integral calculator to get the value of improper integral.

## Steps

Step 1 : Note down the given function with their limits.

Step 2 : The given integral is integrated and their limits are applied to get the value of the given integral.

## Problems

Below are some of the problem based on improper integral.

### Solved Examples

Question 1: Evaluate $\int_{2}^{\infty}\frac{1}{x^{2}}$ $dx$.

Solution:

Step 1 : Given : f(x) = $\frac{1}{x^{2}}$ with their limits [2, $\infty$].

Step 2 : $\int_{2}^{\infty}\frac{1}{x^{2}}$ $dx$ = $\lim_{u \to \infty}$$\int_{2}^{u}\frac{1}{x^{2}} du =\lim_{u \to \infty }$$\frac{-1}{x}]_2^u$

=$\lim_{u \to \infty }$$(\frac{-1}{u} + \frac{1}{2}) = 0 + \frac{1}{2} =\frac{1}{2} Therefore, \int_{2}^{\infty}\frac{1}{x^{2}} dx = \frac{1}{2} Question 2: Evaluate \int_{-\infty}^{0}e^{x} dx. Solution: Step 1 : Given : f(x) = e^{x} with their limits [-\infty, 0]. Step 2 : \int_{-\infty}^{0}e^{x} dx = \lim_{u \to -\infty }$$\int_{0}^{u}e^{x}$ $du$

=$\lim_{u \to -\infty }$$e^{x}]_u^0 =\lim_{u \to -\infty }$$(e^{0} - e^{u})$

=(1 - 0)

=1

Therefore, $\int_{-\infty}^{0}e^{x}$ $dx$ = 1