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Improper Integral Calculator

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$


$\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.


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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.


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Below are some of the problem based on improper integral.

Solved Examples

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

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 + $\frac{1}{2}$


Therefore, $\int_{2}^{\infty}\frac{1}{x^{2}}$ $dx$ = $\frac{1}{2}$

Question 2: Evaluate $\int_{-\infty}^{0}e^{x}$ dx.
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)


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