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## Improper integral calculator

Improper integral calculator is used to measure the definite integrals with defined limits. This convergent or divergent integral calculator can measure the convergence or divergence of the function.

Our integral convergence calculator finds the area under the curve from the lower limit to the upper limit.

## How does this improper integral calculator work?

Follow the below steps to measure the convergence or divergence of the function.

• Input the improper function.
• Use the keypad icon to enter mathematics keys.
• Write the upper limit and the lower limit. Write “inf” for infinity and “pi” for π.
• Choose the variable.
• Press the calculate key to get the result.
• Click the show more button to view the solution with steps.
• To calculate a new improper function hit the reset button.

## What are improper integrals?

Improper integrals are definite integrals where one or both limit values are at infinity, or where the integrand has a vertical asymptote in the vertical of integration.

Improper integral is used to measure the function value whether it exists or not at the defined limits. The improper integral convergence test can be used to check if the function is convergent or divergent.

### Types of improper integral

There are three ways to write an improper integral equation.

1. Form a point to infinity [b, ∞)

$\int _b^{\infty }\:f\left(x\right)dx=\lim _{a\to \infty }\left(\int _b^{a\:}\:f\left(x\right)dx\right)$

1. From negative infinity to a point (-∞, b]

$\int _{-\infty \:}^b\:f\left(x\right)dx=\lim _{a\to -\infty }\left(\int _a^{b\:}\:f\left(x\right)dx\right)$

1. From negative infinity to infinity (-∞, ∞)

$\int _{-\infty \:}^∞\:f\left(x\right)dx=\int _{-∞}^{t\:}\:f\left(x\right)dx+\int _t^{∞\:}\:f\left(x\right)dx$

## How to evaluate improper integral problems?

Following is an example of an improper integral solved by our definite and improper integral calculator.

Example

Evaluate the improper integral of $\frac{1}{\:x^2}$ from 1 to infinity.

Solution

Step 1: Write the improper integral along with the upper and lower limits.

$\int _1^{\infty }\frac{1}{\:x^2}dx$

Step 2: Take the general equation to solve the above expression.

$\int _b^{\infty }\:f\left(x\right)dx=\lim _{a\to \infty }\left(\int _b^{a\:}\:f\left(x\right)dx\right)$

Step 3: Now write the given expression according to the above equation.

$\:\int _1^{\infty \:}\frac{1}{x^2}dx=\lim \:_{a\to \:\infty \:}\left(\int _1^{a\:}\:\frac{1}{x^2}dx\right)$

Step 4: Now integrate the above expression.

$\int _1^{\infty \:}\frac{1}{x^2}dx=\lim \:_{a\to \:\infty \:}\left(\int _1^{a\:}\:x^{-2}dx\right)$

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=\lim \:\:_{a\to \:\:\infty \:\:}\left[\frac{\:x^{-2+1}}{-2+1}\right]^a_1$

$\int _1^{\infty \:\:}\frac{1}{x^2}dx=\lim \:\:_{a\to \:\:\infty \:\:}\left[\frac{\:x^{-1}}{-1}\right]^a_1$

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=\lim _{a\to \infty }\left[-\frac{1}{x}\right]^a_1$

By using fundamental theorem of calculus.

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=\lim _{a\to \infty }\left[-\frac{1}{a}-\left(-\frac{1}{1}\right)\right]$

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=\lim _{a\to \infty }\left[-\frac{1}{a}+1\right]$

Step 5: Now apply the limit value.

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=\left[-\frac{1}{\infty }+1\right]$

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=-0+1$

$\:\int _1^{\infty \:\:}\frac{1}{x^2}dx=1$

### References

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