Improper Integral Calculator + Online Solver With Free Steps

An improper integral calculator is an online tool specifically built to calculate the integral with given limits. In this calculator, we can enter the function, upper and lower bounds, and then can evaluate the improper integral’s value.

Reversing the differentiation process results in an improper integral. Having a higher limit and a lower limit defines an improper integral. We can determine the region below the curve between the lower and upper limits by using the improper integral.

improper integral calculator

What Is an Improper Integral Calculator?

An Improper Integral sometimes referred to as a definite integral in calculus, is a calculator in which one or both limits approach infinity.

Additionally, at one or more places in the integration range, the integrand also approaches infinity. The normal Riemann Integral can be used to calculate the improper integrals. Improper integrals come in two different varieties. They are:

  • The bounds ‘a’ and ‘b’ are both infinite.
  • In the range [a, b], f(x) has one or more discontinuity points.

How To Use an Improper Integral Calculator?

You can use the Improper Integral Calculator by following the given detailed guidelines, and the calculator will provide you with the results you seek. You can now follow the given instructions to get the value of the variable for the given equation.

Step 1

In the “input function” box, type the function. Additionally, you can load samples to test the calculator. This incredible calculator contains a wide variety of examples of all kinds.

Step 2

From the list of X, Y, and Z variables, select the desired variables.

Step 3

Limits are quite important in this case to define the function precisely. Before calculating, you must add the lower and higher bound limitations.

Step 4

Click on the “SUBMIT” button to determine the series for a given function and also the whole step-by-step solution for the Improper Integral Calculator will be displayed.

Additionally, this tool ascertains whether or not the function converges.

How Does Improper Integral Calculator Work?

Improper Integral Calculator works by integrating the definite integrals with one or both boundaries at infinity $\infty$.  Integral calculations that calculate the area between curves are known as improper integrals. There is an upper limit and a lower limit to this form of integral. An example of a definite integral is an inappropriate integral.

A reversal of differentiation is said to occur in an incorrect integral. One of the most effective ways to solve an improper integral is to subject it to an online improper integral calculator.

Types of Improper Integrals

There are two different sorts of improper integrals, depending on the constraints we apply.

Integration Over an Infinite Domain, Type 1

We characterize improper integrals of type one as infinity when they have upper and lower bounds. We must remember that infinity is a process that never ends and cannot be seen as a number.

Assume we have a function f(x) that is specified for the range [a, $\infty$). Now, if we consider integrating over a finite domain, the limits are as follows:

\[ \int_{a}^{\infty} f\left( x \right)dx = \lim\limits_{n \to \infty } \int\limits_a^n f\left( x \right) dx\]

If the function is specified for the range $ (-\infty,b] $, then the integral is as follows:

\[\int\limits_{ – \infty }^b f\left( x \right)dx = \lim\limits_{n \to – \infty } \int\limits_n^b {f\left( x \right)dx} \]

It should be kept in mind that the improper integral is convergent if the limits are finite and produce a number. But the given integral is divergent if limits are not a number.

If we talk about the case where an incorrect integral has two infinite boundaries. In this case, the integral is broken at a random location that we have chosen. The result is two integrals with one of the two bounds being infinite.

\[\int\limits_{ – \infty }^\infty f\left( x \right)dx = \int\limits_{ – \infty }^c f\left( x \right)dx + \int\limits_c^\infty f\left( x \right)dx .\]

With the use of a free online improper integral calculator, these types of integrals can be quickly evaluated.

Integration Over an Infinite Discontinuity, Type 2

At one or more sites of integration, these integrals have integrands that are not specified.

Let f(x) be a function that is continuous between [a, b) and discontinuous at x = b.

\[\int\limits_a^b f\left( x \right)dx= \lim\limits_{\tau \to 0 + } \int\limits_a^{b – \tau } f\left( x \right)dx \]

Like before, we assume that our function is discontinuous at x = a and continuous between (a, b).

\[\int\limits_a^b f\left( x \right)dx= \lim\limits_{\tau \to 0 + } \int\limits_{a + \tau}^{b } f\left( x \right)dx \]

Now suppose that the function has a discontinuity at x = c and is continuous between $(a, c] \cup (c, b]$.

\[\int\limits_a^b f\left( x \right)dx = \int\limits_a^c f\left( x \right)dx+ \int\limits_c^b f\left( x \right)dx \]

To find the integration, we follow a set of standard procedures and guidelines.

$ \frac{d}{dx} (\frac{x^(n+1)}{n+1}) = X^n $$\int_{}^{} x^n \cdot dx = (\frac{x^(n+1)}{n+1}) + C $
$ \frac{d}{dx} (X)= 1 $$\int_{}^{} dx = X + C $
$ \frac{d}{dx} (\sin X)= \cos X $$\int_{}^{} \cos X dX = \sin X + C $
$ \frac{d}{dx} (-\cos X)= \sin X $$\int_{}^{} \sin X dX = -\cos X + C $
$ \frac{d}{dx} (\tan X)= \sec ^2 X $$\int_{}^{} \sec ^2 X dX = \tan X + C $
$ \frac{d}{dx} (-\cot X)= \csc ^2 X $$\int_{}^{} \ csc ^2 X dX = -\cot X + C $
$ \frac{d}{dx} (-\sec X)= \ sec X \cdot \tan x $$\int_{}^{} \sec X \cdot \tan x dX = \ sec X + C $

Solved Examples

Let’s explore some examples to better understand the working of the Improper Integral Calculator.

Example 1

Calculate \[ \int_{0}^{2}\left( 3 x^{2} + x – 1 \right)dx \]


First, calculate the corresponding indefinite integral:

\[\int{\left(3 x^{2} + x – 1\right)d x}=x^{3} + \frac{x^{2}}{2} – x \](for steps, see indefinite integral calculator)

As it states in the Fundamental Theorem of Calculus, \[\int_a^b F(x) dx=f(b)-f(a)\], so just evaluate the integral at the endpoints, and that’s the answer.

\[\left(x^{3} + \frac{x^{2}}{2} – x\right)|_{\left(x=2\right)}=8 \]

\[\left(x^{3} + \frac{x^{2}}{2} – x\right)|_{\left(x=0\right)}=0 \]

\[\int_{0}^{2}\left( 3 x^{2} + x – 1 \right)dx=\left(x^{3} + \frac{x^{2}}{2} – x\right)|_{\left(x=2\right)}-\left(x^{3} + \frac{x^{2}}{2} – x\right)|_{\left(x=0\right)}=8 \]

Answer: \[\int_{0}^{2}\left( 3 x^{2} + x – 1 \right)dx=8\]

Example 2

Calculate \[ \int_{2}^{-2}\left( 4 x^{3} + x^{2} + x – 1 \right)dx \]


First, calculate the corresponding indefinite integral:

\[\int{\left(4 x^{3} + x^{2} + x – 1\right)d x}=x \left(x^{3} + \frac{x^{2}}{3} + \frac{x}{2} – 1\right)\] (for steps, see indefinite integral calculator)

As it states in the Fundamental Theorem of Calculus, \[\int_a^b F(x) dx=f(b)-f(a)\]

So just evaluate the integral at the endpoints, and that’s the answer.

\[\left(x \left(x^{3} + \frac{x^{2}}{3} + \frac{x}{2} – 1\right)\right)|_{\left(x=-2\right)}=\frac{52}{3}\]

\[\left(x \left(x^{3} + \frac{x^{2}}{3} + \frac{x}{2} – 1\right)\right)|_{\left(x=2\right)}=\frac{56}{3}\]

\[\int_{2}^{-2}\left( 4 x^{3} + x^{2} + x – 1 \right)dx=\left(x \left(x^{3} + \frac{x^{2}}{3} + \frac{x}{2} – 1\right)\right)|_{\left(x=-2\right)}-\left(x \left(x^{3} + \frac{x^{2}}{3} + \frac{x}{2} – 1\right)\right)|_{\left(x=2\right)}=- \frac{4}{3} \]

Answer: \[\int_{2}^{-2}\left( 4 x^{3} + x^{2} + x – 1 \right)dx=- \frac{4}{3}\approx -1.33333333333333 \]

Example 3

Determine the improper integral given these values:

\[\int\limits_{0}^\infty \frac{1}{x} dx\]


Your input is:

\[\int\limits_{0}^{\infty} \frac{1}{x}\, dx\]

First, we will need to determine the definite integral:

\[\int \frac{1}{x}\, dx = \log{\left(x \right)}\]

(for the complete steps, see Integral Calculator section).

\[\left(\log{\left(x \right)}\right)|_{x=0}=- f i n \]

\[\lim_{x \to \infty}\left(\log{\left(x \right)}\right)=\infty \]

\[\int\limits_{0}^{\infty} \frac{1}{x}\, dx = \left(\left(\log{\left(x \right)}\right)|_{x=0} \right) – \left(\lim_{x \to \infty}\left(\log{\left(x \right)}\right(\right) = \infty \]

\[\int\limits_{0}^{\infty} \frac{1}{x}\, dx=\infty \]

Because the value of the integral is not a finite number, the integral is now divergent. Furthermore, the integral convergence calculator is definitely the best option to get more precise results.

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