Calculatrice d’intégrales définies et impropres

La calculatrice essaiera d'évaluer l'intégrale définie (c.-à-d. avec bornes), y compris les intégrales impropres, en affichant les étapes.

Enter a function:

Integrate with respect to:

Enter a lower limit:

If you need `-oo`, type -inf.

Enter an upper limit:

If you need `oo`, type inf.

Please write without any differentials such as `dx`, `dy` etc.

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Solution

Your input: calculate $$$\int_{0}^{\infty}\left( \frac{1}{x^{2}} \right)dx$$$

First, calculate the corresponding indefinite integral: $$$\int{\frac{1}{x^{2}} d x}=- \frac{1}{x}$$$ (for steps, see indefinite integral calculator)

Since there is infinity in the upper bound, this is improper integral of type 1.

The interval of integration contains the point $$$0$$$, which is not in the domain of the integrand, so this is an improper integral of type 2.

To evaluate an integral over an interval, we use the Fundamental Theorem of Calculus. However, we need to use limit if an endpoint of the interval is special (infinite or is not in the domain of the function).

$$$\int_{0}^{\infty}\left( \frac{1}{x^{2}} \right)dx=\lim_{x \to \infty}\left(- \frac{1}{x}\right)-\lim_{x \to 0}\left(- \frac{1}{x}\right)=\infty$$$

Since the value of the integral is not finite, then it is divergent.

Answer: the integral is divergent.

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Calculatrice d’intégrales définies et impropres

Calculez des intégrales définies et impropres étape par étape

Solution