Namely, integrals form I and form II, respectively. Generally, both varieties are fixed in the same means using limits.

You are watching: Type 1 and type 2 improper integrals

But consider the following integral:

$\int_0^\infty \frac1\sqrt<3>x dx$

In this case, ns would continue to set the integral $\int_0^t \frac1\sqrt<3>x dx$ to $\displaystyle\lim_t\to \infty$. However, $\frac1\sqrt<3>x$ is discontinuous in ~ $x=0$.

Why is this disregarded during integration??

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inquiry Apr 23 "18 at 20:23

rainierrainier
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## 1 prize 1

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i think in that situation you division the integral in two since it have two cases of wrong integral, so you make

$$\int_0^+\infty\fracdx\sqrt<3>x=\int_0^1\fracdx\sqrt<3>x+\int_1^+\infty\fracdx\sqrt<3>x$$so you examine both part separated and in order to the initial integral converge both integrals must converge. So it will be other like

$$\int_0^1\fracdx\sqrt<3>x=\lim_t\to0^+\int_t^1\fracdx\sqrt<3>x$$

and

$$\int_1^\infty\fracdx\sqrt<3>x=\lim_t\to\infty\int_1^t\fracdx\sqrt<3>x$$

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answer Apr 23 "18 in ~ 21:43

candcand
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