Comparing 2 Kinds of Integrals - Help Understanding Algebraically

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SUMMARY

The discussion clarifies the distinction between the definite integral \(\int_{0}^b x^2 dx\) and the improper integral \(\lim_{b\rightarrow \infty} \int_{0}^b x^2 dx\). The first integral calculates the area under the curve of the function \(y=x^2\) from \(x=0\) to \(x=b\), while the second integral represents the area under the curve as \(b\) approaches infinity, indicating that the function is unbounded. The convergence of the improper integral determines whether the area under the curve from \(0\) to \(\infty\) is finite.

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balaaditya
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Hi all,

I need help.

What is the difference between

[tex] \int_{0}^b x^2 dx[/tex]

with

[tex] lim_{b\rightarrow \ \infty} \int_{0}^b x^2 dx[/tex]

?

Can someone please show me algebraically for its clarity?

I don’t understand of the 2nd integral equation means.
Especially the appears of limit as b approach to infinity.

Thanks in advance
 
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one is a proper integral and the other one is an improper integral
 
The first integral is a regular definite integral. It basically just measures the area that the parabola [tex]y=x^2[/tex] encloses with the x-axis from the point x=0 to another point x=b.


The second one is an improper integral. THati is:

[tex]\int_{0}^{\infty}x^2dx=\lim_{b\to\infty}\int_{0}^{b}x^2dx[/tex]

SO it means that the function x^2 is unbounde from the above. And if that integral converges, than it means that we are calculating the area that the function [tex]f(x)=x^2[/tex] encloses with the x-axis from x=0 to infinity. In other words the area that the right wing of the parabola encloses with the x-axis.
 

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