High School Integration from "Area Under Curve" Perspective: Explained

[songoku]

TL;DR

$$\int_\frac{1}{2}^1 \frac{x}{\sqrt{1-x^2}}dx$$

I can calculate the value of the integration, it will be $\frac{\sqrt{3}}{2}$

But if I draw the function and consider the area bounded by the curve and x-axis from x = 0.5 to x = 1, it seems that the area will be infinite because x = 1 is vertical asymptote.

Why can't I consider from "area under curve" perspective to calculate this integration? Thanks

 

[etotheipi]

You have an improper integral, you need to consider limits. Define$$I(a) = \int_{\frac{1}{2}}^a \frac{x}{\sqrt{1-x^2}} = \left[ -\sqrt{1-x^2} \right]_{\frac{1}{2}}^{a} = \frac{\sqrt{3}}{2} - \sqrt{1-a^2}$$Then take the limit$$\lim_{a \rightarrow 1} I(a) = \lim_{a\rightarrow 1} \left( \frac{\sqrt{3}}{2} - \sqrt{1-a^2} \right) = \frac{\sqrt{3}}{2}$$

 

[PeroK]

it seems that the area will be infinite because x = 1 is vertical asymptote.

As Hamlet said I know not "seems". The integral tells you that the area is in fact finite.

 

[songoku]

Thank you very much etotheipi and Perok

 

[FactChecker]

Summary:: $$\int_\frac{1}{2}^1 \frac{x}{\sqrt{1-x^2}}dx$$

I can calculate the value of the integration, it will be $\frac{\sqrt{3}}{2}$

But if I draw the function and consider the area bounded by the curve and x-axis from x = 0.5 to x = 1, it seems that the area will be infinite because x = 1 is vertical asymptote.

Why can't I consider from "area under curve" perspective to calculate this integration? Thanks

Intuitively, the graph gets thin faster than it goes up. The result of the integration shows that the area is finite. That happens a lot.

 

[songoku]

Intuitively, the graph gets thin faster than it goes up. The result of the integration shows that the area is finite. That happens a lot.

I always thought the area of such case would always be infinite. Now I think maybe the case is similar to sum to infinity of convergent geometric series.

Thank you very much FactChecker

 

[Office_Shredder]

I always thought the area of such case would always be infinite. Now I think maybe the case is similar to sum to infinity of convergent geometric series.

Thank you very much FactChecker

Remember integrals are approximated above and below by infinite series. Here, make a substition $u=1-x^2$ (let's not worry about whether this calculation is actually valid), to get

$$\int_{3/4}^{0} \frac{-\frac{1}{2} du}{\sqrt{u}} $$
$$\frac{1}{2} \int_{0}^{3/4} \frac{du}{\sqrt{u}}.$$

Now do the substition $v=1/u$
$$\frac{1}{2} \int_{4/3}^{\infty} \sqrt{v} \frac{dv}{v^2}$$

This last integral is approximated by Riemann sums that look like (when bounding both from below and above) $\sum \frac{1}{n^{3/2}}$ and therefore it converges.

 

[FactChecker]

Now I think maybe the case is similar to sum to infinity of convergent geometric series.

Exactly.

 

[lurflurf]

Remember integrals are approximated above and below by infinite series. Here, make a substition $u=1-x^2$ (let's not worry about whether this calculation is actually valid), to get

It is valid but $u=\sqrt{1-x^2}$ is nice as well.

What infinite series are you bounding this integral with and why?

 

[Office_Shredder]

It is valid but $u=\sqrt{1-x^2}$ is nice as well.

What infinite series are you bounding this integral with and why?

You can draw rectangles of width 1 above the graph of $1/x^{3/2}$ to show that if $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges, then so does $\int_1^{\infty} \frac{1}{x^{3/2}} dx$. This is equivalent to the convergence of the integral in the original post. My point was that the op said something about how integrals converging feels like infinite sums converging, and to demonstrate how the convergence of the integral in this thread is literally equivalent to the convergence of an infinite series.

 

[PeroK]

I always thought the area of such case would always be infinite. Now I think maybe the case is similar to sum to infinity of convergent geometric series.

One of the most amazing results in mathematics is that the area under the infinite Gaussian curve is: $$\int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt \pi$$

 

[songoku]

Thank you very much for all the help and explanation etotheipi, Perok, FactChecker, Office_Shredder, lurflurf