Integral Is Area
I learned of this example from Kunihiko Chikaya's post on facebook. The task is to compute the integral
$\displaystyle\int_{a}^{b}\sqrt{(x-a)(b-x)}dx.$
The solution is to recollect the geometric meaning of the definite integral.
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Copyright © 1996-2018 Alexander Bogomolny
Compute the followng integral
$\displaystyle\int_{a}^{b}\sqrt{(x-a)(b-x)}dx.$
Solution
Let $y=\sqrt{(x-a)(b-x)}.$ For $y\gt 0,\;$ this is equivalent to $\displaystyle\left(x-\frac{a+b}{2}\right)^2+y^2=\left(\frac{b-a}{2}\right)^2\;$ which is a semicircle with center at $\displaystyle (\frac{a+b}{2},0)\;$ and radius $\displaystyle\frac{b-a}{2}:$
Thus, the integral is half the area of a circle with radius $\displaystyle\frac{b-a}{2}:$
$\displaystyle\frac{1}{2}\pi\left(\frac{b-a}{2}\right)^2=\frac{\pi (b-a)^2}{8}.$
- A simple integral, I
 - A simple integral, II
 - A simple integral, III
 - Another simple integral
 - Yet Another Simple Integral
 - Integral in Normal Distribution
 - Sum Without Adding
 - An Integral Inequality from the RMM
 - Another Integral Inequality from the RMM
 - A Triple Integral Inequality
 - An Integral from the RMM
 - An Integral Identity with Two Functions
 - A Simple Integral of a Peculiar Function
 
|Contact| |Front page| |Contents| |Algebra| |Up|
Copyright © 1996-2018 Alexander Bogomolny
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