Beautiful Integral

Calculus Level 3

∫ a b ( x − a ) ( b − x ) d x = ? \large\int\limits_{a}^{b}\sqrt{(x-a)(b-x)}\,\mathrm dx~=~?

Assume that b > a b>a .

( a − b ) 2 (a-b)^{2} π ( a + b ) 2 8 \frac{\pi (a+b)^{2}}{8} π ( a − b ) 2 8 \frac{\pi (a-b)^{2}}{8} ( a + b ) 2 (a+b)^{2}

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Jessica Wang by Jessica Wang , 22, United Kingdom

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7 solutions

Jessica Wang
Oct 28, 2015

43 Helpful 0 Interesting 0 Brilliant 0 Confused

That's one of the beautiful solution i seen on brilliant.
For those who didn't get how you made that semicircle...
y = ( x − a ) ( b − x ) y = \sqrt{(x - a)(b - x)}
Squaring both sides,

⇒ y 2 = x b − x 2 − a b + a x \Rightarrow y^2 = xb - x^2 - ab + ax .
⇒ x 2 + y 2 − ( a + b ) x + a b = 0 \Rightarrow x^2 + y^2 - (a+b)x + ab= 0 .
Above equation, is an equation of cicle with center at ( a + b 2 , 0 ) \left(\dfrac{a + b}{2},0\right) and the radius b − a 2 \dfrac{b - a}{2} (from diagram) ,but y can't be negative. hence, y = ( x − a ) ( b − x ) y = \sqrt{(x - a)(b - x)} is an equation of half circle lying above x - axis.

Akhil Bansal - 5 years, 7 months ago

Nice geometrical approach @Jessica Wang !!

A Former Brilliant Member - 5 years, 7 months ago

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Thank you Abhi @Abhineet Nayyar :)

Jessica Wang - 5 years, 7 months ago

Nice solution.

Kushal Patankar - 5 years, 7 months ago

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The way I solved it was by substituting x = a ( s i n x ) 2 + b ( c o s x ) 2 a(sinx)^2 + b( cosx)^2 But, this method is simply more elegant and beautiful!

A Former Brilliant Member - 5 years, 7 months ago

I noticed the symmetry of the integrand about x = 1 2 ( a + b ) x=\frac 12(a+b) while solving the problem but didn't think that it was gonna be a nice geometric figure (semicircle in this case). Ended up bashing the integral with basic properties of definite integrals and some known closed forms. Nice problem.

Prasun Biswas - 5 years, 7 months ago

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I found the same. I knew that there was symmetry but I didn't realize I was dealing with a semicircle until I got to the step of evaluating

( b − a ) 2 4 ∫ − 1 1 1 − u 2 d u . \dfrac{(b - a)^{2}}{4} \displaystyle\int_{-1}^{1} \sqrt{1 - u^{2}} du.

Nice problem and an elegant solution, @Jessica Wang . :)

Brian Charlesworth - 5 years, 7 months ago

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Thank you sir :) @Brian Charlesworth

Jessica Wang - 5 years, 7 months ago

wat ıntegrate formula

Soner Karaca - 5 years, 7 months ago

Nice picture~<3

展豪 張 - 5 years, 7 months ago
Michael Mendrin
Oct 29, 2015

I hate to be a party pooper, but the correct answer is actually

1 8 π ( b − a ) ( b − a ) 2 \dfrac { 1 }{ 8 } \pi \left( b-a \right) \sqrt { { \left( b-a \right) }^{ 2 } }

Otherwise, if a > b a>b , the sign would be wrong. Work it out. The hangup is that the limits of the definite integral can actually result in a negative value if a > b a>b , but if the limits were to be reversed, i.e, integrate from b b to a a if a > b a>b , then it would return a positive value.

This problem, as posted, should have included the condition that b > a b>a .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

@Michael Mendrin Thank you very much for pointing this out -- I have edited the problem!

Jessica Wang - 5 years, 7 months ago
Will Jain
Nov 1, 2015

That's a really beautiful Sol by Jessica Wang

2 Helpful 1 Interesting 0 Brilliant 0 Confused
Arjen Vreugdenhil
Oct 28, 2015

A slightly different approach: rewrite the expression under the square root by "completing the square": ( x − a ) ( b − x ) = − ( x − c ) 2 + d . (x-a)(b-x) = -(x-c)^2 + d. Since the linear term is − ( a + b ) x -(a+b)x , we find c = 1 2 ( a + b ) c = \tfrac12(a+b) ; and because the constant term is − a b -ab , d = c 2 − a b = ( 1 2 ( a + b ) ) 2 − a b = ( 1 2 ( a − b ) ) 2 = : r 2 . d = c^2-ab = (\tfrac12(a+b))^2 - ab = (\tfrac12(a-b))^2 =: r^2. Thus we find ( x − a ) ( b − x ) = r 2 − ( x − c ) 2 , \sqrt{(x-a)(b-x)} = \sqrt{r^2-(x-c)^2}, which we recognize the equation for a circle with center ( c , 0 ) (c,0) and radius r r .

It makes sense that the center of the circle is precisely the average of a a and b b , and that the radius is precisely half of the distance between a a and b b ... The line from ( a , 0 ) (a,0) to ( b , 0 ) (b,0) is a diameter of the circle.

The integral gives, naturally, the area of a semicircle: A = 1 2 π r 2 = 1 8 ( a − b ) 2 . A = \tfrac12 \pi r^2 = \tfrac18 (a-b)^2.

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Kevin Yang
Dec 28, 2018

According to Wolframalpha it's 1 8 π ( a − b ) 2 \frac{1}{8}\pi(a-b)^2

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Ramez Hindi
Nov 4, 2015

my tr y

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Righved K
Oct 29, 2015

did it by beta function..done with ease

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I am not very familiar with using special functions. Would you mind posting your solution so I can learn? If it is too much trouble, then it is fine.

Varun Gudibanda - 5 years, 7 months ago

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Sorry john for looking late but please tell me how to share the solution with you,I have a snap of the solution

Righved K - 5 years, 7 months ago

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