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Proof of this integration shortcut: $int_a^b fracdxsqrt(x-a)(b-x)=pi$
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
I came across this as one of the shortcuts in my textbook without any proof.
When $bgt a$,
$$intlimits_a^b dfracdxsqrt(x-a)(b-x)=pi$$
My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfracdtsqrtt(1-t)$$
This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?
calculus integration definite-integrals
edited Jul 16 at 18:31
Nosrati
19.8k41644
asked Jul 16 at 17:57
rsadhvika
1,4891026
 |Â
show 2 more comments
up vote
9
down vote
favorite
I came across this as one of the shortcuts in my textbook without any proof.
When $bgt a$,
$$intlimits_a^b dfracdxsqrt(x-a)(b-x)=pi$$
My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfracdtsqrtt(1-t)$$
This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?
calculus integration definite-integrals
edited Jul 16 at 18:31
Nosrati
19.8k41644
asked Jul 16 at 17:57
rsadhvika
1,4891026
2
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
– Lord Shark the Unknown
Jul 16 at 17:59
@InterstellarProbe here is an example
– rsadhvika
Jul 16 at 18:04
@InterstellarProbe So much the worse for Wolfie!
– Lord Shark the Unknown
Jul 16 at 18:06
1
$intlimits_0^1 dfracdtsqrtt(1-t)=beta(dfrac14,dfrac14)=pi$
– Nosrati
Jul 16 at 18:16
1
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
– usr0192
Jul 16 at 18:18
 |Â
show 2 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I came across this as one of the shortcuts in my textbook without any proof.
When $bgt a$,
$$intlimits_a^b dfracdxsqrt(x-a)(b-x)=pi$$
My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfracdtsqrtt(1-t)$$
This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?
calculus integration definite-integrals
edited Jul 16 at 18:31
Nosrati
19.8k41644
asked Jul 16 at 17:57
rsadhvika
1,4891026
I came across this as one of the shortcuts in my textbook without any proof.
When $bgt a$,
$$intlimits_a^b dfracdxsqrt(x-a)(b-x)=pi$$
My attempt :
I notice that the the denominator is $0$ at both the bounds. I thought of substituting $x=a+(b-a)t$ so that the integral becomes
$$intlimits_0^1 dfracdtsqrtt(1-t)$$
This doesn't look simple, but I'm wondering if the answer can be seen using symmetry/geometry ?
calculus integration definite-integrals
edited Jul 16 at 18:31
Nosrati
19.8k41644
asked Jul 16 at 17:57
rsadhvika
1,4891026
edited Jul 16 at 18:31
Nosrati
19.8k41644
edited Jul 16 at 18:31
Nosrati
19.8k41644
edited Jul 16 at 18:31
Nosrati
19.8k41644
19.8k41644
asked Jul 16 at 17:57
rsadhvika
1,4891026
asked Jul 16 at 17:57
rsadhvika
1,4891026
asked Jul 16 at 17:57
rsadhvika
1,4891026
1,4891026
2
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
– Lord Shark the Unknown
Jul 16 at 17:59
@InterstellarProbe here is an example
– rsadhvika
Jul 16 at 18:04
@InterstellarProbe So much the worse for Wolfie!
– Lord Shark the Unknown
Jul 16 at 18:06
1
$intlimits_0^1 dfracdtsqrtt(1-t)=beta(dfrac14,dfrac14)=pi$
– Nosrati
Jul 16 at 18:16
1
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
– usr0192
Jul 16 at 18:18
 |Â
show 2 more comments
2
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
– Lord Shark the Unknown
Jul 16 at 17:59
@InterstellarProbe here is an example
– rsadhvika
Jul 16 at 18:04
@InterstellarProbe So much the worse for Wolfie!
– Lord Shark the Unknown
Jul 16 at 18:06
1
$intlimits_0^1 dfracdtsqrtt(1-t)=beta(dfrac14,dfrac14)=pi$
– Nosrati
Jul 16 at 18:16
1
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
– usr0192
Jul 16 at 18:18
2
2
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
– Lord Shark the Unknown
Jul 16 at 17:59
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
– Lord Shark the Unknown
Jul 16 at 17:59
@InterstellarProbe here is an example
– rsadhvika
Jul 16 at 18:04
@InterstellarProbe here is an example
– rsadhvika
Jul 16 at 18:04
@InterstellarProbe So much the worse for Wolfie!
– Lord Shark the Unknown
Jul 16 at 18:06
@InterstellarProbe So much the worse for Wolfie!
– Lord Shark the Unknown
Jul 16 at 18:06
1
1
$intlimits_0^1 dfracdtsqrtt(1-t)=beta(dfrac14,dfrac14)=pi$
– Nosrati
Jul 16 at 18:16
$intlimits_0^1 dfracdtsqrtt(1-t)=beta(dfrac14,dfrac14)=pi$
– Nosrati
Jul 16 at 18:16
1
1
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
– usr0192
Jul 16 at 18:18
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
– usr0192
Jul 16 at 18:18
 |Â
show 2 more comments
5 Answers
5
active
oldest
votes
up vote
16
down vote
Other way is substitution $t=sin^2theta$ so
$$intlimits_0^1 dfracdtsqrtt(1-t)=intlimits_0^fracpi2 2dt=pi$$
answered Jul 16 at 18:19
Nosrati
19.8k41644
add a comment |Â
up vote
4
down vote
It's called an Abel Integral ( at least in my language ). You can write that
$$
frac1sqrtleft(x-aright)left(b-xright)=frac2a-bfrac1sqrt1-left(frac2a-bleft(x-fracb+a2right)right)^2$$
that goes into arcsinus
$$int_a^bfractextdxsqrtleft(x-aright)left(b-xright)=textarcsinleft(frac2b-afracb-a2right)+textarcsinleft(frac2a-bfraca-b2right)=2textarcsinleft(1right)=pi$$
answered Jul 16 at 18:22
Atmos
4,660119
add a comment |Â
up vote
4
down vote
beginalign
tan^2 theta &= fracx-ab-x \
2tan theta sec^2 theta , dtheta &=
fracb-a(b-x)^2 , dx \
2sqrtfracx-ab-x times frac(x-a)+(b-x)b-x , dtheta &=
fracb-a(b-x)^2 , dx \
2, dtheta &= fracdxsqrt(x-a)(b-x) \
int fracdxsqrt(x-a)(b-x) &=
2tan^-1 sqrtfracx-ab-x
endalign
The singularity in Wolfram Alpha comes from the upper limit $b$.
Geometrical interpretation
Considering circular arc $(x,y)=(sqrtb-u,sqrtu-a)$
beginalign
ds &= fracsqrtb-a , du2sqrt(u-a)(b-u) \
tan theta &= sqrtfracu-ab-u \
beginpmatrix
x \ y
endpmatrix &=
beginpmatrix
sqrtb-a cos theta \
sqrtb-a sin theta
endpmatrix \
ds &= sqrtb-a , dtheta
endalign
See also another integral here.
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
add a comment |Â
up vote
2
down vote
I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$
answered Jul 17 at 6:41
Hari Shankar
2,121139
add a comment |Â
up vote
1
down vote
Let $m = fracb+a2$ and $r = fracb-a2$. Consider the circle
$$ (x - m)^2 + y^2 = r^2. $$
Part of this locus with $y geq 0$ is given by $y = sqrtr^2 - (x-m)^2 = sqrt(x-a)(b-x)$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence
$$ fracdydx = -fracx-my. $$
So the length of the upper-circular arc is
$$ pi r = int_a^b sqrt1+left(fracdydxright)^2 , dx = int_a^b sqrtfrac(x-m)^2 + y^2y^2 , dx = int_a^b fracrsqrt(x-a)(b-x) , dx. $$
Dividing both sides by $r$ gives the desired answer.
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
Other way is substitution $t=sin^2theta$ so
$$intlimits_0^1 dfracdtsqrtt(1-t)=intlimits_0^fracpi2 2dt=pi$$
answered Jul 16 at 18:19
Nosrati
19.8k41644
add a comment |Â
up vote
16
down vote
Other way is substitution $t=sin^2theta$ so
$$intlimits_0^1 dfracdtsqrtt(1-t)=intlimits_0^fracpi2 2dt=pi$$
answered Jul 16 at 18:19
Nosrati
19.8k41644
add a comment |Â
up vote
16
down vote
up vote
16
down vote
Other way is substitution $t=sin^2theta$ so
$$intlimits_0^1 dfracdtsqrtt(1-t)=intlimits_0^fracpi2 2dt=pi$$
answered Jul 16 at 18:19
Nosrati
19.8k41644
Other way is substitution $t=sin^2theta$ so
$$intlimits_0^1 dfracdtsqrtt(1-t)=intlimits_0^fracpi2 2dt=pi$$
answered Jul 16 at 18:19
Nosrati
19.8k41644
answered Jul 16 at 18:19
Nosrati
19.8k41644
answered Jul 16 at 18:19
Nosrati
19.8k41644
answered Jul 16 at 18:19
Nosrati
19.8k41644
19.8k41644
add a comment |Â
add a comment |Â
up vote
4
down vote
It's called an Abel Integral ( at least in my language ). You can write that
$$
frac1sqrtleft(x-aright)left(b-xright)=frac2a-bfrac1sqrt1-left(frac2a-bleft(x-fracb+a2right)right)^2$$
that goes into arcsinus
$$int_a^bfractextdxsqrtleft(x-aright)left(b-xright)=textarcsinleft(frac2b-afracb-a2right)+textarcsinleft(frac2a-bfraca-b2right)=2textarcsinleft(1right)=pi$$
answered Jul 16 at 18:22
Atmos
4,660119
add a comment |Â
up vote
4
down vote
It's called an Abel Integral ( at least in my language ). You can write that
$$
frac1sqrtleft(x-aright)left(b-xright)=frac2a-bfrac1sqrt1-left(frac2a-bleft(x-fracb+a2right)right)^2$$
that goes into arcsinus
$$int_a^bfractextdxsqrtleft(x-aright)left(b-xright)=textarcsinleft(frac2b-afracb-a2right)+textarcsinleft(frac2a-bfraca-b2right)=2textarcsinleft(1right)=pi$$
answered Jul 16 at 18:22
Atmos
4,660119
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It's called an Abel Integral ( at least in my language ). You can write that
$$
frac1sqrtleft(x-aright)left(b-xright)=frac2a-bfrac1sqrt1-left(frac2a-bleft(x-fracb+a2right)right)^2$$
that goes into arcsinus
$$int_a^bfractextdxsqrtleft(x-aright)left(b-xright)=textarcsinleft(frac2b-afracb-a2right)+textarcsinleft(frac2a-bfraca-b2right)=2textarcsinleft(1right)=pi$$
answered Jul 16 at 18:22
Atmos
4,660119
It's called an Abel Integral ( at least in my language ). You can write that
$$
frac1sqrtleft(x-aright)left(b-xright)=frac2a-bfrac1sqrt1-left(frac2a-bleft(x-fracb+a2right)right)^2$$
that goes into arcsinus
$$int_a^bfractextdxsqrtleft(x-aright)left(b-xright)=textarcsinleft(frac2b-afracb-a2right)+textarcsinleft(frac2a-bfraca-b2right)=2textarcsinleft(1right)=pi$$
answered Jul 16 at 18:22
Atmos
4,660119
answered Jul 16 at 18:22
Atmos
4,660119
answered Jul 16 at 18:22
Atmos
4,660119
answered Jul 16 at 18:22
Atmos
4,660119
4,660119
add a comment |Â
add a comment |Â
up vote
4
down vote
beginalign
tan^2 theta &= fracx-ab-x \
2tan theta sec^2 theta , dtheta &=
fracb-a(b-x)^2 , dx \
2sqrtfracx-ab-x times frac(x-a)+(b-x)b-x , dtheta &=
fracb-a(b-x)^2 , dx \
2, dtheta &= fracdxsqrt(x-a)(b-x) \
int fracdxsqrt(x-a)(b-x) &=
2tan^-1 sqrtfracx-ab-x
endalign
The singularity in Wolfram Alpha comes from the upper limit $b$.
Geometrical interpretation
Considering circular arc $(x,y)=(sqrtb-u,sqrtu-a)$
beginalign
ds &= fracsqrtb-a , du2sqrt(u-a)(b-u) \
tan theta &= sqrtfracu-ab-u \
beginpmatrix
x \ y
endpmatrix &=
beginpmatrix
sqrtb-a cos theta \
sqrtb-a sin theta
endpmatrix \
ds &= sqrtb-a , dtheta
endalign
See also another integral here.
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
add a comment |Â
up vote
4
down vote
beginalign
tan^2 theta &= fracx-ab-x \
2tan theta sec^2 theta , dtheta &=
fracb-a(b-x)^2 , dx \
2sqrtfracx-ab-x times frac(x-a)+(b-x)b-x , dtheta &=
fracb-a(b-x)^2 , dx \
2, dtheta &= fracdxsqrt(x-a)(b-x) \
int fracdxsqrt(x-a)(b-x) &=
2tan^-1 sqrtfracx-ab-x
endalign
The singularity in Wolfram Alpha comes from the upper limit $b$.
Geometrical interpretation
Considering circular arc $(x,y)=(sqrtb-u,sqrtu-a)$
beginalign
ds &= fracsqrtb-a , du2sqrt(u-a)(b-u) \
tan theta &= sqrtfracu-ab-u \
beginpmatrix
x \ y
endpmatrix &=
beginpmatrix
sqrtb-a cos theta \
sqrtb-a sin theta
endpmatrix \
ds &= sqrtb-a , dtheta
endalign
See also another integral here.
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
add a comment |Â
up vote
4
down vote
up vote
4
down vote
beginalign
tan^2 theta &= fracx-ab-x \
2tan theta sec^2 theta , dtheta &=
fracb-a(b-x)^2 , dx \
2sqrtfracx-ab-x times frac(x-a)+(b-x)b-x , dtheta &=
fracb-a(b-x)^2 , dx \
2, dtheta &= fracdxsqrt(x-a)(b-x) \
int fracdxsqrt(x-a)(b-x) &=
2tan^-1 sqrtfracx-ab-x
endalign
The singularity in Wolfram Alpha comes from the upper limit $b$.
Geometrical interpretation
Considering circular arc $(x,y)=(sqrtb-u,sqrtu-a)$
beginalign
ds &= fracsqrtb-a , du2sqrt(u-a)(b-u) \
tan theta &= sqrtfracu-ab-u \
beginpmatrix
x \ y
endpmatrix &=
beginpmatrix
sqrtb-a cos theta \
sqrtb-a sin theta
endpmatrix \
ds &= sqrtb-a , dtheta
endalign
See also another integral here.
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
beginalign
tan^2 theta &= fracx-ab-x \
2tan theta sec^2 theta , dtheta &=
fracb-a(b-x)^2 , dx \
2sqrtfracx-ab-x times frac(x-a)+(b-x)b-x , dtheta &=
fracb-a(b-x)^2 , dx \
2, dtheta &= fracdxsqrt(x-a)(b-x) \
int fracdxsqrt(x-a)(b-x) &=
2tan^-1 sqrtfracx-ab-x
endalign
The singularity in Wolfram Alpha comes from the upper limit $b$.
Geometrical interpretation
Considering circular arc $(x,y)=(sqrtb-u,sqrtu-a)$
beginalign
ds &= fracsqrtb-a , du2sqrt(u-a)(b-u) \
tan theta &= sqrtfracu-ab-u \
beginpmatrix
x \ y
endpmatrix &=
beginpmatrix
sqrtb-a cos theta \
sqrtb-a sin theta
endpmatrix \
ds &= sqrtb-a , dtheta
endalign
See also another integral here.
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
edited Jul 17 at 3:37
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
answered Jul 16 at 18:19
Ng Chung Tak
13k31130
13k31130
add a comment |Â
add a comment |Â
up vote
2
down vote
I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$
answered Jul 17 at 6:41
Hari Shankar
2,121139
add a comment |Â
up vote
2
down vote
I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$
answered Jul 17 at 6:41
Hari Shankar
2,121139
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$
answered Jul 17 at 6:41
Hari Shankar
2,121139
I was taught to use the substitution $x=a sin^2 theta+b cos^2 theta$
answered Jul 17 at 6:41
Hari Shankar
2,121139
answered Jul 17 at 6:41
Hari Shankar
2,121139
answered Jul 17 at 6:41
Hari Shankar
2,121139
answered Jul 17 at 6:41
Hari Shankar
2,121139
2,121139
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $m = fracb+a2$ and $r = fracb-a2$. Consider the circle
$$ (x - m)^2 + y^2 = r^2. $$
Part of this locus with $y geq 0$ is given by $y = sqrtr^2 - (x-m)^2 = sqrt(x-a)(b-x)$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence
$$ fracdydx = -fracx-my. $$
So the length of the upper-circular arc is
$$ pi r = int_a^b sqrt1+left(fracdydxright)^2 , dx = int_a^b sqrtfrac(x-m)^2 + y^2y^2 , dx = int_a^b fracrsqrt(x-a)(b-x) , dx. $$
Dividing both sides by $r$ gives the desired answer.
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
add a comment |Â
up vote
1
down vote
Let $m = fracb+a2$ and $r = fracb-a2$. Consider the circle
$$ (x - m)^2 + y^2 = r^2. $$
Part of this locus with $y geq 0$ is given by $y = sqrtr^2 - (x-m)^2 = sqrt(x-a)(b-x)$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence
$$ fracdydx = -fracx-my. $$
So the length of the upper-circular arc is
$$ pi r = int_a^b sqrt1+left(fracdydxright)^2 , dx = int_a^b sqrtfrac(x-m)^2 + y^2y^2 , dx = int_a^b fracrsqrt(x-a)(b-x) , dx. $$
Dividing both sides by $r$ gives the desired answer.
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $m = fracb+a2$ and $r = fracb-a2$. Consider the circle
$$ (x - m)^2 + y^2 = r^2. $$
Part of this locus with $y geq 0$ is given by $y = sqrtr^2 - (x-m)^2 = sqrt(x-a)(b-x)$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence
$$ fracdydx = -fracx-my. $$
So the length of the upper-circular arc is
$$ pi r = int_a^b sqrt1+left(fracdydxright)^2 , dx = int_a^b sqrtfrac(x-m)^2 + y^2y^2 , dx = int_a^b fracrsqrt(x-a)(b-x) , dx. $$
Dividing both sides by $r$ gives the desired answer.
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
Let $m = fracb+a2$ and $r = fracb-a2$. Consider the circle
$$ (x - m)^2 + y^2 = r^2. $$
Part of this locus with $y geq 0$ is given by $y = sqrtr^2 - (x-m)^2 = sqrt(x-a)(b-x)$ for $a leq x leq b$. By the implicit differentiation, this function satisfies $ 2(x - m) dx + 2ydy = 0 $ and hence
$$ fracdydx = -fracx-my. $$
So the length of the upper-circular arc is
$$ pi r = int_a^b sqrt1+left(fracdydxright)^2 , dx = int_a^b sqrtfrac(x-m)^2 + y^2y^2 , dx = int_a^b fracrsqrt(x-a)(b-x) , dx. $$
Dividing both sides by $r$ gives the desired answer.
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
answered Jul 17 at 7:01
Sangchul Lee
85.6k12155253
85.6k12155253
add a comment |Â
add a comment |Â
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2
It's better to reduce to the $(a,b)=(-1,1)$ case rather than the $(0,1)$ case.
– Lord Shark the Unknown
Jul 16 at 17:59
@InterstellarProbe here is an example
– rsadhvika
Jul 16 at 18:04
@InterstellarProbe So much the worse for Wolfie!
– Lord Shark the Unknown
Jul 16 at 18:06
1
$intlimits_0^1 dfracdtsqrtt(1-t)=beta(dfrac14,dfrac14)=pi$
– Nosrati
Jul 16 at 18:16
1
maybe try the substitution $t=cos theta$ or $t=sintheta$? I haven't tried it but the denominator would simplify
– usr0192
Jul 16 at 18:18