Convergent property of integrals Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Numerical Integration of Irregular Temperature ReadingsLooking for an accessible explanation of Henstock–Kurzweil (gauge) integrallooking for advice on studying calculusIntegral of $xe^-ax^2-bx^-1$Book for studying Calculus IIAre there real definite integrals which can only be evaluated by contour integration?General approach to apply Fubini's theorem to integrals of the form $int_x=a^b int_y=alpha(x)^beta(x)$Explanation of spectral theorem for reals to high schoolers who have only done computational single-variable calculus.Special functions for non-elementry antiderivatives.Types of integration
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Convergent property of integrals
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Numerical Integration of Irregular Temperature ReadingsLooking for an accessible explanation of Henstock–Kurzweil (gauge) integrallooking for advice on studying calculusIntegral of $xe^-ax^2-bx^-1$Book for studying Calculus IIAre there real definite integrals which can only be evaluated by contour integration?General approach to apply Fubini's theorem to integrals of the form $int_x=a^b int_y=alpha(x)^beta(x)$Explanation of spectral theorem for reals to high schoolers who have only done computational single-variable calculus.Special functions for non-elementry antiderivatives.Types of integration
$begingroup$
When studying various functions and proofs, I've often seen the following...
.
Which is the reason why some functions such as the Beta function can be written as...
$$B(x,y) = B(y,x)$$
What I want to know is why this proof holds true? What is a good explanation for someone that is relatively new to integrals? I would like to be able to prove this to myself and visualize it better.
Looking at it alone doesn't make sense to me, but for some reason the rule holds.
calculus integration definite-integrals proof-writing
asked Apr 1 at 23:29
BolboaBolboa
408616
$endgroup$
add a comment |
$begingroup$
When studying various functions and proofs, I've often seen the following...
.
Which is the reason why some functions such as the Beta function can be written as...
$$B(x,y) = B(y,x)$$
What I want to know is why this proof holds true? What is a good explanation for someone that is relatively new to integrals? I would like to be able to prove this to myself and visualize it better.
Looking at it alone doesn't make sense to me, but for some reason the rule holds.
calculus integration definite-integrals proof-writing
asked Apr 1 at 23:29
BolboaBolboa
408616
$endgroup$
1
$begingroup$
It is a simple application of the variable transformation $t=a-u$.
$endgroup$
– Stan Tendijck
Apr 1 at 23:33
add a comment |
$begingroup$
When studying various functions and proofs, I've often seen the following...
.
Which is the reason why some functions such as the Beta function can be written as...
$$B(x,y) = B(y,x)$$
What I want to know is why this proof holds true? What is a good explanation for someone that is relatively new to integrals? I would like to be able to prove this to myself and visualize it better.
Looking at it alone doesn't make sense to me, but for some reason the rule holds.
calculus integration definite-integrals proof-writing
asked Apr 1 at 23:29
BolboaBolboa
408616
$endgroup$
When studying various functions and proofs, I've often seen the following...
.
Which is the reason why some functions such as the Beta function can be written as...
$$B(x,y) = B(y,x)$$
What I want to know is why this proof holds true? What is a good explanation for someone that is relatively new to integrals? I would like to be able to prove this to myself and visualize it better.
Looking at it alone doesn't make sense to me, but for some reason the rule holds.
calculus integration definite-integrals proof-writing
calculus integration definite-integrals proof-writing
asked Apr 1 at 23:29
BolboaBolboa
408616
asked Apr 1 at 23:29
BolboaBolboa
408616
asked Apr 1 at 23:29
BolboaBolboa
408616
asked Apr 1 at 23:29
BolboaBolboa
408616
asked Apr 1 at 23:29
BolboaBolboa
408616
408616
1
$begingroup$
It is a simple application of the variable transformation $t=a-u$.
$endgroup$
– Stan Tendijck
Apr 1 at 23:33
add a comment |
1
$begingroup$
It is a simple application of the variable transformation $t=a-u$.
$endgroup$
– Stan Tendijck
Apr 1 at 23:33
1
1
$begingroup$
It is a simple application of the variable transformation $t=a-u$.
$endgroup$
– Stan Tendijck
Apr 1 at 23:33
$begingroup$
It is a simple application of the variable transformation $t=a-u$.
$endgroup$
– Stan Tendijck
Apr 1 at 23:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Starting with $$int_0^af(t)dt, $$ consider the substitution $t = a-u.$ Then $dt = -du$ and the limits $t=0$ and $t=a$ become the limits $u = a$ and $u = 0$. Hence
$$int_0^af(t)dt = int_a^0f(a-u)(-du) = -int_a^0f(a-u)du = int_0^af(a-u)du.$$ Since $u$ is a dummy variable in this last step, we may simply change each $u$ to a $t$ and we end up with the form you have.
answered Apr 1 at 23:45
user328442user328442
1,9431616
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Starting with $$int_0^af(t)dt, $$ consider the substitution $t = a-u.$ Then $dt = -du$ and the limits $t=0$ and $t=a$ become the limits $u = a$ and $u = 0$. Hence
$$int_0^af(t)dt = int_a^0f(a-u)(-du) = -int_a^0f(a-u)du = int_0^af(a-u)du.$$ Since $u$ is a dummy variable in this last step, we may simply change each $u$ to a $t$ and we end up with the form you have.
answered Apr 1 at 23:45
user328442user328442
1,9431616
$endgroup$
add a comment |
$begingroup$
Starting with $$int_0^af(t)dt, $$ consider the substitution $t = a-u.$ Then $dt = -du$ and the limits $t=0$ and $t=a$ become the limits $u = a$ and $u = 0$. Hence
$$int_0^af(t)dt = int_a^0f(a-u)(-du) = -int_a^0f(a-u)du = int_0^af(a-u)du.$$ Since $u$ is a dummy variable in this last step, we may simply change each $u$ to a $t$ and we end up with the form you have.
answered Apr 1 at 23:45
user328442user328442
1,9431616
$endgroup$
add a comment |
$begingroup$
Starting with $$int_0^af(t)dt, $$ consider the substitution $t = a-u.$ Then $dt = -du$ and the limits $t=0$ and $t=a$ become the limits $u = a$ and $u = 0$. Hence
$$int_0^af(t)dt = int_a^0f(a-u)(-du) = -int_a^0f(a-u)du = int_0^af(a-u)du.$$ Since $u$ is a dummy variable in this last step, we may simply change each $u$ to a $t$ and we end up with the form you have.
answered Apr 1 at 23:45
user328442user328442
1,9431616
$endgroup$
Starting with $$int_0^af(t)dt, $$ consider the substitution $t = a-u.$ Then $dt = -du$ and the limits $t=0$ and $t=a$ become the limits $u = a$ and $u = 0$. Hence
$$int_0^af(t)dt = int_a^0f(a-u)(-du) = -int_a^0f(a-u)du = int_0^af(a-u)du.$$ Since $u$ is a dummy variable in this last step, we may simply change each $u$ to a $t$ and we end up with the form you have.
answered Apr 1 at 23:45
user328442user328442
1,9431616
answered Apr 1 at 23:45
user328442user328442
1,9431616
answered Apr 1 at 23:45
user328442user328442
1,9431616
answered Apr 1 at 23:45
user328442user328442
1,9431616
1,9431616
add a comment |
add a comment |
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$begingroup$
It is a simple application of the variable transformation $t=a-u$.
$endgroup$
– Stan Tendijck
Apr 1 at 23:33