Mixing
A question about integration by substitution
$begingroup$
$f(x)$ is defined on $[a, b]$.
$phi(t)$ is defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi'(t)$ is countinuous on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $x_1 = phi(t_1)$, $x_2 = phi(t_2)$.
Then, the following equation is true:
$$int_{x_1}^{x_2} f(x) dx = int_{t_1}^{t_2} f(phi(t))phi'(t)dt.$$
Let $F(x)$ be a differentiable function defined on $[a,b]$ and $F'(x) = f(x)$ on $[a,b]$.
Let $phi(t)$ be a function defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $phi'(t)$ be a function which is not continuous on $[alpha, beta]$.
Then, $frac{d}{dt} F(phi(t)) = f(phi(t))phi'(t)$ on $[alpha, beta]$.
Now my question is here:
Is there a function $g(x)$ such that $g(x)$ is a function defined on $[a, b]$ and $g(x)$ is not a continuous function on $[a,b]$ and has a primitive function $G(x)$ on $[a, b]$ and $g(x)$ is integrable on $[a,b]$ and $int_a^b g(x) dx neq G(b) - G(a)$ ?
calculus integration definite-integrals
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
$endgroup$
add a comment |
$begingroup$
$f(x)$ is defined on $[a, b]$.
$phi(t)$ is defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi'(t)$ is countinuous on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $x_1 = phi(t_1)$, $x_2 = phi(t_2)$.
Then, the following equation is true:
$$int_{x_1}^{x_2} f(x) dx = int_{t_1}^{t_2} f(phi(t))phi'(t)dt.$$
Let $F(x)$ be a differentiable function defined on $[a,b]$ and $F'(x) = f(x)$ on $[a,b]$.
Let $phi(t)$ be a function defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $phi'(t)$ be a function which is not continuous on $[alpha, beta]$.
Then, $frac{d}{dt} F(phi(t)) = f(phi(t))phi'(t)$ on $[alpha, beta]$.
Now my question is here:
Is there a function $g(x)$ such that $g(x)$ is a function defined on $[a, b]$ and $g(x)$ is not a continuous function on $[a,b]$ and has a primitive function $G(x)$ on $[a, b]$ and $g(x)$ is integrable on $[a,b]$ and $int_a^b g(x) dx neq G(b) - G(a)$ ?
calculus integration definite-integrals
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
$endgroup$
add a comment |
$begingroup$
$f(x)$ is defined on $[a, b]$.
$phi(t)$ is defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi'(t)$ is countinuous on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $x_1 = phi(t_1)$, $x_2 = phi(t_2)$.
Then, the following equation is true:
$$int_{x_1}^{x_2} f(x) dx = int_{t_1}^{t_2} f(phi(t))phi'(t)dt.$$
Let $F(x)$ be a differentiable function defined on $[a,b]$ and $F'(x) = f(x)$ on $[a,b]$.
Let $phi(t)$ be a function defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $phi'(t)$ be a function which is not continuous on $[alpha, beta]$.
Then, $frac{d}{dt} F(phi(t)) = f(phi(t))phi'(t)$ on $[alpha, beta]$.
Now my question is here:
Is there a function $g(x)$ such that $g(x)$ is a function defined on $[a, b]$ and $g(x)$ is not a continuous function on $[a,b]$ and has a primitive function $G(x)$ on $[a, b]$ and $g(x)$ is integrable on $[a,b]$ and $int_a^b g(x) dx neq G(b) - G(a)$ ?
calculus integration definite-integrals
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
$endgroup$
$f(x)$ is defined on $[a, b]$.
$phi(t)$ is defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi'(t)$ is countinuous on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $x_1 = phi(t_1)$, $x_2 = phi(t_2)$.
Then, the following equation is true:
$$int_{x_1}^{x_2} f(x) dx = int_{t_1}^{t_2} f(phi(t))phi'(t)dt.$$
Let $F(x)$ be a differentiable function defined on $[a,b]$ and $F'(x) = f(x)$ on $[a,b]$.
Let $phi(t)$ be a function defined on $[alpha, beta]$ and differentiable on $[alpha, beta]$ and $phi(t) in [a, b]$ for all $t in [alpha, beta]$.
Let $phi'(t)$ be a function which is not continuous on $[alpha, beta]$.
Then, $frac{d}{dt} F(phi(t)) = f(phi(t))phi'(t)$ on $[alpha, beta]$.
Now my question is here:
Is there a function $g(x)$ such that $g(x)$ is a function defined on $[a, b]$ and $g(x)$ is not a continuous function on $[a,b]$ and has a primitive function $G(x)$ on $[a, b]$ and $g(x)$ is integrable on $[a,b]$ and $int_a^b g(x) dx neq G(b) - G(a)$ ?
calculus integration definite-integrals
calculus integration definite-integrals
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
edited Jan 27 at 4:15
tchappy ha
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
asked Jan 27 at 3:41
tchappy hatchappy ha
773412
773412
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
If $g$ has a primitive $G$ then for any partition $P = (x_0,x_1,ldots,x_n)$ of $[a,b]$ we have by the mean value theorem,
$$G(b) - G(a) = sum_{k=1}^n left( ,G(x_k) - G(x_{k-1}), right) =sum_{k=1}^n G'(t_k)(x_k -x_{k-1}) \ = sum_{k=1}^n g(t_k)(x_k -x_{k-1}) $$
Since $g$ is integrable the Riemann sum on the RHS converges as $|P| to 0$ to the integral and the LHS is unchanged.
Thus,
$$G(b) - G(a) = int_a^b g(x) , dx$$
The answer to your question is no, and this is the fundamental theorem of calculus.
answered Jan 27 at 4:17
RRLRRL
52.9k42573
$endgroup$
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
add a comment |
$begingroup$
Without the condition that $g(x)$ be Riemann integrable, then such functions $g(x)$ do exist. Take for example the derivative of Volterra's function, $V'$, which admits a primitive $V$ differentiable everywhere, but $V'$ is not Riemann integrable, so technically $int_a^b V'(x)dxneq V(b)-V(a)$, since $int_a^b V'(x)dx$ does not exist.
With the condition that $g(x)$ be Riemann integrable, if we take your assertion that $g(x)$ admits a primitive $G(x)$ to mean that $G(x)$ is differentiable on $[a,b]$ and that $G'(x)=g(x)$ for all $xin[a,b]$, or even that $G(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=g(x)$ for all $xin(a,b)$, then no such function $g(x)$ where $int_a^b g(x)dxneq G(b)-G(a)$ exists. You will note that this is true regardless of the continuity of $g$. This is forced by the fundamental theorem of calculus.
answered Jan 27 at 4:15
Harry QuHarry Qu
986
$endgroup$
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
If $g$ has a primitive $G$ then for any partition $P = (x_0,x_1,ldots,x_n)$ of $[a,b]$ we have by the mean value theorem,
$$G(b) - G(a) = sum_{k=1}^n left( ,G(x_k) - G(x_{k-1}), right) =sum_{k=1}^n G'(t_k)(x_k -x_{k-1}) \ = sum_{k=1}^n g(t_k)(x_k -x_{k-1}) $$
Since $g$ is integrable the Riemann sum on the RHS converges as $|P| to 0$ to the integral and the LHS is unchanged.
Thus,
$$G(b) - G(a) = int_a^b g(x) , dx$$
The answer to your question is no, and this is the fundamental theorem of calculus.
answered Jan 27 at 4:17
RRLRRL
52.9k42573
$endgroup$
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
add a comment |
$begingroup$
If $g$ has a primitive $G$ then for any partition $P = (x_0,x_1,ldots,x_n)$ of $[a,b]$ we have by the mean value theorem,
$$G(b) - G(a) = sum_{k=1}^n left( ,G(x_k) - G(x_{k-1}), right) =sum_{k=1}^n G'(t_k)(x_k -x_{k-1}) \ = sum_{k=1}^n g(t_k)(x_k -x_{k-1}) $$
Since $g$ is integrable the Riemann sum on the RHS converges as $|P| to 0$ to the integral and the LHS is unchanged.
Thus,
$$G(b) - G(a) = int_a^b g(x) , dx$$
The answer to your question is no, and this is the fundamental theorem of calculus.
answered Jan 27 at 4:17
RRLRRL
52.9k42573
$endgroup$
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
add a comment |
$begingroup$
If $g$ has a primitive $G$ then for any partition $P = (x_0,x_1,ldots,x_n)$ of $[a,b]$ we have by the mean value theorem,
$$G(b) - G(a) = sum_{k=1}^n left( ,G(x_k) - G(x_{k-1}), right) =sum_{k=1}^n G'(t_k)(x_k -x_{k-1}) \ = sum_{k=1}^n g(t_k)(x_k -x_{k-1}) $$
Since $g$ is integrable the Riemann sum on the RHS converges as $|P| to 0$ to the integral and the LHS is unchanged.
Thus,
$$G(b) - G(a) = int_a^b g(x) , dx$$
The answer to your question is no, and this is the fundamental theorem of calculus.
answered Jan 27 at 4:17
RRLRRL
52.9k42573
$endgroup$
If $g$ has a primitive $G$ then for any partition $P = (x_0,x_1,ldots,x_n)$ of $[a,b]$ we have by the mean value theorem,
$$G(b) - G(a) = sum_{k=1}^n left( ,G(x_k) - G(x_{k-1}), right) =sum_{k=1}^n G'(t_k)(x_k -x_{k-1}) \ = sum_{k=1}^n g(t_k)(x_k -x_{k-1}) $$
Since $g$ is integrable the Riemann sum on the RHS converges as $|P| to 0$ to the integral and the LHS is unchanged.
Thus,
$$G(b) - G(a) = int_a^b g(x) , dx$$
The answer to your question is no, and this is the fundamental theorem of calculus.
answered Jan 27 at 4:17
RRLRRL
52.9k42573
answered Jan 27 at 4:17
RRLRRL
52.9k42573
answered Jan 27 at 4:17
RRLRRL
52.9k42573
answered Jan 27 at 4:17
RRLRRL
52.9k42573
52.9k42573
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
add a comment |
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
$begingroup$
Thank you very much for the proof and sorry for my stupid question, RRL.
$endgroup$
– tchappy ha
Jan 27 at 5:02
add a comment |
$begingroup$
Without the condition that $g(x)$ be Riemann integrable, then such functions $g(x)$ do exist. Take for example the derivative of Volterra's function, $V'$, which admits a primitive $V$ differentiable everywhere, but $V'$ is not Riemann integrable, so technically $int_a^b V'(x)dxneq V(b)-V(a)$, since $int_a^b V'(x)dx$ does not exist.
With the condition that $g(x)$ be Riemann integrable, if we take your assertion that $g(x)$ admits a primitive $G(x)$ to mean that $G(x)$ is differentiable on $[a,b]$ and that $G'(x)=g(x)$ for all $xin[a,b]$, or even that $G(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=g(x)$ for all $xin(a,b)$, then no such function $g(x)$ where $int_a^b g(x)dxneq G(b)-G(a)$ exists. You will note that this is true regardless of the continuity of $g$. This is forced by the fundamental theorem of calculus.
answered Jan 27 at 4:15
Harry QuHarry Qu
986
$endgroup$
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
add a comment |
$begingroup$
Without the condition that $g(x)$ be Riemann integrable, then such functions $g(x)$ do exist. Take for example the derivative of Volterra's function, $V'$, which admits a primitive $V$ differentiable everywhere, but $V'$ is not Riemann integrable, so technically $int_a^b V'(x)dxneq V(b)-V(a)$, since $int_a^b V'(x)dx$ does not exist.
With the condition that $g(x)$ be Riemann integrable, if we take your assertion that $g(x)$ admits a primitive $G(x)$ to mean that $G(x)$ is differentiable on $[a,b]$ and that $G'(x)=g(x)$ for all $xin[a,b]$, or even that $G(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=g(x)$ for all $xin(a,b)$, then no such function $g(x)$ where $int_a^b g(x)dxneq G(b)-G(a)$ exists. You will note that this is true regardless of the continuity of $g$. This is forced by the fundamental theorem of calculus.
answered Jan 27 at 4:15
Harry QuHarry Qu
986
$endgroup$
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
add a comment |
$begingroup$
Without the condition that $g(x)$ be Riemann integrable, then such functions $g(x)$ do exist. Take for example the derivative of Volterra's function, $V'$, which admits a primitive $V$ differentiable everywhere, but $V'$ is not Riemann integrable, so technically $int_a^b V'(x)dxneq V(b)-V(a)$, since $int_a^b V'(x)dx$ does not exist.
With the condition that $g(x)$ be Riemann integrable, if we take your assertion that $g(x)$ admits a primitive $G(x)$ to mean that $G(x)$ is differentiable on $[a,b]$ and that $G'(x)=g(x)$ for all $xin[a,b]$, or even that $G(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=g(x)$ for all $xin(a,b)$, then no such function $g(x)$ where $int_a^b g(x)dxneq G(b)-G(a)$ exists. You will note that this is true regardless of the continuity of $g$. This is forced by the fundamental theorem of calculus.
answered Jan 27 at 4:15
Harry QuHarry Qu
986
$endgroup$
Without the condition that $g(x)$ be Riemann integrable, then such functions $g(x)$ do exist. Take for example the derivative of Volterra's function, $V'$, which admits a primitive $V$ differentiable everywhere, but $V'$ is not Riemann integrable, so technically $int_a^b V'(x)dxneq V(b)-V(a)$, since $int_a^b V'(x)dx$ does not exist.
With the condition that $g(x)$ be Riemann integrable, if we take your assertion that $g(x)$ admits a primitive $G(x)$ to mean that $G(x)$ is differentiable on $[a,b]$ and that $G'(x)=g(x)$ for all $xin[a,b]$, or even that $G(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=g(x)$ for all $xin(a,b)$, then no such function $g(x)$ where $int_a^b g(x)dxneq G(b)-G(a)$ exists. You will note that this is true regardless of the continuity of $g$. This is forced by the fundamental theorem of calculus.
answered Jan 27 at 4:15
Harry QuHarry Qu
986
answered Jan 27 at 4:15
Harry QuHarry Qu
986
answered Jan 27 at 4:15
Harry QuHarry Qu
986
answered Jan 27 at 4:15
Harry QuHarry Qu
986
986
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
add a comment |
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
$begingroup$
Thank you very much, Harry Qu for your information about Volterra's function. And my question was stupid, sorry.
$endgroup$
– tchappy ha
Jan 27 at 5:01
add a comment |
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