Proof of an integral property [closed]
Clash Royale CLAN TAG#URR8PPP
$begingroup$
$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$
calculus integration definite-integrals
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
asked Feb 21 at 1:36
adam hanyadam hany
212
$endgroup$
closed as off-topic by RRL, stressed out, Song, Cesareo, Kemono Chen Feb 22 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, stressed out, Song, Cesareo, Kemono Chen
add a comment |
$begingroup$
$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$
calculus integration definite-integrals
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
asked Feb 21 at 1:36
adam hanyadam hany
212
$endgroup$
closed as off-topic by RRL, stressed out, Song, Cesareo, Kemono Chen Feb 22 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, stressed out, Song, Cesareo, Kemono Chen
8
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
Feb 21 at 1:37
add a comment |
$begingroup$
$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$
calculus integration definite-integrals
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
asked Feb 21 at 1:36
adam hanyadam hany
212
$endgroup$
$$int_0^1 f(x)g'(x)dx=pi$$ If $f(1)g(1)=f(0)g(0)$ then $$int_0^1 f'(x)g(x)dx= -pi$$
So I have to prove this and I have absolutely no idea how to do it. I am guessing I will have to use the fundamental theorem of calculus and it show that the rate of change is $1$ because it didn't change from $f(1)g(1)$ to $f(0)g(0)$
calculus integration definite-integrals
calculus integration definite-integrals
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
asked Feb 21 at 1:36
adam hanyadam hany
212
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
asked Feb 21 at 1:36
adam hanyadam hany
212
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
edited Feb 21 at 1:41
Eevee Trainer
7,96821439
7,96821439
asked Feb 21 at 1:36
adam hanyadam hany
212
asked Feb 21 at 1:36
adam hanyadam hany
212
asked Feb 21 at 1:36
adam hanyadam hany
212
212
closed as off-topic by RRL, stressed out, Song, Cesareo, Kemono Chen Feb 22 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, stressed out, Song, Cesareo, Kemono Chen
closed as off-topic by RRL, stressed out, Song, Cesareo, Kemono Chen Feb 22 at 0:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, stressed out, Song, Cesareo, Kemono Chen
8
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
Feb 21 at 1:37
add a comment |
8
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
Feb 21 at 1:37
8
8
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
Feb 21 at 1:37
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
Feb 21 at 1:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)mathrm dx = f(x)g(x) - int f'(x)g(x) mathrm dx$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)mathrm dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x) mathrm dx$$
edited Feb 21 at 6:42
mrtaurho
6,00551641
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
$endgroup$
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
add a comment |
$begingroup$
Hint:
$$dfracd(f(x)cdot g(x))dx=?$$
Integrate both sides with respect to $x$ between $[0,1]$
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)mathrm dx = f(x)g(x) - int f'(x)g(x) mathrm dx$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)mathrm dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x) mathrm dx$$
edited Feb 21 at 6:42
mrtaurho
6,00551641
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
$endgroup$
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
add a comment |
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)mathrm dx = f(x)g(x) - int f'(x)g(x) mathrm dx$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)mathrm dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x) mathrm dx$$
edited Feb 21 at 6:42
mrtaurho
6,00551641
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
$endgroup$
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
add a comment |
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)mathrm dx = f(x)g(x) - int f'(x)g(x) mathrm dx$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)mathrm dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x) mathrm dx$$
edited Feb 21 at 6:42
mrtaurho
6,00551641
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
$endgroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)mathrm dx = f(x)g(x) - int f'(x)g(x) mathrm dx$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)mathrm dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x) mathrm dx$$
edited Feb 21 at 6:42
mrtaurho
6,00551641
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
edited Feb 21 at 6:42
mrtaurho
6,00551641
edited Feb 21 at 6:42
mrtaurho
6,00551641
edited Feb 21 at 6:42
mrtaurho
6,00551641
6,00551641
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
answered Feb 21 at 1:39
Eevee TrainerEevee Trainer
7,96821439
7,96821439
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
add a comment |
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
1
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
Feb 21 at 2:01
add a comment |
$begingroup$
Hint:
$$dfracd(f(x)cdot g(x))dx=?$$
Integrate both sides with respect to $x$ between $[0,1]$
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
$$dfracd(f(x)cdot g(x))dx=?$$
Integrate both sides with respect to $x$ between $[0,1]$
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
$$dfracd(f(x)cdot g(x))dx=?$$
Integrate both sides with respect to $x$ between $[0,1]$
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
Hint:
$$dfracd(f(x)cdot g(x))dx=?$$
Integrate both sides with respect to $x$ between $[0,1]$
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
edited Feb 21 at 6:09
Eevee Trainer
7,96821439
7,96821439
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
answered Feb 21 at 2:05
lab bhattacharjeelab bhattacharjee
227k15158275
227k15158275
add a comment |
add a comment |
8
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
Feb 21 at 1:37