Proof of an integral property
$begingroup$
$$int_0^1 f(x)g'(x)dx=pi$$ 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
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$endgroup$
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 2 hours ago
Eevee Trainer
6,39311237
asked 2 hours ago
adam hanyadam hany
161
New contributor
adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago
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 2 hours ago
Eevee Trainer
6,39311237
asked 2 hours ago
adam hanyadam hany
161
New contributor
adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$$int_0^1 f(x)g'(x)dx=pi$$ 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 2 hours ago
Eevee Trainer
6,39311237
asked 2 hours ago
adam hanyadam hany
161
New contributor
adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago
Eevee Trainer
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asked 2 hours ago
adam hanyadam hany
161
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adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago
Eevee Trainer
6,39311237
edited 2 hours ago
Eevee Trainer
6,39311237
edited 2 hours ago
Eevee Trainer
6,39311237
6,39311237
asked 2 hours ago
adam hanyadam hany
161
New contributor
adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago
adam hanyadam hany
161
asked 2 hours ago
adam hanyadam hany
161
161
New contributor
adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
adam hany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
5
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago
add a comment |
5
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago
5
5
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
$endgroup$
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
add a comment |
$begingroup$
Hint
$dfrac{d(f(x)cdot g(x))}{dx}=?$
Integrate both sides wrt $x$ between $[0,1]$
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
$endgroup$
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
add a comment |
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
$endgroup$
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
add a comment |
$begingroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
$endgroup$
Hint:
Utilize integration by parts:
$$int f(x)g'(x)dx = f(x)g(x) - int f'(x)g(x)$$
If we have a definite integral, then this formula becomes
$$int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - int_a^b f'(x)g(x)$$
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
answered 2 hours ago
Eevee TrainerEevee Trainer
6,39311237
6,39311237
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
add a comment |
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
1
1
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
$begingroup$
thank you so much for the answer ∫baf(x)g′(x)dx=−∫baf′(x)g(x)
$endgroup$
– adam hany
2 hours ago
add a comment |
$begingroup$
Hint
$dfrac{d(f(x)cdot g(x))}{dx}=?$
Integrate both sides wrt $x$ between $[0,1]$
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Hint
$dfrac{d(f(x)cdot g(x))}{dx}=?$
Integrate both sides wrt $x$ between $[0,1]$
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Hint
$dfrac{d(f(x)cdot g(x))}{dx}=?$
Integrate both sides wrt $x$ between $[0,1]$
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint
$dfrac{d(f(x)cdot g(x))}{dx}=?$
Integrate both sides wrt $x$ between $[0,1]$
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
answered 2 hours ago
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
adam hany is a new contributor. Be nice, and check out our Code of Conduct.
adam hany is a new contributor. Be nice, and check out our Code of Conduct.
adam hany is a new contributor. Be nice, and check out our Code of Conduct.
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5
$begingroup$
Do you know integration by parts?
$endgroup$
– Minus One-Twelfth
2 hours ago