Mixing
Fundamental Theorem of Calculus with Functions
$begingroup$
Please excuse the weird title. I was unsure about how to summarize this.
So lets say we have this integral:
$ int_{kx}^{cx} t^2 dt = f(x)$
If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$
or
$ int_{2x}^{2x} t^2 dt = f(2x)$
Edit:
If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$
calculus integration definite-integrals
asked Jan 29 at 5:17
user2793618user2793618
1047
$endgroup$
add a comment |
$begingroup$
Please excuse the weird title. I was unsure about how to summarize this.
So lets say we have this integral:
$ int_{kx}^{cx} t^2 dt = f(x)$
If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$
or
$ int_{2x}^{2x} t^2 dt = f(2x)$
Edit:
If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$
calculus integration definite-integrals
asked Jan 29 at 5:17
user2793618user2793618
1047
$endgroup$
$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26
$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26
$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27
$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29
$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40
add a comment |
$begingroup$
Please excuse the weird title. I was unsure about how to summarize this.
So lets say we have this integral:
$ int_{kx}^{cx} t^2 dt = f(x)$
If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$
or
$ int_{2x}^{2x} t^2 dt = f(2x)$
Edit:
If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$
calculus integration definite-integrals
asked Jan 29 at 5:17
user2793618user2793618
1047
$endgroup$
Please excuse the weird title. I was unsure about how to summarize this.
So lets say we have this integral:
$ int_{kx}^{cx} t^2 dt = f(x)$
If we wanted to apply $f(x)$ with $2x$ instead of $x$ then would the integral bound change like so:
$ int_{2xk}^{2xc} t^2 dt = f(2x)$
or
$ int_{2x}^{2x} t^2 dt = f(2x)$
Edit:
If the above is true, then would differentiating f result like so:
$-(2xk)^2 + (2xc)^2 = f'(2x)$
calculus integration definite-integrals
calculus integration definite-integrals
asked Jan 29 at 5:17
user2793618user2793618
1047
asked Jan 29 at 5:17
user2793618user2793618
1047
edited Jan 29 at 5:30
user2793618
asked Jan 29 at 5:17
user2793618user2793618
1047
asked Jan 29 at 5:17
user2793618user2793618
1047
asked Jan 29 at 5:17
user2793618user2793618
1047
1047
$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26
$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26
$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27
$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29
$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40
add a comment |
$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26
$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26
$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27
$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29
$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40
$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26
$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26
$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26
$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26
$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27
$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27
$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29
$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29
$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40
$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$f(2x)=int_{2xk}^{2xc} t^2 dt$
But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.
The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$
But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
$f(2x)=int_{2xk}^{2xc} t^2 dt$
But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.
The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$
But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
$endgroup$
add a comment |
$begingroup$
$f(2x)=int_{2xk}^{2xc} t^2 dt$
But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.
The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$
But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
$endgroup$
add a comment |
$begingroup$
$f(2x)=int_{2xk}^{2xc} t^2 dt$
But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.
The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$
But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
$endgroup$
$f(2x)=int_{2xk}^{2xc} t^2 dt$
But the derivative is wrong, since the boundaries of the integral both contain the variable $x$.
The lower bound must be constant to apply the rule that you applied, i.e. $$frac{d}{dx} int_{a}^{x} f(t) dt = f(x)$$
But instead you can use this $$frac{d}{dx}int_{g(x)}^{f(x)}h(t),dt=h(f(x))cdot f'(x)-h(g(x))cdot g'(x) $$
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
edited Feb 1 at 9:17
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
answered Jan 29 at 6:38
Fareed AFFareed AF
63212
63212
add a comment |
add a comment |
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$begingroup$
what is wrong with the result? Please explain your problem a little bit more detailed.
$endgroup$
– Evan William Chandra
Jan 29 at 5:26
$begingroup$
What's your question?
$endgroup$
– Stefan Lafon
Jan 29 at 5:26
$begingroup$
I'm just asking to confirm if what I have is right.
$endgroup$
– user2793618
Jan 29 at 5:27
$begingroup$
I've updated for clarity.
$endgroup$
– user2793618
Jan 29 at 5:29
$begingroup$
In your last integral you have both your bounds being $2x$; is this okay? I mean, wouldn't that integral be $0$ for all $x$?
$endgroup$
– Ryan Goulden
Jan 29 at 7:40