Fundamental Theorem of Calculus: Derivative of integral with respect to parameter in integration limits
$begingroup$
I have the following function:
$$F(t) = int_{a-t}^{b+t} f(x) dx$$
In my notes when this is differentiated with respect to $t$ we get:
$$frac{d F}{d t} = f(b + t) + f(a - t)$$
How can we obtain this result?
My approach
The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:
$$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$
So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.
calculus integration ordinary-differential-equations derivatives
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
I have the following function:
$$F(t) = int_{a-t}^{b+t} f(x) dx$$
In my notes when this is differentiated with respect to $t$ we get:
$$frac{d F}{d t} = f(b + t) + f(a - t)$$
How can we obtain this result?
My approach
The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:
$$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$
So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.
calculus integration ordinary-differential-equations derivatives
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
I have the following function:
$$F(t) = int_{a-t}^{b+t} f(x) dx$$
In my notes when this is differentiated with respect to $t$ we get:
$$frac{d F}{d t} = f(b + t) + f(a - t)$$
How can we obtain this result?
My approach
The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:
$$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$
So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.
calculus integration ordinary-differential-equations derivatives
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
$endgroup$
I have the following function:
$$F(t) = int_{a-t}^{b+t} f(x) dx$$
In my notes when this is differentiated with respect to $t$ we get:
$$frac{d F}{d t} = f(b + t) + f(a - t)$$
How can we obtain this result?
My approach
The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:
$$F_1(t) = int_{a-t}^c f(x) dx quad quad F_2(t) = int_{c}^{b+t} f(x)dx$$
So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.
calculus integration ordinary-differential-equations derivatives
calculus integration ordinary-differential-equations derivatives
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
asked Dec 1 '18 at 19:48
Euler_SalterEuler_Salter
2,0671336
2,0671336
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You're on the right track. You can proceed by using the chain rule.
Observe that (with your notation)
$$ F_1(t) = G(H(t)), $$
where
$$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
Now
$$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
where we invoked FTC when differentiating $G$.
Hence, the chain rule gives
$$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
Can you deal with $F_2$ on your own?
P.S. You could also directly refer to the more general Leibniz integral rule.
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
$endgroup$
1
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
add a comment |
$begingroup$
If we have a function defined by $$Q'(x)=f(x)$$
Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
$$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
$$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
$$F'(t)=Q'(b+t)+Q'(a-t)$$
$$F'(t)=f(b+t)+f(a-t)$$
QED
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
$endgroup$
1
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
2
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
1
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
add a comment |
$begingroup$
$$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
$endgroup$
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
1
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're on the right track. You can proceed by using the chain rule.
Observe that (with your notation)
$$ F_1(t) = G(H(t)), $$
where
$$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
Now
$$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
where we invoked FTC when differentiating $G$.
Hence, the chain rule gives
$$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
Can you deal with $F_2$ on your own?
P.S. You could also directly refer to the more general Leibniz integral rule.
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
$endgroup$
1
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
add a comment |
$begingroup$
You're on the right track. You can proceed by using the chain rule.
Observe that (with your notation)
$$ F_1(t) = G(H(t)), $$
where
$$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
Now
$$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
where we invoked FTC when differentiating $G$.
Hence, the chain rule gives
$$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
Can you deal with $F_2$ on your own?
P.S. You could also directly refer to the more general Leibniz integral rule.
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
$endgroup$
1
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
add a comment |
$begingroup$
You're on the right track. You can proceed by using the chain rule.
Observe that (with your notation)
$$ F_1(t) = G(H(t)), $$
where
$$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
Now
$$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
where we invoked FTC when differentiating $G$.
Hence, the chain rule gives
$$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
Can you deal with $F_2$ on your own?
P.S. You could also directly refer to the more general Leibniz integral rule.
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
$endgroup$
You're on the right track. You can proceed by using the chain rule.
Observe that (with your notation)
$$ F_1(t) = G(H(t)), $$
where
$$ H(t) = a-t, quad text{and} quad G(t) = int_t^c f(x) , mathrm dx = -int_c^t f(x) , mathrm dx. $$
Now
$$ H'(t) = -1, quad text{and} quad G'(t) = -f(t), $$
where we invoked FTC when differentiating $G$.
Hence, the chain rule gives
$$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
Can you deal with $F_2$ on your own?
P.S. You could also directly refer to the more general Leibniz integral rule.
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
edited Dec 1 '18 at 20:03
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
answered Dec 1 '18 at 19:52
MisterRiemannMisterRiemann
5,8021625
5,8021625
1
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
add a comment |
1
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
1
1
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
Thanks for directing me to Leibniz integral rule, seems super handy!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Indeed!
$endgroup$
– MisterRiemann
Dec 1 '18 at 20:32
add a comment |
$begingroup$
If we have a function defined by $$Q'(x)=f(x)$$
Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
$$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
$$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
$$F'(t)=Q'(b+t)+Q'(a-t)$$
$$F'(t)=f(b+t)+f(a-t)$$
QED
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
$endgroup$
1
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
2
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
1
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
add a comment |
$begingroup$
If we have a function defined by $$Q'(x)=f(x)$$
Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
$$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
$$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
$$F'(t)=Q'(b+t)+Q'(a-t)$$
$$F'(t)=f(b+t)+f(a-t)$$
QED
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
$endgroup$
1
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
2
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
1
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
add a comment |
$begingroup$
If we have a function defined by $$Q'(x)=f(x)$$
Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
$$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
$$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
$$F'(t)=Q'(b+t)+Q'(a-t)$$
$$F'(t)=f(b+t)+f(a-t)$$
QED
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
$endgroup$
If we have a function defined by $$Q'(x)=f(x)$$
Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
Taking $frac{mathrm d}{mathrm dt}(cdot)$ on both sides:
$$F'(t)=frac{mathrm d}{mathrm dt}Q(b+t)-frac{mathrm d}{mathrm dt}Q(a-t)$$
$$F'(t)=Q'(b+t)frac{mathrm d}{mathrm dt}(b+t)-Q'(a-t)frac{mathrm d}{mathrm dt}(a-t)$$
$$F'(t)=Q'(b+t)+Q'(a-t)$$
$$F'(t)=f(b+t)+f(a-t)$$
QED
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
answered Dec 1 '18 at 20:23
clathratusclathratus
4,555337
4,555337
1
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
2
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
1
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
add a comment |
1
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
2
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
1
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
1
1
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
$begingroup$
This is amazingly simple!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:29
2
2
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
$begingroup$
@Euler_Salter Thank you very much. I have often found that many seemingly complex problems in calculus have amazingly simple results.
$endgroup$
– clathratus
Dec 1 '18 at 20:32
1
1
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
$begingroup$
True! I think in this case I was just subconsciously deceived by the usual notation F on the LHS of the Fundamental Theorem of Calculus
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:34
add a comment |
$begingroup$
$$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
$endgroup$
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
1
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
add a comment |
$begingroup$
$$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
$endgroup$
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
1
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
add a comment |
$begingroup$
$$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
$endgroup$
$$frac{dF}{dt}=frac{d(b+t)}{dt}frac{d}{d(b+t)}int_c^{b+t}f(x)dx-frac{d(a-t)}{dt}frac{d}{d(a-t)}int_c^{a-t}f(x)dx\=1cdot f(b+t)-(-1)cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
edited Dec 1 '18 at 20:29
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
answered Dec 1 '18 at 20:16
J.G.J.G.
26.7k22742
26.7k22742
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
1
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
add a comment |
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
1
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
$begingroup$
can you please state which results you're using?
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:23
1
1
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
@Euler_Salter See edit.
$endgroup$
– J.G.
Dec 1 '18 at 20:29
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
$begingroup$
Thank you for explaining it more!
$endgroup$
– Euler_Salter
Dec 1 '18 at 20:30
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