What's the derivative of an integral?
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2
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Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$ mathcal F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$mathcal F'(x) = mathcal f(x)$$
My question is the following: What will happen in this case?
$$mathcal H(x) = int_0^x e^{t^2} dt$$
Would the derivative be:
$$mathcal H'(x) = mathcal e^{x^2}$$
or
$$mathcal H'(x) = mathcal e^{x^2}-1$$
calculus
New contributor
add a comment |
up vote
2
down vote
favorite
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$ mathcal F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$mathcal F'(x) = mathcal f(x)$$
My question is the following: What will happen in this case?
$$mathcal H(x) = int_0^x e^{t^2} dt$$
Would the derivative be:
$$mathcal H'(x) = mathcal e^{x^2}$$
or
$$mathcal H'(x) = mathcal e^{x^2}-1$$
calculus
New contributor
Welcome to MSE; please proofread your question before posting, and use the formatting guide to fix this.
– T. Bongers
3 hours ago
2
Why would it be $mathcal H'(x) = e^{x^2} - 1$? Where would the $-1$ come from?
– littleO
3 hours ago
$e^{t^2}$ you replace t with x - $e^{t^2}$ you replace t with 0.
– Ryk
3 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$ mathcal F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$mathcal F'(x) = mathcal f(x)$$
My question is the following: What will happen in this case?
$$mathcal H(x) = int_0^x e^{t^2} dt$$
Would the derivative be:
$$mathcal H'(x) = mathcal e^{x^2}$$
or
$$mathcal H'(x) = mathcal e^{x^2}-1$$
calculus
New contributor
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$ mathcal F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$mathcal F'(x) = mathcal f(x)$$
My question is the following: What will happen in this case?
$$mathcal H(x) = int_0^x e^{t^2} dt$$
Would the derivative be:
$$mathcal H'(x) = mathcal e^{x^2}$$
or
$$mathcal H'(x) = mathcal e^{x^2}-1$$
calculus
calculus
New contributor
New contributor
edited 3 hours ago
Toby Mak
3,32811128
3,32811128
New contributor
asked 3 hours ago
Ryk
132
132
New contributor
New contributor
Welcome to MSE; please proofread your question before posting, and use the formatting guide to fix this.
– T. Bongers
3 hours ago
2
Why would it be $mathcal H'(x) = e^{x^2} - 1$? Where would the $-1$ come from?
– littleO
3 hours ago
$e^{t^2}$ you replace t with x - $e^{t^2}$ you replace t with 0.
– Ryk
3 hours ago
add a comment |
Welcome to MSE; please proofread your question before posting, and use the formatting guide to fix this.
– T. Bongers
3 hours ago
2
Why would it be $mathcal H'(x) = e^{x^2} - 1$? Where would the $-1$ come from?
– littleO
3 hours ago
$e^{t^2}$ you replace t with x - $e^{t^2}$ you replace t with 0.
– Ryk
3 hours ago
Welcome to MSE; please proofread your question before posting, and use the formatting guide to fix this.
– T. Bongers
3 hours ago
Welcome to MSE; please proofread your question before posting, and use the formatting guide to fix this.
– T. Bongers
3 hours ago
2
2
Why would it be $mathcal H'(x) = e^{x^2} - 1$? Where would the $-1$ come from?
– littleO
3 hours ago
Why would it be $mathcal H'(x) = e^{x^2} - 1$? Where would the $-1$ come from?
– littleO
3 hours ago
$e^{t^2}$ you replace t with x - $e^{t^2}$ you replace t with 0.
– Ryk
3 hours ago
$e^{t^2}$ you replace t with x - $e^{t^2}$ you replace t with 0.
– Ryk
3 hours ago
add a comment |
4 Answers
4
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up vote
1
down vote
accepted
If we use the idea in your comment (to the question) then $mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).
Part 1 of FTC deals with an integral of the form $int_{a} ^{x} f(t) , dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $mathcal{F} :[a, b] tomathbb{R} $ via the relation $$mathcal {F} (x) =int_{a} ^{x} f(t) , dttag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $mathcal{F} $ in terms of properties of $f$. And it says that $mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $cin[a, b] $ then $mathcal{F} $ is differentiable at $c$ and $mathcal{F} '(c) =f(c) $.
You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $mathcal{F}' (x) $ depends on $f$ and $x$ only.
Part 2 of FTC deals with the evaluation of $int_{a} ^{b} f(x) , dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $mathcal{F} $ so that $$mathcal{F} '(x) =f(x), forall xin[a, b] $$ and then says that $$int_{a} ^{b} f(x) , dx=mathcal{F} (b) - mathcal {F} (a) tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.
Note that the $mathcal{F} $ in both parts of FTC are different and in particular the $mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.
add a comment |
up vote
3
down vote
Note that $F(x+h)= F(x) + int_x^{x+h} f(t)dt$ and for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.
It is straightforward to make this argument rigorous.
1
Very nice (+1).
– gimusi
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
add a comment |
up vote
1
down vote
The simplest when applying a new formula is to identify each component:
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$F'(x) = mathcal f(x)$$
For $H(x) = int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.
It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $mathbb R$, so their composition $f=gcirc h$ is continuous on $mathbb R$, hence on $[a,b]$.
Now, we can safely apply the formula: $H'(x) = f(x)=mathcal e^{x^2}$.
Edit: to answer a comment
What would happen if $a$ is not $0$?
Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=int_1^{x} f(t); dt= int_1^x e^{t^2}; dt$. Then, all of the above applies here and we have
$$H_2'(x) = f(x) = e^{x^2} $$
Wait! Why do the functions $H_2$ and $H_1(x)=int^x_0 f(t); dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that
$$H_1(x)=int_0^x f(t); dt=int_0^1 f(t); dt + int_1^x f(t); dt= C + H_2(x)$$
where $C=int_0^1 f(t); dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.
What happens when a is not zero then ?
– Ryk
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
add a comment |
up vote
0
down vote
Recall that in general by Leibniz integral rule the following holds
$$F(x)=int_{a(x)}^{b(x)}g(u) duimplies F'(x)=g(b(x))cdot b'(x)-g(a(x))cdot a'(x)$$
therefore
$$mathcal H(x) = int_0^x e^{t^2} dtimplies mathcal H'(x)=e^{x^2}$$
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
1
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If we use the idea in your comment (to the question) then $mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).
Part 1 of FTC deals with an integral of the form $int_{a} ^{x} f(t) , dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $mathcal{F} :[a, b] tomathbb{R} $ via the relation $$mathcal {F} (x) =int_{a} ^{x} f(t) , dttag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $mathcal{F} $ in terms of properties of $f$. And it says that $mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $cin[a, b] $ then $mathcal{F} $ is differentiable at $c$ and $mathcal{F} '(c) =f(c) $.
You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $mathcal{F}' (x) $ depends on $f$ and $x$ only.
Part 2 of FTC deals with the evaluation of $int_{a} ^{b} f(x) , dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $mathcal{F} $ so that $$mathcal{F} '(x) =f(x), forall xin[a, b] $$ and then says that $$int_{a} ^{b} f(x) , dx=mathcal{F} (b) - mathcal {F} (a) tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.
Note that the $mathcal{F} $ in both parts of FTC are different and in particular the $mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.
add a comment |
up vote
1
down vote
accepted
If we use the idea in your comment (to the question) then $mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).
Part 1 of FTC deals with an integral of the form $int_{a} ^{x} f(t) , dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $mathcal{F} :[a, b] tomathbb{R} $ via the relation $$mathcal {F} (x) =int_{a} ^{x} f(t) , dttag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $mathcal{F} $ in terms of properties of $f$. And it says that $mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $cin[a, b] $ then $mathcal{F} $ is differentiable at $c$ and $mathcal{F} '(c) =f(c) $.
You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $mathcal{F}' (x) $ depends on $f$ and $x$ only.
Part 2 of FTC deals with the evaluation of $int_{a} ^{b} f(x) , dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $mathcal{F} $ so that $$mathcal{F} '(x) =f(x), forall xin[a, b] $$ and then says that $$int_{a} ^{b} f(x) , dx=mathcal{F} (b) - mathcal {F} (a) tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.
Note that the $mathcal{F} $ in both parts of FTC are different and in particular the $mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If we use the idea in your comment (to the question) then $mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).
Part 1 of FTC deals with an integral of the form $int_{a} ^{x} f(t) , dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $mathcal{F} :[a, b] tomathbb{R} $ via the relation $$mathcal {F} (x) =int_{a} ^{x} f(t) , dttag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $mathcal{F} $ in terms of properties of $f$. And it says that $mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $cin[a, b] $ then $mathcal{F} $ is differentiable at $c$ and $mathcal{F} '(c) =f(c) $.
You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $mathcal{F}' (x) $ depends on $f$ and $x$ only.
Part 2 of FTC deals with the evaluation of $int_{a} ^{b} f(x) , dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $mathcal{F} $ so that $$mathcal{F} '(x) =f(x), forall xin[a, b] $$ and then says that $$int_{a} ^{b} f(x) , dx=mathcal{F} (b) - mathcal {F} (a) tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.
Note that the $mathcal{F} $ in both parts of FTC are different and in particular the $mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.
If we use the idea in your comment (to the question) then $mathcal{F} '(x) $ should equal $f(x) - f(a) $ and not just $f(x) $ as mentioned in your question. I guess the confusion comes from mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC).
Part 1 of FTC deals with an integral of the form $int_{a} ^{x} f(t) , dt$ where the lower limit of integral is a constant $a$ and upper limit $x$ is a variable. This then defines a new function, say $mathcal{F} :[a, b] tomathbb{R} $ via the relation $$mathcal {F} (x) =int_{a} ^{x} f(t) , dttag{1}$$ The goal of part 1 of FTC is to study the properties of this new function $mathcal{F} $ in terms of properties of $f$. And it says that $mathcal{F} $ is continuous on $[a, b] $ and if $f$ is continuous at some $cin[a, b] $ then $mathcal{F} $ is differentiable at $c$ and $mathcal{F} '(c) =f(c) $.
You should notice that the lower limit $a$ does not figure out in conclusion of the theorem. The value $mathcal{F} (x) $ depends on $f, a$ and $x$ but the value $mathcal{F}' (x) $ depends on $f$ and $x$ only.
Part 2 of FTC deals with the evaluation of $int_{a} ^{b} f(x) , dx$ under certain conditions. It assumes that $f$ is Riemann integrable on $[a, b] $ and possesses an anti-derivative $mathcal{F} $ so that $$mathcal{F} '(x) =f(x), forall xin[a, b] $$ and then says that $$int_{a} ^{b} f(x) , dx=mathcal{F} (b) - mathcal {F} (a) tag{2}$$ It is here that both the upper and lower limits of integration play key role and the integral is expressed as difference between the values of the anti-derivative.
Note that the $mathcal{F} $ in both parts of FTC are different and in particular the $mathcal{F} $ in part 1 is not necessarily an anti-derivative of $f$.
edited 1 hour ago
answered 1 hour ago
Paramanand Singh
48.4k555156
48.4k555156
add a comment |
add a comment |
up vote
3
down vote
Note that $F(x+h)= F(x) + int_x^{x+h} f(t)dt$ and for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.
It is straightforward to make this argument rigorous.
1
Very nice (+1).
– gimusi
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
add a comment |
up vote
3
down vote
Note that $F(x+h)= F(x) + int_x^{x+h} f(t)dt$ and for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.
It is straightforward to make this argument rigorous.
1
Very nice (+1).
– gimusi
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Note that $F(x+h)= F(x) + int_x^{x+h} f(t)dt$ and for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.
It is straightforward to make this argument rigorous.
Note that $F(x+h)= F(x) + int_x^{x+h} f(t)dt$ and for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$. Hence we expect $F'(x) = f(x)$.
It is straightforward to make this argument rigorous.
answered 3 hours ago
copper.hat
125k559159
125k559159
1
Very nice (+1).
– gimusi
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
add a comment |
1
Very nice (+1).
– gimusi
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
1
1
Very nice (+1).
– gimusi
2 hours ago
Very nice (+1).
– gimusi
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
You said for small $h$ we have $int_x^{x+h} f(t)dt approx int_x^{x+h} f(x)dt = f(x) h$ why are you not getting rid of the h from the second integral and say something like this given that h is very small, approaching zero. $int_x^{x+h} f(t)dt approx int_x^x f(x)dt = f(x) h$
– Ryk
2 hours ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
@Ryk: You are trying to compute the limit as $h to 0$ of ${F(x+h)-F(x) over h}$. The above does is not divided across by $h$.
– copper.hat
26 mins ago
add a comment |
up vote
1
down vote
The simplest when applying a new formula is to identify each component:
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$F'(x) = mathcal f(x)$$
For $H(x) = int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.
It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $mathbb R$, so their composition $f=gcirc h$ is continuous on $mathbb R$, hence on $[a,b]$.
Now, we can safely apply the formula: $H'(x) = f(x)=mathcal e^{x^2}$.
Edit: to answer a comment
What would happen if $a$ is not $0$?
Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=int_1^{x} f(t); dt= int_1^x e^{t^2}; dt$. Then, all of the above applies here and we have
$$H_2'(x) = f(x) = e^{x^2} $$
Wait! Why do the functions $H_2$ and $H_1(x)=int^x_0 f(t); dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that
$$H_1(x)=int_0^x f(t); dt=int_0^1 f(t); dt + int_1^x f(t); dt= C + H_2(x)$$
where $C=int_0^1 f(t); dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.
What happens when a is not zero then ?
– Ryk
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
add a comment |
up vote
1
down vote
The simplest when applying a new formula is to identify each component:
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$F'(x) = mathcal f(x)$$
For $H(x) = int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.
It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $mathbb R$, so their composition $f=gcirc h$ is continuous on $mathbb R$, hence on $[a,b]$.
Now, we can safely apply the formula: $H'(x) = f(x)=mathcal e^{x^2}$.
Edit: to answer a comment
What would happen if $a$ is not $0$?
Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=int_1^{x} f(t); dt= int_1^x e^{t^2}; dt$. Then, all of the above applies here and we have
$$H_2'(x) = f(x) = e^{x^2} $$
Wait! Why do the functions $H_2$ and $H_1(x)=int^x_0 f(t); dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that
$$H_1(x)=int_0^x f(t); dt=int_0^1 f(t); dt + int_1^x f(t); dt= C + H_2(x)$$
where $C=int_0^1 f(t); dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.
What happens when a is not zero then ?
– Ryk
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
The simplest when applying a new formula is to identify each component:
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$F'(x) = mathcal f(x)$$
For $H(x) = int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.
It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $mathbb R$, so their composition $f=gcirc h$ is continuous on $mathbb R$, hence on $[a,b]$.
Now, we can safely apply the formula: $H'(x) = f(x)=mathcal e^{x^2}$.
Edit: to answer a comment
What would happen if $a$ is not $0$?
Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=int_1^{x} f(t); dt= int_1^x e^{t^2}; dt$. Then, all of the above applies here and we have
$$H_2'(x) = f(x) = e^{x^2} $$
Wait! Why do the functions $H_2$ and $H_1(x)=int^x_0 f(t); dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that
$$H_1(x)=int_0^x f(t); dt=int_0^1 f(t); dt + int_1^x f(t); dt= C + H_2(x)$$
where $C=int_0^1 f(t); dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.
The simplest when applying a new formula is to identify each component:
Let f be a continuous function on the interval $[a, b]$. The function F defined by
$$F(x) = int_a^x f(t)dt $$
is continuous on $[a,b]$, differentiable on $(a,b)$ and has derivative
$$F'(x) = mathcal f(x)$$
For $H(x) = int_0^x e^{t^2} dt$ we have $a=0$ (lower limit of integration) and $f(t)=e^{t^2}$.
It is also important to check that all conditions of the theorem are satisfied: here, the functions $g(t)=e^{t}$ and $h(t)=t^2$ are continuous on $mathbb R$, so their composition $f=gcirc h$ is continuous on $mathbb R$, hence on $[a,b]$.
Now, we can safely apply the formula: $H'(x) = f(x)=mathcal e^{x^2}$.
Edit: to answer a comment
What would happen if $a$ is not $0$?
Note that the formula depends only on $f$ and its continuity and not really on $a$. For example, consider $H_2(x)=int_1^{x} f(t); dt= int_1^x e^{t^2}; dt$. Then, all of the above applies here and we have
$$H_2'(x) = f(x) = e^{x^2} $$
Wait! Why do the functions $H_2$ and $H_1(x)=int^x_0 f(t); dt$ have the same derivative $f(x)=e^{x^2}$? That's clear when you remark that
$$H_1(x)=int_0^x f(t); dt=int_0^1 f(t); dt + int_1^x f(t); dt= C + H_2(x)$$
where $C=int_0^1 f(t); dt$. Since $C$ is a constant, we have $H_1'(x)=H_2'(x)$.
edited 2 hours ago
answered 3 hours ago
Taladris
4,63731832
4,63731832
What happens when a is not zero then ?
– Ryk
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
add a comment |
What happens when a is not zero then ?
– Ryk
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
What happens when a is not zero then ?
– Ryk
2 hours ago
What happens when a is not zero then ?
– Ryk
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
Nice and simple approach! (+1)
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
@Ryk For the general case refer to the link I've given for Leibniz's rule.
– gimusi
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
Thank you, super approach! @Taladris
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
@gimusi, I will take a look into it, thank you!
– Ryk
2 hours ago
add a comment |
up vote
0
down vote
Recall that in general by Leibniz integral rule the following holds
$$F(x)=int_{a(x)}^{b(x)}g(u) duimplies F'(x)=g(b(x))cdot b'(x)-g(a(x))cdot a'(x)$$
therefore
$$mathcal H(x) = int_0^x e^{t^2} dtimplies mathcal H'(x)=e^{x^2}$$
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
1
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
add a comment |
up vote
0
down vote
Recall that in general by Leibniz integral rule the following holds
$$F(x)=int_{a(x)}^{b(x)}g(u) duimplies F'(x)=g(b(x))cdot b'(x)-g(a(x))cdot a'(x)$$
therefore
$$mathcal H(x) = int_0^x e^{t^2} dtimplies mathcal H'(x)=e^{x^2}$$
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
1
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Recall that in general by Leibniz integral rule the following holds
$$F(x)=int_{a(x)}^{b(x)}g(u) duimplies F'(x)=g(b(x))cdot b'(x)-g(a(x))cdot a'(x)$$
therefore
$$mathcal H(x) = int_0^x e^{t^2} dtimplies mathcal H'(x)=e^{x^2}$$
Recall that in general by Leibniz integral rule the following holds
$$F(x)=int_{a(x)}^{b(x)}g(u) duimplies F'(x)=g(b(x))cdot b'(x)-g(a(x))cdot a'(x)$$
therefore
$$mathcal H(x) = int_0^x e^{t^2} dtimplies mathcal H'(x)=e^{x^2}$$
answered 3 hours ago
gimusi
89.1k74495
89.1k74495
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
1
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
add a comment |
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
1
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
+1, very nice answer. I was not aware about Leibniz integral rule. Maybe I used it many times, but I was not aware that it was due to Leibniz.
– the_candyman
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
@the_candyman Thanks, much appreciative! Bye
– gimusi
2 hours ago
1
1
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
That's correct but I feel that's killing a bird with a bazooka. The OP has obviously problems with the application of a simple formula, why would a more general formula be helpful?
– Taladris
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
@Taladris Yes you are right but also I want to give a more general reference for more general cases.
– gimusi
2 hours ago
add a comment |
Ryk is a new contributor. Be nice, and check out our Code of Conduct.
Ryk is a new contributor. Be nice, and check out our Code of Conduct.
Ryk is a new contributor. Be nice, and check out our Code of Conduct.
Ryk is a new contributor. Be nice, and check out our Code of Conduct.
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– T. Bongers
3 hours ago
2
Why would it be $mathcal H'(x) = e^{x^2} - 1$? Where would the $-1$ come from?
– littleO
3 hours ago
$e^{t^2}$ you replace t with x - $e^{t^2}$ you replace t with 0.
– Ryk
3 hours ago