How to find the derivative of a definite integral that has unusual lower and upper bounds?












-1














I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.



The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $frac{d}{dx} int f(x)dx = f(x)$, i.e. the first part of the fundamental theorem of calculus, to answer the question.



The function is:



$$int_{sqrt{x}}^{pi/4} theta cdot tantheta cdot dtheta = g(x)$$



In words: if a function corresponds to an integral where the upper bound on the integral is $pi/4$, the lower bound is $sqrt{x}$, and the function being integrated is $theta cdot tantheta$ with respect to $theta$, then what is the derivative of the function?



I've tried setting $u = pi/4$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $sqrt{x}$, in my solution somehow, but I have no idea how.



Any help would be greatly appreciated.










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  • 3




    Hint: chain rule.
    – GEdgar
    Nov 24 at 22:34










  • As a suggested reading, check out Leibniz Rule for differentiation of integrals.
    – Shubham Johri
    Nov 24 at 22:44










  • Hint: $int_{-infty}^x f(x) dx = f(x)$, so $int_a^bf(x) dx = int_{-infty}^b f(x) dx - int_{-infty}^a f(x) dx$
    – eSurfsnake
    Nov 25 at 7:19


















-1














I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.



The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $frac{d}{dx} int f(x)dx = f(x)$, i.e. the first part of the fundamental theorem of calculus, to answer the question.



The function is:



$$int_{sqrt{x}}^{pi/4} theta cdot tantheta cdot dtheta = g(x)$$



In words: if a function corresponds to an integral where the upper bound on the integral is $pi/4$, the lower bound is $sqrt{x}$, and the function being integrated is $theta cdot tantheta$ with respect to $theta$, then what is the derivative of the function?



I've tried setting $u = pi/4$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $sqrt{x}$, in my solution somehow, but I have no idea how.



Any help would be greatly appreciated.










share|cite|improve this question




















  • 3




    Hint: chain rule.
    – GEdgar
    Nov 24 at 22:34










  • As a suggested reading, check out Leibniz Rule for differentiation of integrals.
    – Shubham Johri
    Nov 24 at 22:44










  • Hint: $int_{-infty}^x f(x) dx = f(x)$, so $int_a^bf(x) dx = int_{-infty}^b f(x) dx - int_{-infty}^a f(x) dx$
    – eSurfsnake
    Nov 25 at 7:19
















-1












-1








-1







I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.



The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $frac{d}{dx} int f(x)dx = f(x)$, i.e. the first part of the fundamental theorem of calculus, to answer the question.



The function is:



$$int_{sqrt{x}}^{pi/4} theta cdot tantheta cdot dtheta = g(x)$$



In words: if a function corresponds to an integral where the upper bound on the integral is $pi/4$, the lower bound is $sqrt{x}$, and the function being integrated is $theta cdot tantheta$ with respect to $theta$, then what is the derivative of the function?



I've tried setting $u = pi/4$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $sqrt{x}$, in my solution somehow, but I have no idea how.



Any help would be greatly appreciated.










share|cite|improve this question















I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.



The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $frac{d}{dx} int f(x)dx = f(x)$, i.e. the first part of the fundamental theorem of calculus, to answer the question.



The function is:



$$int_{sqrt{x}}^{pi/4} theta cdot tantheta cdot dtheta = g(x)$$



In words: if a function corresponds to an integral where the upper bound on the integral is $pi/4$, the lower bound is $sqrt{x}$, and the function being integrated is $theta cdot tantheta$ with respect to $theta$, then what is the derivative of the function?



I've tried setting $u = pi/4$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $sqrt{x}$, in my solution somehow, but I have no idea how.



Any help would be greatly appreciated.







calculus integration definite-integrals






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edited Nov 24 at 22:42









Eevee Trainer

4,3531631




4,3531631










asked Nov 24 at 22:30









James Ronald

977




977








  • 3




    Hint: chain rule.
    – GEdgar
    Nov 24 at 22:34










  • As a suggested reading, check out Leibniz Rule for differentiation of integrals.
    – Shubham Johri
    Nov 24 at 22:44










  • Hint: $int_{-infty}^x f(x) dx = f(x)$, so $int_a^bf(x) dx = int_{-infty}^b f(x) dx - int_{-infty}^a f(x) dx$
    – eSurfsnake
    Nov 25 at 7:19
















  • 3




    Hint: chain rule.
    – GEdgar
    Nov 24 at 22:34










  • As a suggested reading, check out Leibniz Rule for differentiation of integrals.
    – Shubham Johri
    Nov 24 at 22:44










  • Hint: $int_{-infty}^x f(x) dx = f(x)$, so $int_a^bf(x) dx = int_{-infty}^b f(x) dx - int_{-infty}^a f(x) dx$
    – eSurfsnake
    Nov 25 at 7:19










3




3




Hint: chain rule.
– GEdgar
Nov 24 at 22:34




Hint: chain rule.
– GEdgar
Nov 24 at 22:34












As a suggested reading, check out Leibniz Rule for differentiation of integrals.
– Shubham Johri
Nov 24 at 22:44




As a suggested reading, check out Leibniz Rule for differentiation of integrals.
– Shubham Johri
Nov 24 at 22:44












Hint: $int_{-infty}^x f(x) dx = f(x)$, so $int_a^bf(x) dx = int_{-infty}^b f(x) dx - int_{-infty}^a f(x) dx$
– eSurfsnake
Nov 25 at 7:19






Hint: $int_{-infty}^x f(x) dx = f(x)$, so $int_a^bf(x) dx = int_{-infty}^b f(x) dx - int_{-infty}^a f(x) dx$
– eSurfsnake
Nov 25 at 7:19












2 Answers
2






active

oldest

votes


















5














So,$$g(x)=int_{sqrt x}^{fracpi4}thetatantheta,mathrm dtheta=-int_{fracpi4}^{sqrt x}thetatantheta,mathrm dtheta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?






share|cite|improve this answer





















  • Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
    – James Ronald
    Nov 24 at 22:42










  • No, the lower bound does not matter.
    – José Carlos Santos
    Nov 24 at 22:45



















1














After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $x$ and the lower limit is not a constant.



So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.



Suppose the lower limit is $L(x)$ and the upper limit is $U(x)$ of the integral. Define



$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}ttext{.}$$



Suppose $F$ is an antiderivative of $f$. By part II of the Fundamental Theorem of Calculus, you know that
$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}t = F(U(x)) - F(L(x))text{.}$$
Then, the derivative of $g$ is given by, assuming differentiability of $U$ and $L$,



$$dfrac{text{d}}{text{d}x}[g(x)] = F^{prime}(U(x))U^{prime}(x)-F^{prime}(L(x))L^{prime}(x)$$
after making use of the chain rule for derivatives. But, $F$ is an antiderivative of $f$, so $F^{prime} = f$, hence
$$dfrac{text{d}}{text{d}x}[g(x)] = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}$$
In other words, the main result is
$$boxed{ dfrac{text{d}}{text{d}x}int_{L(x)}^{U(x)}f(t)text{ d}t = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}}$$





Applying to this problem, we have $f(theta) = theta tan(theta)$, $U(x) = dfrac{pi}{4}$, and $L(x) = sqrt{x}$. The derivatives are $U^{prime}(x) = 0$ and $L^{prime}(x) = dfrac{1}{2sqrt{x}}$. Hence, the derivative of $g$ is
$$g^{prime}(x) = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x) = fleft(dfrac{pi}{4}right)(0) - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}}$$
which simplifies to
$$g^{prime}(x) = - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}} = -sqrt{x}tan(sqrt{x}) cdot dfrac{1}{2sqrt{x}} = -dfrac{1}{2}tan(sqrt{x})text{.}$$






share|cite|improve this answer





















  • Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
    – James Ronald
    Nov 24 at 23:11











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2 Answers
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2 Answers
2






active

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active

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active

oldest

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5














So,$$g(x)=int_{sqrt x}^{fracpi4}thetatantheta,mathrm dtheta=-int_{fracpi4}^{sqrt x}thetatantheta,mathrm dtheta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?






share|cite|improve this answer





















  • Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
    – James Ronald
    Nov 24 at 22:42










  • No, the lower bound does not matter.
    – José Carlos Santos
    Nov 24 at 22:45
















5














So,$$g(x)=int_{sqrt x}^{fracpi4}thetatantheta,mathrm dtheta=-int_{fracpi4}^{sqrt x}thetatantheta,mathrm dtheta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?






share|cite|improve this answer





















  • Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
    – James Ronald
    Nov 24 at 22:42










  • No, the lower bound does not matter.
    – José Carlos Santos
    Nov 24 at 22:45














5












5








5






So,$$g(x)=int_{sqrt x}^{fracpi4}thetatantheta,mathrm dtheta=-int_{fracpi4}^{sqrt x}thetatantheta,mathrm dtheta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?






share|cite|improve this answer












So,$$g(x)=int_{sqrt x}^{fracpi4}thetatantheta,mathrm dtheta=-int_{fracpi4}^{sqrt x}thetatantheta,mathrm dtheta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 22:34









José Carlos Santos

150k22120221




150k22120221












  • Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
    – James Ronald
    Nov 24 at 22:42










  • No, the lower bound does not matter.
    – José Carlos Santos
    Nov 24 at 22:45


















  • Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
    – James Ronald
    Nov 24 at 22:42










  • No, the lower bound does not matter.
    – José Carlos Santos
    Nov 24 at 22:45
















Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
– James Ronald
Nov 24 at 22:42




Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC?
– James Ronald
Nov 24 at 22:42












No, the lower bound does not matter.
– José Carlos Santos
Nov 24 at 22:45




No, the lower bound does not matter.
– José Carlos Santos
Nov 24 at 22:45











1














After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $x$ and the lower limit is not a constant.



So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.



Suppose the lower limit is $L(x)$ and the upper limit is $U(x)$ of the integral. Define



$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}ttext{.}$$



Suppose $F$ is an antiderivative of $f$. By part II of the Fundamental Theorem of Calculus, you know that
$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}t = F(U(x)) - F(L(x))text{.}$$
Then, the derivative of $g$ is given by, assuming differentiability of $U$ and $L$,



$$dfrac{text{d}}{text{d}x}[g(x)] = F^{prime}(U(x))U^{prime}(x)-F^{prime}(L(x))L^{prime}(x)$$
after making use of the chain rule for derivatives. But, $F$ is an antiderivative of $f$, so $F^{prime} = f$, hence
$$dfrac{text{d}}{text{d}x}[g(x)] = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}$$
In other words, the main result is
$$boxed{ dfrac{text{d}}{text{d}x}int_{L(x)}^{U(x)}f(t)text{ d}t = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}}$$





Applying to this problem, we have $f(theta) = theta tan(theta)$, $U(x) = dfrac{pi}{4}$, and $L(x) = sqrt{x}$. The derivatives are $U^{prime}(x) = 0$ and $L^{prime}(x) = dfrac{1}{2sqrt{x}}$. Hence, the derivative of $g$ is
$$g^{prime}(x) = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x) = fleft(dfrac{pi}{4}right)(0) - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}}$$
which simplifies to
$$g^{prime}(x) = - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}} = -sqrt{x}tan(sqrt{x}) cdot dfrac{1}{2sqrt{x}} = -dfrac{1}{2}tan(sqrt{x})text{.}$$






share|cite|improve this answer





















  • Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
    – James Ronald
    Nov 24 at 23:11
















1














After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $x$ and the lower limit is not a constant.



So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.



Suppose the lower limit is $L(x)$ and the upper limit is $U(x)$ of the integral. Define



$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}ttext{.}$$



Suppose $F$ is an antiderivative of $f$. By part II of the Fundamental Theorem of Calculus, you know that
$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}t = F(U(x)) - F(L(x))text{.}$$
Then, the derivative of $g$ is given by, assuming differentiability of $U$ and $L$,



$$dfrac{text{d}}{text{d}x}[g(x)] = F^{prime}(U(x))U^{prime}(x)-F^{prime}(L(x))L^{prime}(x)$$
after making use of the chain rule for derivatives. But, $F$ is an antiderivative of $f$, so $F^{prime} = f$, hence
$$dfrac{text{d}}{text{d}x}[g(x)] = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}$$
In other words, the main result is
$$boxed{ dfrac{text{d}}{text{d}x}int_{L(x)}^{U(x)}f(t)text{ d}t = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}}$$





Applying to this problem, we have $f(theta) = theta tan(theta)$, $U(x) = dfrac{pi}{4}$, and $L(x) = sqrt{x}$. The derivatives are $U^{prime}(x) = 0$ and $L^{prime}(x) = dfrac{1}{2sqrt{x}}$. Hence, the derivative of $g$ is
$$g^{prime}(x) = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x) = fleft(dfrac{pi}{4}right)(0) - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}}$$
which simplifies to
$$g^{prime}(x) = - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}} = -sqrt{x}tan(sqrt{x}) cdot dfrac{1}{2sqrt{x}} = -dfrac{1}{2}tan(sqrt{x})text{.}$$






share|cite|improve this answer





















  • Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
    – James Ronald
    Nov 24 at 23:11














1












1








1






After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $x$ and the lower limit is not a constant.



So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.



Suppose the lower limit is $L(x)$ and the upper limit is $U(x)$ of the integral. Define



$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}ttext{.}$$



Suppose $F$ is an antiderivative of $f$. By part II of the Fundamental Theorem of Calculus, you know that
$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}t = F(U(x)) - F(L(x))text{.}$$
Then, the derivative of $g$ is given by, assuming differentiability of $U$ and $L$,



$$dfrac{text{d}}{text{d}x}[g(x)] = F^{prime}(U(x))U^{prime}(x)-F^{prime}(L(x))L^{prime}(x)$$
after making use of the chain rule for derivatives. But, $F$ is an antiderivative of $f$, so $F^{prime} = f$, hence
$$dfrac{text{d}}{text{d}x}[g(x)] = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}$$
In other words, the main result is
$$boxed{ dfrac{text{d}}{text{d}x}int_{L(x)}^{U(x)}f(t)text{ d}t = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}}$$





Applying to this problem, we have $f(theta) = theta tan(theta)$, $U(x) = dfrac{pi}{4}$, and $L(x) = sqrt{x}$. The derivatives are $U^{prime}(x) = 0$ and $L^{prime}(x) = dfrac{1}{2sqrt{x}}$. Hence, the derivative of $g$ is
$$g^{prime}(x) = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x) = fleft(dfrac{pi}{4}right)(0) - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}}$$
which simplifies to
$$g^{prime}(x) = - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}} = -sqrt{x}tan(sqrt{x}) cdot dfrac{1}{2sqrt{x}} = -dfrac{1}{2}tan(sqrt{x})text{.}$$






share|cite|improve this answer












After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $x$ and the lower limit is not a constant.



So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.



Suppose the lower limit is $L(x)$ and the upper limit is $U(x)$ of the integral. Define



$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}ttext{.}$$



Suppose $F$ is an antiderivative of $f$. By part II of the Fundamental Theorem of Calculus, you know that
$$g(x) = int_{L(x)}^{U(x)}f(t)text{ d}t = F(U(x)) - F(L(x))text{.}$$
Then, the derivative of $g$ is given by, assuming differentiability of $U$ and $L$,



$$dfrac{text{d}}{text{d}x}[g(x)] = F^{prime}(U(x))U^{prime}(x)-F^{prime}(L(x))L^{prime}(x)$$
after making use of the chain rule for derivatives. But, $F$ is an antiderivative of $f$, so $F^{prime} = f$, hence
$$dfrac{text{d}}{text{d}x}[g(x)] = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}$$
In other words, the main result is
$$boxed{ dfrac{text{d}}{text{d}x}int_{L(x)}^{U(x)}f(t)text{ d}t = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x)text{.}}$$





Applying to this problem, we have $f(theta) = theta tan(theta)$, $U(x) = dfrac{pi}{4}$, and $L(x) = sqrt{x}$. The derivatives are $U^{prime}(x) = 0$ and $L^{prime}(x) = dfrac{1}{2sqrt{x}}$. Hence, the derivative of $g$ is
$$g^{prime}(x) = f(U(x))U^{prime}(x)-f(L(x))L^{prime}(x) = fleft(dfrac{pi}{4}right)(0) - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}}$$
which simplifies to
$$g^{prime}(x) = - fleft(sqrt{x}right) cdot dfrac{1}{2sqrt{x}} = -sqrt{x}tan(sqrt{x}) cdot dfrac{1}{2sqrt{x}} = -dfrac{1}{2}tan(sqrt{x})text{.}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 22:47









Clarinetist

10.8k42778




10.8k42778












  • Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
    – James Ronald
    Nov 24 at 23:11


















  • Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
    – James Ronald
    Nov 24 at 23:11
















Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
– James Ronald
Nov 24 at 23:11




Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again!
– James Ronald
Nov 24 at 23:11


















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