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Derivative of product of polynomial and “unsolvable integralâ€
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While studying for an exam I came across an old exam question;
Consider the function
$ F(x) = x^2 int_x^3^1e^t^2 dt $
and compute $ dF/dx $.
As far as I know, this is an unsolvable integral but since the question is asking for the derivative, I figured you wouldn't need to determine the integral at all. The problem I'm having is the factor $ x^2 $.
I know how to solve this if the factor $ x^2 $ was not in there, you swap the lower and upper limit and change the sign in front of the integral, then you just apply the fundamental theorem of calculus (with the chain rule) and get $ dF/dx = -3x^2 * e^x^6 $. However, with the factor $ x^2 $ present, I have no idea how to solve it. It feels like you would need to use the product rule, which in this case would have the disastrous consequence of having to determine the integral. Is there any smart tricks to solving this?
calculus integration derivatives
asked Jul 25 at 1:03
oloens
61
add a comment |Â
up vote
1
down vote
favorite
While studying for an exam I came across an old exam question;
Consider the function
$ F(x) = x^2 int_x^3^1e^t^2 dt $
and compute $ dF/dx $.
As far as I know, this is an unsolvable integral but since the question is asking for the derivative, I figured you wouldn't need to determine the integral at all. The problem I'm having is the factor $ x^2 $.
I know how to solve this if the factor $ x^2 $ was not in there, you swap the lower and upper limit and change the sign in front of the integral, then you just apply the fundamental theorem of calculus (with the chain rule) and get $ dF/dx = -3x^2 * e^x^6 $. However, with the factor $ x^2 $ present, I have no idea how to solve it. It feels like you would need to use the product rule, which in this case would have the disastrous consequence of having to determine the integral. Is there any smart tricks to solving this?
calculus integration derivatives
asked Jul 25 at 1:03
oloens
61
3
The last operation done to compute that expression is the multiplication of $x^2$ times $int_x^2^1e^t^2dt$. Then, when you apply derivative it faces that multiplication first. Apply the product rule $(fg)'=f'g+fg'$. The final result will have $int_x^2^1e^t^2dt$ as a sub-expression, specifically $2xint_x^2^1e^t^2dt+x^2(-3x^2e^x^6)$ and that is OK.
– user578878
Jul 25 at 1:06
Compare to writing $sqrt2$, for which there is no simpler expression using finitely many additions, subtractions, multiplications and divisions of natural numbers.
– user578878
Jul 25 at 1:13
Looking at the graph, the result looks weird enough to not have any "clean" formula
– Sudix
Jul 25 at 2:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
While studying for an exam I came across an old exam question;
Consider the function
$ F(x) = x^2 int_x^3^1e^t^2 dt $
and compute $ dF/dx $.
As far as I know, this is an unsolvable integral but since the question is asking for the derivative, I figured you wouldn't need to determine the integral at all. The problem I'm having is the factor $ x^2 $.
I know how to solve this if the factor $ x^2 $ was not in there, you swap the lower and upper limit and change the sign in front of the integral, then you just apply the fundamental theorem of calculus (with the chain rule) and get $ dF/dx = -3x^2 * e^x^6 $. However, with the factor $ x^2 $ present, I have no idea how to solve it. It feels like you would need to use the product rule, which in this case would have the disastrous consequence of having to determine the integral. Is there any smart tricks to solving this?
calculus integration derivatives
asked Jul 25 at 1:03
oloens
61
While studying for an exam I came across an old exam question;
Consider the function
$ F(x) = x^2 int_x^3^1e^t^2 dt $
and compute $ dF/dx $.
As far as I know, this is an unsolvable integral but since the question is asking for the derivative, I figured you wouldn't need to determine the integral at all. The problem I'm having is the factor $ x^2 $.
I know how to solve this if the factor $ x^2 $ was not in there, you swap the lower and upper limit and change the sign in front of the integral, then you just apply the fundamental theorem of calculus (with the chain rule) and get $ dF/dx = -3x^2 * e^x^6 $. However, with the factor $ x^2 $ present, I have no idea how to solve it. It feels like you would need to use the product rule, which in this case would have the disastrous consequence of having to determine the integral. Is there any smart tricks to solving this?
calculus integration derivatives
asked Jul 25 at 1:03
oloens
61
asked Jul 25 at 1:03
oloens
61
asked Jul 25 at 1:03
oloens
61
asked Jul 25 at 1:03
oloens
61
61
3
The last operation done to compute that expression is the multiplication of $x^2$ times $int_x^2^1e^t^2dt$. Then, when you apply derivative it faces that multiplication first. Apply the product rule $(fg)'=f'g+fg'$. The final result will have $int_x^2^1e^t^2dt$ as a sub-expression, specifically $2xint_x^2^1e^t^2dt+x^2(-3x^2e^x^6)$ and that is OK.
– user578878
Jul 25 at 1:06
Compare to writing $sqrt2$, for which there is no simpler expression using finitely many additions, subtractions, multiplications and divisions of natural numbers.
– user578878
Jul 25 at 1:13
Looking at the graph, the result looks weird enough to not have any "clean" formula
– Sudix
Jul 25 at 2:23
add a comment |Â
3
The last operation done to compute that expression is the multiplication of $x^2$ times $int_x^2^1e^t^2dt$. Then, when you apply derivative it faces that multiplication first. Apply the product rule $(fg)'=f'g+fg'$. The final result will have $int_x^2^1e^t^2dt$ as a sub-expression, specifically $2xint_x^2^1e^t^2dt+x^2(-3x^2e^x^6)$ and that is OK.
– user578878
Jul 25 at 1:06
Compare to writing $sqrt2$, for which there is no simpler expression using finitely many additions, subtractions, multiplications and divisions of natural numbers.
– user578878
Jul 25 at 1:13
Looking at the graph, the result looks weird enough to not have any "clean" formula
– Sudix
Jul 25 at 2:23
3
3
The last operation done to compute that expression is the multiplication of $x^2$ times $int_x^2^1e^t^2dt$. Then, when you apply derivative it faces that multiplication first. Apply the product rule $(fg)'=f'g+fg'$. The final result will have $int_x^2^1e^t^2dt$ as a sub-expression, specifically $2xint_x^2^1e^t^2dt+x^2(-3x^2e^x^6)$ and that is OK.
– user578878
Jul 25 at 1:06
The last operation done to compute that expression is the multiplication of $x^2$ times $int_x^2^1e^t^2dt$. Then, when you apply derivative it faces that multiplication first. Apply the product rule $(fg)'=f'g+fg'$. The final result will have $int_x^2^1e^t^2dt$ as a sub-expression, specifically $2xint_x^2^1e^t^2dt+x^2(-3x^2e^x^6)$ and that is OK.
– user578878
Jul 25 at 1:06
Compare to writing $sqrt2$, for which there is no simpler expression using finitely many additions, subtractions, multiplications and divisions of natural numbers.
– user578878
Jul 25 at 1:13
Compare to writing $sqrt2$, for which there is no simpler expression using finitely many additions, subtractions, multiplications and divisions of natural numbers.
– user578878
Jul 25 at 1:13
Looking at the graph, the result looks weird enough to not have any "clean" formula
– Sudix
Jul 25 at 2:23
Looking at the graph, the result looks weird enough to not have any "clean" formula
– Sudix
Jul 25 at 2:23
add a comment |Â
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3
The last operation done to compute that expression is the multiplication of $x^2$ times $int_x^2^1e^t^2dt$. Then, when you apply derivative it faces that multiplication first. Apply the product rule $(fg)'=f'g+fg'$. The final result will have $int_x^2^1e^t^2dt$ as a sub-expression, specifically $2xint_x^2^1e^t^2dt+x^2(-3x^2e^x^6)$ and that is OK.
– user578878
Jul 25 at 1:06
Compare to writing $sqrt2$, for which there is no simpler expression using finitely many additions, subtractions, multiplications and divisions of natural numbers.
– user578878
Jul 25 at 1:13
Looking at the graph, the result looks weird enough to not have any "clean" formula
– Sudix
Jul 25 at 2:23