Multiplying Integrals with Different Bounds
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The question I have is two integrals mutliplied with different bounds
$$alpha_0(x)=int_x^1expleft(frac{-varphilambda^2}2right)dlambdacdotleft(int_0^1expleft(frac{-varphilambda^2}2right)dlambdaright)$$
What I would like to do is differentiate this with respect to $x$, but am not sure how to do so considering that I have the two integrals multiplied by eachother.
What is f '(alpha0(x))
I understand that the problem would be simple if they were added/subtracted.
I understand from the Fundamental Law of Calculus that the $x$ will replace the $lambda$ when I differentiate a single integral, just not sure how to approach with the double integral.
I know that an analytical solution is definitely possible.
Thank you.
calculus integration differential-equations derivatives
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The question I have is two integrals mutliplied with different bounds
$$alpha_0(x)=int_x^1expleft(frac{-varphilambda^2}2right)dlambdacdotleft(int_0^1expleft(frac{-varphilambda^2}2right)dlambdaright)$$
What I would like to do is differentiate this with respect to $x$, but am not sure how to do so considering that I have the two integrals multiplied by eachother.
What is f '(alpha0(x))
I understand that the problem would be simple if they were added/subtracted.
I understand from the Fundamental Law of Calculus that the $x$ will replace the $lambda$ when I differentiate a single integral, just not sure how to approach with the double integral.
I know that an analytical solution is definitely possible.
Thank you.
calculus integration differential-equations derivatives
1
See that the term in brackets is a constant, since the bounds of integration are constant. You need only to differentiate the leftmost integral.
– rafa11111
Nov 17 at 14:05
1
Just one of the integrals depends on $x$. The other one is a constant. So proceed with the Fundamental Theorem of Calculus; be sure to account for the fact that the variable is the lower limit of integration. (I have just reprased @rafa11111 's comment, which you seem to have ignored.)
– Ethan Bolker
Nov 17 at 16:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question I have is two integrals mutliplied with different bounds
$$alpha_0(x)=int_x^1expleft(frac{-varphilambda^2}2right)dlambdacdotleft(int_0^1expleft(frac{-varphilambda^2}2right)dlambdaright)$$
What I would like to do is differentiate this with respect to $x$, but am not sure how to do so considering that I have the two integrals multiplied by eachother.
What is f '(alpha0(x))
I understand that the problem would be simple if they were added/subtracted.
I understand from the Fundamental Law of Calculus that the $x$ will replace the $lambda$ when I differentiate a single integral, just not sure how to approach with the double integral.
I know that an analytical solution is definitely possible.
Thank you.
calculus integration differential-equations derivatives
The question I have is two integrals mutliplied with different bounds
$$alpha_0(x)=int_x^1expleft(frac{-varphilambda^2}2right)dlambdacdotleft(int_0^1expleft(frac{-varphilambda^2}2right)dlambdaright)$$
What I would like to do is differentiate this with respect to $x$, but am not sure how to do so considering that I have the two integrals multiplied by eachother.
What is f '(alpha0(x))
I understand that the problem would be simple if they were added/subtracted.
I understand from the Fundamental Law of Calculus that the $x$ will replace the $lambda$ when I differentiate a single integral, just not sure how to approach with the double integral.
I know that an analytical solution is definitely possible.
Thank you.
calculus integration differential-equations derivatives
calculus integration differential-equations derivatives
edited Nov 17 at 16:22
asked Nov 17 at 14:00
Emma Houchell
61
61
1
See that the term in brackets is a constant, since the bounds of integration are constant. You need only to differentiate the leftmost integral.
– rafa11111
Nov 17 at 14:05
1
Just one of the integrals depends on $x$. The other one is a constant. So proceed with the Fundamental Theorem of Calculus; be sure to account for the fact that the variable is the lower limit of integration. (I have just reprased @rafa11111 's comment, which you seem to have ignored.)
– Ethan Bolker
Nov 17 at 16:27
add a comment |
1
See that the term in brackets is a constant, since the bounds of integration are constant. You need only to differentiate the leftmost integral.
– rafa11111
Nov 17 at 14:05
1
Just one of the integrals depends on $x$. The other one is a constant. So proceed with the Fundamental Theorem of Calculus; be sure to account for the fact that the variable is the lower limit of integration. (I have just reprased @rafa11111 's comment, which you seem to have ignored.)
– Ethan Bolker
Nov 17 at 16:27
1
1
See that the term in brackets is a constant, since the bounds of integration are constant. You need only to differentiate the leftmost integral.
– rafa11111
Nov 17 at 14:05
See that the term in brackets is a constant, since the bounds of integration are constant. You need only to differentiate the leftmost integral.
– rafa11111
Nov 17 at 14:05
1
1
Just one of the integrals depends on $x$. The other one is a constant. So proceed with the Fundamental Theorem of Calculus; be sure to account for the fact that the variable is the lower limit of integration. (I have just reprased @rafa11111 's comment, which you seem to have ignored.)
– Ethan Bolker
Nov 17 at 16:27
Just one of the integrals depends on $x$. The other one is a constant. So proceed with the Fundamental Theorem of Calculus; be sure to account for the fact that the variable is the lower limit of integration. (I have just reprased @rafa11111 's comment, which you seem to have ignored.)
– Ethan Bolker
Nov 17 at 16:27
add a comment |
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See that the term in brackets is a constant, since the bounds of integration are constant. You need only to differentiate the leftmost integral.
– rafa11111
Nov 17 at 14:05
1
Just one of the integrals depends on $x$. The other one is a constant. So proceed with the Fundamental Theorem of Calculus; be sure to account for the fact that the variable is the lower limit of integration. (I have just reprased @rafa11111 's comment, which you seem to have ignored.)
– Ethan Bolker
Nov 17 at 16:27