Why can't I use symmetry to solve this integration?
Given the following equation (in rads):
$f(x) = cos(x^2-2x)$
The graph of its derivative will look like this:
Graph
I would like to find the total area between limited by the graph and the $x$ axis where $0leq xleq 2$.
I integrated the derivative:
$$int_0 ^2 f(x)'dx = f(x) |^2 _0 = cos(2^2 -2cdot 2) - cos(0^2 - 2cdot 0) = 0$$
This tells me that both the negative and positive part of the graph at the given range are equal to one another. Now, here's what I don't understand:
At first, I assumed that since the two parts are equal:
$$cos(2^2 -2cdot 2) = cos(0^2 - 2cdot 0)$$
I could calculate either one of them, then multiply by 2 to get the answer to my question. This is incorrect, and I would like to know why. Both parts are equal to 1 (which already looks wrong if you eyeball the area in the graph), while the correct answer is ~0.46 for both the negative and positive part, which was derived by integrating $0leq x leq 1$ or $1leq x leq 2$ separately. Multiplying this by 2 gives us the area of the graph in the requested range.
Why does my first method fail?
Apologies in advance if my question is not clear enough.
calculus real-analysis integration trigonometry
asked Oct 30 at 15:01
daedsidog
1614
add a comment |
Given the following equation (in rads):
$f(x) = cos(x^2-2x)$
The graph of its derivative will look like this:
Graph
I would like to find the total area between limited by the graph and the $x$ axis where $0leq xleq 2$.
I integrated the derivative:
$$int_0 ^2 f(x)'dx = f(x) |^2 _0 = cos(2^2 -2cdot 2) - cos(0^2 - 2cdot 0) = 0$$
This tells me that both the negative and positive part of the graph at the given range are equal to one another. Now, here's what I don't understand:
At first, I assumed that since the two parts are equal:
$$cos(2^2 -2cdot 2) = cos(0^2 - 2cdot 0)$$
I could calculate either one of them, then multiply by 2 to get the answer to my question. This is incorrect, and I would like to know why. Both parts are equal to 1 (which already looks wrong if you eyeball the area in the graph), while the correct answer is ~0.46 for both the negative and positive part, which was derived by integrating $0leq x leq 1$ or $1leq x leq 2$ separately. Multiplying this by 2 gives us the area of the graph in the requested range.
Why does my first method fail?
Apologies in advance if my question is not clear enough.
calculus real-analysis integration trigonometry
asked Oct 30 at 15:01
daedsidog
1614
add a comment |
Given the following equation (in rads):
$f(x) = cos(x^2-2x)$
The graph of its derivative will look like this:
Graph
I would like to find the total area between limited by the graph and the $x$ axis where $0leq xleq 2$.
I integrated the derivative:
$$int_0 ^2 f(x)'dx = f(x) |^2 _0 = cos(2^2 -2cdot 2) - cos(0^2 - 2cdot 0) = 0$$
This tells me that both the negative and positive part of the graph at the given range are equal to one another. Now, here's what I don't understand:
At first, I assumed that since the two parts are equal:
$$cos(2^2 -2cdot 2) = cos(0^2 - 2cdot 0)$$
I could calculate either one of them, then multiply by 2 to get the answer to my question. This is incorrect, and I would like to know why. Both parts are equal to 1 (which already looks wrong if you eyeball the area in the graph), while the correct answer is ~0.46 for both the negative and positive part, which was derived by integrating $0leq x leq 1$ or $1leq x leq 2$ separately. Multiplying this by 2 gives us the area of the graph in the requested range.
Why does my first method fail?
Apologies in advance if my question is not clear enough.
calculus real-analysis integration trigonometry
asked Oct 30 at 15:01
daedsidog
1614
Given the following equation (in rads):
$f(x) = cos(x^2-2x)$
The graph of its derivative will look like this:
Graph
I would like to find the total area between limited by the graph and the $x$ axis where $0leq xleq 2$.
I integrated the derivative:
$$int_0 ^2 f(x)'dx = f(x) |^2 _0 = cos(2^2 -2cdot 2) - cos(0^2 - 2cdot 0) = 0$$
This tells me that both the negative and positive part of the graph at the given range are equal to one another. Now, here's what I don't understand:
At first, I assumed that since the two parts are equal:
$$cos(2^2 -2cdot 2) = cos(0^2 - 2cdot 0)$$
I could calculate either one of them, then multiply by 2 to get the answer to my question. This is incorrect, and I would like to know why. Both parts are equal to 1 (which already looks wrong if you eyeball the area in the graph), while the correct answer is ~0.46 for both the negative and positive part, which was derived by integrating $0leq x leq 1$ or $1leq x leq 2$ separately. Multiplying this by 2 gives us the area of the graph in the requested range.
Why does my first method fail?
Apologies in advance if my question is not clear enough.
calculus real-analysis integration trigonometry
calculus real-analysis integration trigonometry
asked Oct 30 at 15:01
daedsidog
1614
asked Oct 30 at 15:01
daedsidog
1614
edited Nov 26 at 21:55
asked Oct 30 at 15:01
daedsidog
1614
asked Oct 30 at 15:01
daedsidog
1614
asked Oct 30 at 15:01
daedsidog
1614
1614
add a comment |
add a comment |
2 Answers
2
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$$int_0^2f'(x)dx=int_0^2-(2x-2)sin(x^2-2x)dx=left[cos(x^2-2x)right]_0^2=cos(0)-cos(0)=0$$
Whilst the two parts are equal:
$$cos(2^2-2*2)=cos(0^2-2*0)$$
we cannot just multiply one of them by two as there is no logic behind this, the reason why we minus one from the other is because of the fundamental theorem of calculus:
$$int_a^bf(x)dx=F(b)-F(a)$$
and in our case it just so happens that $$F(b)=F(a)$$
The correct way to break up this integral would be:
$$I=left|int_0^1f'(x)dxright|+left|int_1^2f'(x)dxright|$$
or you could write this as:
$$I=int_0^2left|f(x)right|dx$$
answered Oct 30 at 15:55
Henry Lee
1,703218
add a comment |
Yes, the two parts are equal, but when you say you could calculate either and multiply by 2 to get the answer, that is incorrect - the value of the integral is the difference, not the sum, of these values.
Second, you are correct that $cos(2^2-2cdot 2) = cos(0^2-2cdot 0) = 1$, but that does not mean that the integral under the right-hand part of the curve is $1$. The value of that integral is in fact $cos(2^2-2cdot 2) - cos(1^2-2cdot 1)$.
answered Oct 30 at 15:07
rogerl
17.4k22746
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
add a comment |
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2 Answers
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2 Answers
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$$int_0^2f'(x)dx=int_0^2-(2x-2)sin(x^2-2x)dx=left[cos(x^2-2x)right]_0^2=cos(0)-cos(0)=0$$
Whilst the two parts are equal:
$$cos(2^2-2*2)=cos(0^2-2*0)$$
we cannot just multiply one of them by two as there is no logic behind this, the reason why we minus one from the other is because of the fundamental theorem of calculus:
$$int_a^bf(x)dx=F(b)-F(a)$$
and in our case it just so happens that $$F(b)=F(a)$$
The correct way to break up this integral would be:
$$I=left|int_0^1f'(x)dxright|+left|int_1^2f'(x)dxright|$$
or you could write this as:
$$I=int_0^2left|f(x)right|dx$$
answered Oct 30 at 15:55
Henry Lee
1,703218
add a comment |
$$int_0^2f'(x)dx=int_0^2-(2x-2)sin(x^2-2x)dx=left[cos(x^2-2x)right]_0^2=cos(0)-cos(0)=0$$
Whilst the two parts are equal:
$$cos(2^2-2*2)=cos(0^2-2*0)$$
we cannot just multiply one of them by two as there is no logic behind this, the reason why we minus one from the other is because of the fundamental theorem of calculus:
$$int_a^bf(x)dx=F(b)-F(a)$$
and in our case it just so happens that $$F(b)=F(a)$$
The correct way to break up this integral would be:
$$I=left|int_0^1f'(x)dxright|+left|int_1^2f'(x)dxright|$$
or you could write this as:
$$I=int_0^2left|f(x)right|dx$$
answered Oct 30 at 15:55
Henry Lee
1,703218
add a comment |
$$int_0^2f'(x)dx=int_0^2-(2x-2)sin(x^2-2x)dx=left[cos(x^2-2x)right]_0^2=cos(0)-cos(0)=0$$
Whilst the two parts are equal:
$$cos(2^2-2*2)=cos(0^2-2*0)$$
we cannot just multiply one of them by two as there is no logic behind this, the reason why we minus one from the other is because of the fundamental theorem of calculus:
$$int_a^bf(x)dx=F(b)-F(a)$$
and in our case it just so happens that $$F(b)=F(a)$$
The correct way to break up this integral would be:
$$I=left|int_0^1f'(x)dxright|+left|int_1^2f'(x)dxright|$$
or you could write this as:
$$I=int_0^2left|f(x)right|dx$$
answered Oct 30 at 15:55
Henry Lee
1,703218
$$int_0^2f'(x)dx=int_0^2-(2x-2)sin(x^2-2x)dx=left[cos(x^2-2x)right]_0^2=cos(0)-cos(0)=0$$
Whilst the two parts are equal:
$$cos(2^2-2*2)=cos(0^2-2*0)$$
we cannot just multiply one of them by two as there is no logic behind this, the reason why we minus one from the other is because of the fundamental theorem of calculus:
$$int_a^bf(x)dx=F(b)-F(a)$$
and in our case it just so happens that $$F(b)=F(a)$$
The correct way to break up this integral would be:
$$I=left|int_0^1f'(x)dxright|+left|int_1^2f'(x)dxright|$$
or you could write this as:
$$I=int_0^2left|f(x)right|dx$$
answered Oct 30 at 15:55
Henry Lee
1,703218
answered Oct 30 at 15:55
Henry Lee
1,703218
answered Oct 30 at 15:55
Henry Lee
1,703218
answered Oct 30 at 15:55
Henry Lee
1,703218
1,703218
add a comment |
add a comment |
Yes, the two parts are equal, but when you say you could calculate either and multiply by 2 to get the answer, that is incorrect - the value of the integral is the difference, not the sum, of these values.
Second, you are correct that $cos(2^2-2cdot 2) = cos(0^2-2cdot 0) = 1$, but that does not mean that the integral under the right-hand part of the curve is $1$. The value of that integral is in fact $cos(2^2-2cdot 2) - cos(1^2-2cdot 1)$.
answered Oct 30 at 15:07
rogerl
17.4k22746
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
add a comment |
Yes, the two parts are equal, but when you say you could calculate either and multiply by 2 to get the answer, that is incorrect - the value of the integral is the difference, not the sum, of these values.
Second, you are correct that $cos(2^2-2cdot 2) = cos(0^2-2cdot 0) = 1$, but that does not mean that the integral under the right-hand part of the curve is $1$. The value of that integral is in fact $cos(2^2-2cdot 2) - cos(1^2-2cdot 1)$.
answered Oct 30 at 15:07
rogerl
17.4k22746
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
add a comment |
Yes, the two parts are equal, but when you say you could calculate either and multiply by 2 to get the answer, that is incorrect - the value of the integral is the difference, not the sum, of these values.
Second, you are correct that $cos(2^2-2cdot 2) = cos(0^2-2cdot 0) = 1$, but that does not mean that the integral under the right-hand part of the curve is $1$. The value of that integral is in fact $cos(2^2-2cdot 2) - cos(1^2-2cdot 1)$.
answered Oct 30 at 15:07
rogerl
17.4k22746
Yes, the two parts are equal, but when you say you could calculate either and multiply by 2 to get the answer, that is incorrect - the value of the integral is the difference, not the sum, of these values.
Second, you are correct that $cos(2^2-2cdot 2) = cos(0^2-2cdot 0) = 1$, but that does not mean that the integral under the right-hand part of the curve is $1$. The value of that integral is in fact $cos(2^2-2cdot 2) - cos(1^2-2cdot 1)$.
answered Oct 30 at 15:07
rogerl
17.4k22746
answered Oct 30 at 15:07
rogerl
17.4k22746
answered Oct 30 at 15:07
rogerl
17.4k22746
answered Oct 30 at 15:07
rogerl
17.4k22746
17.4k22746
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
add a comment |
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
Thank you for your answer, but I would like to ask for further clarification (first paragraph): if we know that the difference is zero, both parts must equal each other. When I say parts, I am referring to the subtracted & subtrahend, which both compose the integral of the entire graph. So, even if it is the difference, why can't I do this?
– daedsidog
Oct 30 at 15:19
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
When you compute one of them and multiply by 2, you are effectively computing their sum (if $x=y$, then $x+y=2x$, but $x-y=0$). Is that what you are asking?
– rogerl
Oct 30 at 15:20
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
What I am thinking is: $x - x = 0$, then $xcdot 2 = 2x $. Why isn't $2x$ the area of the graph?
– daedsidog
Oct 30 at 15:23
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
$xcdot 2 = 2x$ regardless of whether $x-x=0$ or not! And $2x$ is not the area under the graph for the reasons I outlined in my answer. I'm not sure what it is that you're asking here.
– rogerl
Oct 30 at 20:04
add a comment |
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