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Confusing (1 Viewer)
- Thread starter Rmd_1
- Start date
The area in question is against the y-axis, so the required area is:
int (from 2 to 4) x dy, where y = ln x, and so x = e^y
= int (from 2 to 4) e^y dy
= [ e^y ] (from 2 to 4)
= e^4 - e^2
= 47.209 ... by calculator
= 47.2 (to 3 sig fig).
int (from 2 to 4) x dy, where y = ln x, and so x = e^y
= int (from 2 to 4) e^y dy
= [ e^y ] (from 2 to 4)
= e^4 - e^2
= 47.209 ... by calculator
= 47.2 (to 3 sig fig).
Volume about the y-axis, so the volume is given by int (from a to b) pi * x^2 dy. Here, that is:
int (from 1 to 3) pi * x^2 dy, where y = ln x, so x = e^y
= pi * int (from 1 to 3) (e^y)^2 dy
= pi * int (from 1 to 3) e^2y dy
= pi * [e^2y / 2] (from 1 to 3)
= (pi / 2) * [e^2(3) - e^2(1)]
= (pi / 2) * (e^6 - e^2)
= pi * e^2 (e^4 - 1) / 2 cu units (in exact form)
This is approximately 622 cu units (3 sig fig).
int (from 1 to 3) pi * x^2 dy, where y = ln x, so x = e^y
= pi * int (from 1 to 3) (e^y)^2 dy
= pi * int (from 1 to 3) e^2y dy
= pi * [e^2y / 2] (from 1 to 3)
= (pi / 2) * [e^2(3) - e^2(1)]
= (pi / 2) * (e^6 - e^2)
= pi * e^2 (e^4 - 1) / 2 cu units (in exact form)
This is approximately 622 cu units (3 sig fig).
The integral of e^f(x) is not repeat NOT e^f(x) / f'(x). This is a shortuct that some people use, that happens to work for a variety of HSC questions - including the one that Rmd_1 is aking about. Nevertheless, IT SHOULD NOT BE USED, as it is not mathematically valid - I know some teachers teach it, but that doesn't make it true.
Breathing out, and calming down ...
Sorry Heinz, you've hit on one of my pet hates.
In the above, I am integrating with respect to y, so use y instead of x, with A = 2 and B = 0.
The three integral results you need to know for exponential functions (ignoring arbitrary constants) are:
the integral of e^x with respect to x is e^x
the integral of e^(Ax + B) with respect to x is e^(Ax + B) / A, where A and B are constants
the integral of f'(x) * e^f(x) with respect to x is e^f(x)
Breathing out, and calming down ...
Sorry Heinz, you've hit on one of my pet hates.
I have used the result that the integral of e^(Ax + B) with respect to x is e^(Ax + B) / A, when A and B are constants.
In the above, I am integrating with respect to y, so use y instead of x, with A = 2 and B = 0.
The three integral results you need to know for exponential functions (ignoring arbitrary constants) are:
the integral of e^x with respect to x is e^x
the integral of e^(Ax + B) with respect to x is e^(Ax + B) / A, where A and B are constants
the integral of f'(x) * e^f(x) with respect to x is e^f(x)
I dont like it when teachers yell at meOriginally posted by CM_Tutor
The integral of e^f(x) is not repeat NOT e^f(x) / f'(x). This is a shortuct that some people use, that happens to work for a variety of HSC questions - including the one that Rmd_1 is aking about. Nevertheless, IT SHOULD NOT BE USED, as it is not mathematically valid - I know some teachers teach it, but that doesn't make it true.
Breathing out, and calming down ...
Sorry Heinz, you've hit on one of my pet hates.
Oh and blame J.B. fitzpatrick, thats the text i used when i did 2u. *flicks through text* pg 409 I am not a teacher, and i am trying to help you (and others). The result you claim is a dangerous and (unfortunately) common misconception. It works for most HSC questions, and that just makes it more dangerous, as it allows it to spread. As I said, it is one of my pet hates.
PS: I just looked at p 409 of 2u Fitzpatrick, and I don't see this result claimed.
PS: I just looked at p 409 of 2u Fitzpatrick, and I don't see this result claimed.
The result you claim, that the integral of e^f(x) is e^f(x) / f'(x) is true ONLY if f(x) is a linear function in x, ie if
f(x) = Ax + B, for A and B constants. Fitzpatrick, p. 409, wrote that the integral of e^(kx) is (1 / k) * e^(kx) + c, which is of the same form as your result, but is not the same thing.
Your result is mostly used as a short cut for questions like integrate xe^(x^2). This is not a function of the form e^f(x), but people use your result anyway, and say the answer is xe^(x^2) / 2x + c, and then cancel the x's to get e^(x^2) / 2 + c, which is the correct answer, but the working is fundamentally flawed. Whether this gets penalised in an exam situation depends on the marker.
The correct way to do this type of question is using the third result I stated about, namely that the integral of
f'(x) * e^f(x) is e^f(x). This is done as follows: First, recognise that e^(x^2) is of the form e^f(x), with f(x) = x^2, so this result is probably needed. We need an f'(x), which is 2x, so start by rewriting the question. So ...
int xe^(x^2) dx = (1 / 2) * int 2x * e^(x^2) dx [Notice that we now have an f'(x) * e^f(x)]
= (1/2) * e^(x^2) + C, for some constant C [I integrated f'(x) * e^f(x) to e^f(x)]
= e^(x^2) / 2 + C, as expected.
Now, why is your result so dangerous? Well, it leads people to think things like int e^(x^2) dx = e^(x^2) / 2x, which is totally wrong - and that's just for starters.
f(x) = Ax + B, for A and B constants. Fitzpatrick, p. 409, wrote that the integral of e^(kx) is (1 / k) * e^(kx) + c, which is of the same form as your result, but is not the same thing.
Your result is mostly used as a short cut for questions like integrate xe^(x^2). This is not a function of the form e^f(x), but people use your result anyway, and say the answer is xe^(x^2) / 2x + c, and then cancel the x's to get e^(x^2) / 2 + c, which is the correct answer, but the working is fundamentally flawed. Whether this gets penalised in an exam situation depends on the marker.
The correct way to do this type of question is using the third result I stated about, namely that the integral of
f'(x) * e^f(x) is e^f(x). This is done as follows: First, recognise that e^(x^2) is of the form e^f(x), with f(x) = x^2, so this result is probably needed. We need an f'(x), which is 2x, so start by rewriting the question. So ...
int xe^(x^2) dx = (1 / 2) * int 2x * e^(x^2) dx [Notice that we now have an f'(x) * e^f(x)]
= (1/2) * e^(x^2) + C, for some constant C [I integrated f'(x) * e^f(x) to e^f(x)]
= e^(x^2) / 2 + C, as expected.
Now, why is your result so dangerous? Well, it leads people to think things like int e^(x^2) dx = e^(x^2) / 2x, which is totally wrong - and that's just for starters.
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(1+x)^20 = (1+x)^10*(1+x)^10
LHS =(20C0*x^0 + 20C1*x + 20C@*x^2+20C3*x^3 + 20C4*x^4...) You wont need to worry about the rest of the expansion as its not used. Just go up to the required term.
:. the coefficient of x^4 is 20C4
RHS = (10C0*x^0 + 10C1*x + 10C2*x^2 + 10C3*x^3 +10C4*x^4...)*(10C0*x^0 + 10C1*x + 10C2*x^2 + 10C3*x^3 +10C4*x^4...)
C*x^4 = (10C0*x^0)*(10C4*x^4) + (10C1*x)*(10C3*x^3) + (10C2*x^2)*(10C2*x^2) + (10C3*x^3)*(10C1*x) + (10C4*x^4)*(10C0*x^0)
:. the coefficient of x^4 =2(10C0*10C4 + 10C1*10C3) + (10C2)^2
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