[ tex ] y = 2x^{2} + 5x - 3 [ /tex ]
write that^ without spaces around 'tex' and '/tex' and you get the following:
Quote me to see
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Acceleration assisstance (1 Viewer)
- Thread starter Dragonmaster262
- Start date
[ tex ] y = 2x^{2} + 5x - 3 [ /tex ]
write that^ without spaces around 'tex' and '/tex' and you get the following:
Quote me to see
Dragonmaster262
Unorthodox top student
[ tex ] y = 2x^{2} + 5x - 3 [ /tex ]
write that^ without spaces around 'tex' and '/tex' and you get the following:
Quote me to see
Dragonmaster262
Unorthodox top student
So just what do I do to get the equation on the document?[ tex ] y = 2x^{2} + 5x - 3 [ /tex ]
write that^ without spaces around 'tex' and '/tex' and you get the following:
Quote me to see
Dragonmaster262
Unorthodox top student
[ tex ] y = 2x^{2} + 5x - 3 [ /tex ]
write that^ without spaces around 'tex' and '/tex' and you get the following:
Quote me to see
Last edited:
Timothy.Siu
Prophet 9
no spaces in [ tex] and [ /tex]
Dragonmaster262
Unorthodox top student
Thanks man, could you teach me how to do this equation?
no spaces in [ tex] and [ /tex]
Express n^2 as a quadratic in n - 3
Timothy.Siu
Prophet 9
i dont get the question
Dragonmaster262
Unorthodox top student
Dragonmaster262
Unorthodox top student
Re: If you do not participate in this thread then you will get a Band E1 for your HSC
Can anyone explain to me how to do these equations? Do you use the Product rule or the function of a function rule?
Differentiate with respect to x:
a) )
b)^2 )
c)^2 (x^3 + 2)^4 )
Can anyone explain to me how to do these equations? Do you use the Product rule or the function of a function rule?
Differentiate with respect to x:
a)
b)
c)
Last edited:
Dragonmaster262
Unorthodox top student
Re: If you do not participate in this thread then you will get a Band E1 for your HSC
Can anyone explain to me how you integrate this:
1~2 (x/(x+1)^1/2) dx
My teacher said that you need standard integrals but there aren't any given in the question.
Edit: Sorry I forgot that it was a definite integral.
Can anyone explain to me how you integrate this:
1~2 (x/(x+1)^1/2) dx
My teacher said that you need standard integrals but there aren't any given in the question.
Edit: Sorry I forgot that it was a definite integral.
Last edited:
Dragonmaster262
Unorthodox top student
Re: If you do not participate in this thread then you will get a Band E1 for your HSC
Find using the given substitutions:
1)/(x-1)^1/3 dx )
using u = x-1
2)/(x+2)^3/2 dx )
using u = x+2
Find using the given substitutions:
1)
using u = x-1
2)
using u = x+2
Last edited:
Dragonmaster262
Unorthodox top student
Re: If you do not participate in this thread then you will get a Band E1 for your HSC
Could you also do the second one as well?
In your third step I still don't get how you ened up with the index of 2/3 for x-1. Care to explain?^{\frac{1}{3}}}\ dx\ \ =\ \ \int \frac{x-1\ +\ 2}{(x-1)^{\frac{1}{3}} }\ dx\ \ =\ \ \int \left \{(x-1)^{\frac{2}{3}}\ +\ 2 (x-1)^{ -\frac{1}{3}} \right \}\ dx\\ \\ \\ =\ \ \frac{3}{5}(x-1)^{\frac{5}{3}}\ +\ 3(x-1)^{\frac{2}{3}}\ \ +\ \ \ C)
Could you also do the second one as well?
Dragonmaster262
Unorthodox top student
Re: If you do not participate in this thread then you will get a Band E1 for your HSC
I get it now! Thanks man!}{(x-1)^{\frac{1}{3}}}\ =\ (x-1)^{1-\frac{1}{3}}\ =\ (x-1)^{\frac{2}{3}}\ \ \ (by\ inspection,\ one\ would\ have\ done\ this\ \ straight\ away)\\ \\ \\ Q2)\ I'll\ do\ it\ the\ conventional\ way.\\ \\ Let\ \ u\ =\ x+2\ \ \ \therefore \frac{\mathrm{d} u}{\mathrm{d} x}\ =\ 1\ \ \Rightarrow \ \ dx\ =\ du\\ \\ \therefore \ \ \ integral\ \ =\ \ \int \frac{(u-1)}{u^{\frac{3}{2}}}\ du\ \ =\ \ \int \ u^{- \frac{1}{2}}\ -\ u^{-\frac{3}{2}}\ \ du\ \\ \\ \ \ =\ \ \ \ \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\ -\ \ \frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}\ \ +\ \ C\\ \\ =\ \ \ 2u^{\frac{1}{2}}\ +\ 2 \ u^{-\frac{1}{2}}\ \ +\ \ C\\ \\ =\ \ 2\sqrt{x+2}\ \ +\ \ \frac{2}{\sqrt{x+2}} \ +\ C\\ \\ \\ (Doing\ all\ these\ in\ LaTeX\ is\ killing\ me!))