Integration challenge questions- help (1 Viewer)

[tam89]

There's no answers for these questions and any help would be much appreciated.

See attachment.

 

[acmilan]

These may be useful, see if you can work them out One of the questions (question 6 on page 2) is actually answered in the 2nd link

http://www.maths.usyd.edu.au/u/UG/JM/MATH1903/r/t3solutions.pdf
http://www.maths.usyd.edu.au/u/UG/JM/MATH1903/r/t6-05solutions.pdf

 

[robbo_145]

using question 5 from page2.jpg will allow you to complete question 5 from page 1 and question 6 from page 2

so let I = Integral{a->0} f(x) dx
let u = a - x
-du = dx

so I = Integral{0->a} f(a - u) (-du)
= Integral{a->0} f(a- u) du
now the variable does not matter so replacing u with x
I = Integral{a->0} f(a- x) dx

be aware that they must give you the substitution in 3 unit exams.
also question 4 from page 1 cannot be done using 3 unit methods. Usually they will ask you to find the derivitive of xtan^-1x first
so d/dx (xtan^-1x ) = tan^-1x + x/1+x²
so integrating both sides

Integral (tan^-1x) dx = xtan^-1x - 1/2ln(1+x²)

 

[damo676767]

EDIT: i started to post before robbo finished, so alot of this is repeting him

4) I tan^-1 x dx
= x tan^-1 x - I ( x/(1 + x²)) dx
= x tan^-1 x - log_e(1+x²)/2

first 5)
i) I { a -> 0} ( f(x) / (f(x) - f(a-x))) dx
i will try to get back to you
ii) f(x) = root(x)

I = 3/2

second 5) rtp I {a -> 0} f(x) = I {a -> 0} f(a - x)
just explain why, for the rhs, when x = a, f(a-x) = f(0), which is the lower bound of the other side. when x = o, f(a-x) - f(a) which is the upper bound of the other. seing as these equations are reflections of each other, therfore for them to equal each other for must swap them round. hence its right

6)
you could use the rule you proved in the other part 5, i thinkl thats what they ment.
a = PI/2
f(x) = sinⁿx
f(a-x) = sinⁿ(PI/2 - x)
= cosⁿx

I = PI /2 /2
= PI / 4

 

[lucifel]

wrong forum, no? This is ext 2 stuff. Not Ext 1.

 

[clever angel]

is this 3 unit?

i hope its not cos i can't solve ne of these!!!

 

[lucifel]

i thought i just answerd that question 'clever' angel. TO BE CLEAR: THIS IS NOT 3U MATHS. IT IS 4U maths. So 3u people, if you can't do em, don't worry about it.

 

[damo676767]

yes it is a 4-unit technique put theres nothing to say you cant use that in an exam.\

for 3 unit people the rule for intergration by parts is

I ( f(x)g(x) ) dx = f(x)G(x) - I ( fⁱ(x)G(x)) dx

 

[tam89]

Many thanks for your help.

Errr... i don't know if it's 3/4unit questions. These challenge questions came from my maths tutor and he got it from uni.

 

[Riviet]

Consider these questions harder 3 unit or ones that appear at the end of the 3u hsc paper.

 

[Stan..]

I thought I answered question 6 for you. Oh well.