4u Mathematics Marathon V 1.0 (3 Viewers)

[Sober]

I'll post up the answers, forget about the last 2

Why forget the last two? And I don't think you mean 'undefined', surely you mean non-analytic?

Besides this doesn't seem very 4 unitish.

Spoiler

3) y = |x| + |x-2|

(0, 2) and (2, 2)

4) y = |x| - |x-2|

(0, -2) and (2, 2)

 

[Slidey]

Actually, it is completely within the scope of the 4-unit syllabus, and is taught in both Patel and Cambridge.

 

[Mountain.Dew]

next question:

let In = ∫(pi/2 --> pi/6) cosecnx dx, where n is a positive integer.

(i) using integration, show that:

(n-1)In = 2n-2*sqrt3 + (n-2)In-2

(ii) Evaluate J = ∫(pi/3 --> 0)sec4x dx

 

[Mountain.Dew]

time to revive an old thread...

bump. please have a look at the previous post for the question to get the ball rolling!

remember, when you answer a question, you must provide a question yourself!

 

[icycloud]

Sure Mountain.Dew:

Spoiler

I assume you meant I_n = ∫Cosecⁿ(x)dx {pi/6-->pi/2}

(i) Now,

I_n = ∫Cosecⁿ(x)dx
= ∫Cosec^n-2(x)Cosec²(x)dx

Let u = Cosec^n-2(x)
du = -(n-2)Cosec^n-2(x)Cot(x)dx
v = ∫Cosec²xdx = -Cot(x)

Using IBP,
I_n = -Cot(x)Cosec^n-2(x) - (n-2) ∫Cosec^n-2(x)(Cosec²(x)-1)dx
= -Cot(x)Cosec^n-2(x) - (n-2) I_n + (n-2) I_n-2

Rearranging, (n-1)I_n = -Cot(x)Cosec^n-2(x) + (n-2) I_n-2

Subbing in the limits in the first part,

[-Cot(x)Cosec^n-2(x)] {pi/6 --> pi/2}
= 0 + Cot(pi/6)Cosec^n-2(pi/6)
= Sqrt[3] 2^n-2

Thus, we have (n-1)I_n = Sqrt[3] 2^n-2 + (n-2) I_n-2 as required.

(ii) I assume you meant: Evaluate J = ∫Sec⁴(x)dx {0-->pi/3}

Now, J = ∫Sec⁴(x)dx {0-->pi/3}

Let u = pi/2 - x
du = -dx

Thus, J = -∫Cosec⁴(u)du {pi/2-->pi/6}
= ∫Cosec⁴(u)du {pi/6-->pi/2}
= I₄

Now, I₄ = {2² Sqrt[3] + 2I₂}/3
I₂ = {2^0 Sqrt[3] + 0}/1 = Sqrt[3]

Thus, I₄ = {2² Sqrt[3] + 2Sqrt[3]}/3 = 2 Sqrt[3]

The final answer is J = I₄ = 2 Sqrt[3] #

Next Q
Find lim {n-->+infinity} ∫dx/(1+x^4) {0 --> n}

Big hint:

Spoiler

What happens when you use the substitution x=1/u?

 

[Stan..]

1/u = x
Limits become,
0 and Infinity
I u^2/(u^4 + 1) du Limits( Infinity & 0)
I = 1/2Arctan(u^2) * u - I
2I = Infinity
I = Infinity

EDIT: Are we allowed to use the hint? Partial Fractions become very difficult but eventually yield the same answer.

 

[Gowr]

er....no
I get something completely different

I= integral(u^2/u^4+1)
=integral (u^2)/(u^2+sqr(2)u+1)(u^2-sqr(2)u+1)
= integral ((u^2+1)/(u^2+sqr(2)u+1)(u^2-sqr(2)u+1) - 1/(u^4+1))
= integral (1/2 (1/(u^2+sqr(2)u+1) + 1/(u^2-sqr(2)u+1)) - I
4I=integral (1/((u+1/sqr(2))^2 +1/2) + 1/((u-1/sqr(2)^2 +1/2))
I=1/4 (sqr(2)tan^-1 (sqr(2)(u+1/sqr(2)) + sqr(2)tan^-1(sqr(2)(u-1/sqr(2)))
I= 1/4 (sqr(2) *pi/2 *2)
I=pi*sqr(2)/4

and no, i haven't checked my algebra... so someone tell me if this is even remotely correct please

But i know the answer posted by stan has got to be incorrect, as the limit of integral 1/(u^2+1) from 0 to infinity is pi/2, so it would not make much sense if the limit of the integral of 1/(u^4+1), which is smaller, is infinity, plus, technically infinity is not a limit, so it would be misleading for a limit question to have an answer of infinity

 

[pLuvia]

*Bump* Might as well get this up and running again

Evaluate this integral:
∫e^-xsinxdx

 

[pLuvia]

Bump~

 

[Yip]

∫e^-xsinxdx=-e^-xsinx+ ∫e^-xcosxdx=-e^-xsinx+[-e^-xcosx- ∫e^-xsinxdx]
∫e^-xsinxdx=(-e^-x/2)(sinx+cosx)+C

Q. Evaluate ∫{[ln(x+1)]/(x²+1)}dx with 0 as the bottom limit and 1 as the top


 

[shsshs]

Q: integrate 1/(1+e^x) dx

let y = e^x
then dy/y = dx

int 1/(1+e^x) dx
= int dy/y(1+y)
= int dy/y - int dy/(1+y) by partial fraction
= ln (y/(1+y)) + C
= - ln (1+e^-x) + C


Q: integrate 1/(e^x-e^-x) dx

wth r all u ppl doing.. why not juss (+ e^x - e^x) to the numerator

then its just 1 - (e^X)/(1 + e^X)

 

[Riviet]

wth r all u ppl doing.. why not juss (+ e^x - e^x) to the numerator

then its just 1 - (e^X)/(1 + e^X)

That certainly works, much better than resorting to partial fractions.

 

[darkliight]

Q. Evaluate ∫{[ln(x+1)]/(x²+1)}dx with 0 as the bottom limit and 1 as the top

Maple says you're being a bit harsh ...

int(ln(x+1)/(x^2+1), x=0..1) = (i/2)*dilog(1/2+i/2)-(i/2)*dilog(1/2-i/2)+πln(2)/4-Catalan

Changing it to ∫ xln(x^2 + 1)/(x^2 + 1) dx

 

[Yip]

The q has a nice solution

 

[pLuvia]

The q has a nice solution

Bump, enlighten us Yip

 

[Yip]

Bump, enlighten us Yip

∫₀1{[ln(x+1)]/(x²+1)}dx
Let x=tan(u)
dx=sec²udu
I=∫₀π/4{[ln(tan+1)]/sec²u}sec²udu
=∫₀π/4{[ln(tan+1)]du
=∫₀π/4{[ln|{sin+cos}/cos|}du
=∫₀π/4{[ln|{sin+cos}|-ln|cos|}du
=∫₀π/4{[ln|root2[cos(u-π/4)]|-ln|cos|}du
=∫₀π/4{0.5ln2+ln|cos(u-π/4)]|-ln|cos|]}du
Consider I=∫₀π/4[ln|cos(u-π/4)]|]du=∫₀π/4[ln|cos(π/4-u)]|][cos is even]
Let π/4-u=t
du=-dt
I=-∫_pi/40[ln|cos(t)|]dt
=∫₀π/4{ln|cos(t)|]dt=∫₀π/4ln|cos|]}du

Thus,

∫₀1{[ln(x+1)]/(x²+1)}dx=∫₀π/4(0.5ln2)du
=0.5[ln2(pi/4)}=π/8[ln2]
//

[SIZE=-1]満足が行く[/SIZE][SIZE=-1]の,[/SIZE][SIZE=-1]プルビアさん?[/SIZE]

 

[pLuvia]

Yep, I'm satisfied

 

[.ben]

Q. Evaluate [/B] ∫{[ln(x+1)]/(x²+1)}dx with 0 as the bottom limit and 1 as the top


- did you get this question from Putnam 2005?

 

[Yip]

yup

 

[buchanan]

Yep. It's A5 in the 2005 Putnam:

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005.pdf

And there are 4 methods (one of which is Yip's) in the pdf

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2005s.pdf