ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: squance on May 11, 2008, 10:39:04 pm

Title: Intergrals - Hard for me and giving me headache!! Help me!!!
Post by: squance on May 11, 2008, 10:39:04 pm: Can someone solve these ones for me???

I was going ok with these until I started to get errors....

I only need help with the ones that are circled..

(http://img158.imageshack.us/img158/4126/mathstl0.png)

Thanks!!!!
Title: Re: Intergrals - Hard for me and giving me headache!! Help me!!!
Post by: Mao on May 11, 2008, 10:41:11 pm: b):

$\int_{-2}^{2}\frac{6}{4+x^2}dx$

recognising that: $\frac{d}{dx}\tan^{-1} \frac{x}{a} = \frac{a}{a^2+x^2}$

$\implies 3 \int_{-2}^{2}\frac{2}{2^2+x^2}dx$

$\implies 3 \left[ \tan^{-1}\frac{x}{2}\right] _{-2}^2=3\left( \frac{\pi}{4}-\frac{-\pi}{4}\right) = \frac{3\pi}{2}$

d)
similar deal:

to find this one out, first we need to find when y intersects with x axis:

$0=-1+\frac{2}{x^2+1}$

$\implies 1=\frac{2}{x^2+1}$

$\implies x^2+1=2$

$\implies x=\pm 1$

so, set the integral to:

$\int_{-1}^1 -1+\frac{2}{x^2+1} dx = \left[ -x + 2\tan^{-1}x \right]_{-1}^{1} = -1 + \frac{\pi}{2} - \left(1 -\frac{\pi}{2}\right)=\pi-2$

e)

we'll need partial fractions for this one:

$\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$

$\implies 1=A(x+1)+B(x)$
setting x=0 and -1 respectively:
$A=1, B=-1$

$\int_1^2 \frac{1}{x(x+1)} dx = \int_1^2 \left(\frac{1}{x}-\frac{1}{x+1}\left)dx$

$\implies \left[ log_e x - log_e (x+1)\right]_1^2$

$\implies \left[ log_e\left( \frac{x}{x+1}\right) \right]_1^2=log_e\left(\frac{2}{3}\right) - log_e\left(\frac{1}{2}\right) = -ln\left(\frac{4}{3}\right)$

h)
by simple substitution:

let $u=e^t+1, \frac{du}{dt}=e^t \implies dt=\frac{du}{e^t}$

when $t=1, u=e+1$ , when $t=-1, u=e^{-1}+1$

$\int_{-1}^1 \frac{e^t}{e^t+1} dt$

$\implies \int_{e^{-1}+1}^{e+1} \frac{e^t}{u} \frac{du}{e^t}$

$\implies \int_{e^{-1}+1}^{e+1} \frac{du}{u}$

$\implies \left[log_e u\right]_{e^{-1}+1}^{e+1}$

$\implies log_e (e+1) - log_e(e^{-1}+1) = log_e \left( \frac{e+1}{\dfrac{1}{e}+1}\right) = log_e \left(\frac{e+1}{\dfrac{1}{e}(1+e)}\right) = log_e e = 1$

i) this question is still the substitution method:

let $u=log_e x, \frac{du}{dx}=\frac{1}{x} \implies dx=x\cdot du$

when $x=e, u=1$ , when $x=e^2, u=2$

$\int_e^{e^2} \frac{1}{log_e x}\frac{1}{x} dx$

$\implies \int_1^{2} \frac{1}{u}\frac{1}{x} x\cdot du$

$\implies \int_1^{2} \frac{du}{u}$

$\implies \left[log_e u\right]_1^2 = log_e2 - log_e 1 = ln(2)$

done~~ :D
Title: Re: Intergrals - Hard for me and giving me headache!! Help me!!!
Post by: dcc on May 11, 2008, 10:59:20 pm: i:)

$\int_e^{e^2} \dfrac{1}{log_{e}x} \cdot \dfrac{1}{x} dx\! = \int_{e}^{e^2} \dfrac{\frac{1}{x}}{log_e{x}} dx\!$

Now let $u = log_{e}x \rightarrow \dfrac{du}{dx} = \frac{1}{x}$

$\implies \int_{log_{e}e}^{loge_{e}e^2} \dfrac{\frac{du}{dx}}{u} dx\!$

$\implies \int_{1}^{2} \dfrac{1}{u} du\!$

$\implies \left[ log_{e}u \right]_{1}^{2}$

$\implies \int_e^{e^2} \dfrac{1}{log_{e}x} \cdot \dfrac{1}{x} dx\! = \boxed{log_e{2}}$
Title: Re: Intergrals - Hard for me and giving me headache!! Help me!!!
Post by: squance on May 11, 2008, 11:11:54 pm: Woah!!!
Thanks guys!!
IM not actully doing maths at the moment so i'll look at them tommorow!!

Thanks!!!!

(Um...how do you guys do the maths thing on the forum...like, special maths character thingys that can appear like integrals and stuff??)
Title: Re: Intergrals - Hard for me and giving me headache!! Help me!!!
Post by: Mao on May 11, 2008, 11:13:54 pm: OMG $\log_e\left(\log_ex\right)$

:D

squance: http://vcenotes.com/forum/index.php/topic,3137.0.html

=)