integration siikho

this thread is only for integration sums............

AND YA FORGOT TO TELL ALL TARGETIT USERS, I HAVE COPIED ALL THESE SUMS. SO KISIKO GAALI DENA HAIN YA KUCH KEHNA HAIN, PLEASE VISIT

http://targetiit.com/profile711.html AND POST THEM IN CHATBOX. THANKS IN ADVANCE............

1}\int_{0}^{1}e^{\sqrt{e^{x}}}\ dx+2\int_{e}^{e^{\sqrt{e}}}\ln (\ln x)\ dx

2} \int \frac{x^{3}}{(x-1)^{3}(x-2)}\ dx

3}1985 japan women university

\lim_{a\rightarrow + \infty} \frac {\int_0^a \sin ^ 4 x\ dx}{a}

4} \frac{1}{\displaystyle \int _0^{\frac{\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}

5} \int_0^{\frac {\pi}{2}} \frac {x^2}{(\cos x + x\sin x)^2}\ dx

6} IF F(X)= x2 + |x| then prove that

\int_{0}^{\pi}f(\cos x)\ dx=2\int_{0}^{\frac{\pi}{2}}f(\sin x)\ dx

7}Evaluate the following definite integral.

\int_{e^{2}}^{e^{3}}\frac {\ln x\cdot\ln (x\ln x)\cdot\ln\{x\ln (x\ln x)\} + \ln x + 1}{\ln x\cdot\ln (x\ln x)}\ dx

8}

Let f a nonnegative ,continuous and periodical function defined on the reals, such that the arithmetic mean of the numbers f(1), f(2),...f(n) tends to zero when n tends to infinity. Prove that f(k)=0 for any natural number k.

9}\int_{0}^{1}\frac {x\ dx}{(x^{2} + x + 1)^{\frac {3}{2}}}

10}\int_{2}^{6}\ln\frac { - 1 + \sqrt {1 + 4x}}{2}\ dx

174 Answers

1
Akand ·

mathi......i dint understand u.............and bhaiyya can u help wid tht a power -x/2 wala???

1
Akand ·

the 23rd wala...........i cant get it.......u said to take summation.....i still cant get it

1
Akand ·

28....
I=question............partial frac
I=1/2(lnx/1+x)+lnx/(1-x))
I=1/2(I1+I2)
I1=lnxdx/1+x............by parts
I1=lnxln(1+x)-∫ln(1+x)dx/x
=lnxln(1+x)-{ln(1+x)lnx-∫lnxdx/ln(1+x)}
=lnxln(1+x)-ln(1+x)lnx+∫lnxdx/ln(1+x)
I1=0+I1..............

wat d heck am i getting.....sum1 help plzzzzzzzzzz

11
Mani Pal Singh ·

Q38

11
Mani Pal Singh ·

@nishant sir,prashant sir and manish sir

sir can u please help me with Q2

62
Lokesh Verma ·

in the 2nd question on very bad but possible method is to use x-2 =t

try that.. there is a method by partial fraction, which i think will be more brilliant...

try these 2 approaches

1
Akand ·

sum1 do 23.....plzzzzzzzzz

11
Subash ·

23

take -x/2=t

i get -2a-x/2/loga

11
Mani Pal Singh ·

akand see for Q23
put t=-x/2 and dt =-dx/2
∫-2atdt
=>-2∫atdt

=>-2at/loga

=>-2a-x/2/loga

11
Subash ·

@akand bhaiyya was just saying that you couldnt take that log just like that

11
Subash ·

manipal bhai

it is at/loga

11
Mani Pal Singh ·

bhai pehle dekh liya tha!!!!!!!!

1
betrayed.... always ·

I have been trin so hard to solve dis but all in vain , Mr. Mathie plz help me,

∫√1+x3 dx

1
Akand ·

but bhaiyya said to use sum summation wala........so was confused

1
KR ·

22] first separate it:
\int \frac{1}{cos^{2}x}+\frac{2cos^{2}x}{xcos^{2}x}dx \left\{\in since \left( \right) 1+cos2x=cos^{2}x\left( \right)\ni \right\}tanx+2lnx+C

1
KR ·

tanx+2lnx+C

1
Akand ·

dude KR......22 is already solved.......

39
Dr.House ·

SORRY, i IS JUST A CONSTANT


IT IS NOT√-1

106
Asish Mahapatra ·

b555 is Q45 .. 1/e?

11
Mani Pal Singh ·

NO ASISH TAKE i AS WE TAKE r

39
Dr.House ·

@ash: not having answers.

106
Asish Mahapatra ·

Q41.. zero?

mujhe kich pata nahiin ...
its lim(n-->∞) (1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n)

when n-->∞ all of the individual terms tend to zero..

so limit is zero.

39
Dr.House ·

guys wanted to tell u : nearly 60% of these sums are from www.mathlinks.ro and some are from our assignments and some from papers i solved.

as said u will find nearly 30 of these 50 posted till now from mathlinks.ro but ya, these are from pretty old threads which i solved in september that time. i am making sure, it doe not cross jee limit anywhere.

106
Asish Mahapatra ·

Q43.

lim(n-->∞)(summation of 2/n + 4i/n2)
= lim(n-->∞) [2/n .. n times + 4(1+2+3+...+n)/n2]
= lim(n-->∞) [(2/n).n + 4n(n+1)/2n2]
= lim(n-->∞) [2 + 2(1+1/n)]
= 2+2
= 4

11
Mani Pal Singh ·

Q39
put i/n=t
so 1/n=dt
it becomes 01∫cos∩/2tdt

=>2/∩sin∩/2tI01

=2/∩

106
Asish Mahapatra ·

Q44.

[summation of 4/n + 32i/n2 + 64i2/n3]
= 4/n .. n times + 32(1+2+3+...+n)/n2 + 64(12+22+32+..+n2)/n3
= 4 + 32.n(n+1)/2n2 + 64.n(n+1)(2n+1)/6n3
= 4 + 16 + 16/n + 64/3 + 32/n + 32/3n2

taking limit as n-->∞
we get limit = 4+16+64/3
= 124/3

11
Mani Pal Singh ·

Q42
put i/n->t
so 1/n->dt
it could be written as

01∫2 (2+t)5dt
=>01∫2(t5+10t4+40t3+80t2+80t+32)

now it could be easily solved by opening brackets and integrating it

106
Asish Mahapatra ·

@mani .. how have u used that??
put i/n =t so 1/n = dt
01∫2 (2+t)5dt
=>01∫2(t5+10t4+40t3+80t2+80t+32)

is there any such rule for that?

11
Mani Pal Singh ·

this is a method known as limit of sum
it is used to find summations using integrations
if u have any book of integeration
find the topic integeration as limit of a sum

for more knowledge
this year paper contained question of limit as a sum

106
Asish Mahapatra ·

try doing Q43.. using that method mani.. i think ull get different answer...

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