please tell to how to solve question related to calclus and where &how to use integration &differentiation?

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- anonymous

- schrodinger

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- anonymous

rather a broad question, isn't it?

- hartnn

Calculus Questions are easy to solve once you know the formulas :)
integration is used in finding are of solids, figures, etc.
differentiation can be used to find rate of change of a quantity.
and both of these have many other applications..
if you have any specific question, you please ask :)

- hartnn

finding area*

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## More answers

- anonymous

\[gm/l \int\limits_{-l/2}^{+l/2}(r-x)^2dx\]

- hartnn

you need to first expand (r-x)^2 =.... ?

- anonymous

don't know

- hartnn

\[(a-b)^2=a^2-2ab+b^2 \\ (r-x)^2 =... ?\]

- anonymous

\[r^2-2rx+x^2\]

- hartnn

thats correct, now \[\int x^n dx=x^{n+1}/(n+1) \\ \int x^2dx=... ?\]

- anonymous

x^3/4

- hartnn

it will be x^3/3 ,
as n+1 =3
\[ \int\limits_{-l/2}^{+l/2}(x^2-2rx+r^2)dx\\= \int\limits_{-l/2}^{+l/2}x^2dx-2r\int\limits_{-l/2}^{+l/2}x dx+r^2\int\limits_{-l/2}^{+l/2}1dx = \\ =[x^3/3]_{-l/2}^{+l/2} -2r [x^2/2]_{-l/2}^{+l/2}+r^2[x]_{-l/2}^{+l/2}\]
got this ?

- anonymous

why u hav taken 2r &r out?

- hartnn

because when we are integrating with respect to x, 'r' will be constant and can be taken out of integration and differentiation..

- anonymous

how u will know that x is a constant? or any digit or letter is a constant?

- hartnn

do you see 'dx' in integration, ?
means only 'x' is variable, all other letters are treated to be constant.

- anonymous

the letter attach to d like dx always only that letter r variable?

- anonymous

- hartnn

sorry, yes. only that letter is variable.
so here r is constant, x is variable.

- anonymous

okay after that

- hartnn

now l/2 is upper limit and -l/2 is lower limit.
\[[x^3/3]^{l/2}_{-l/2}\]
is solved by first putting x= upper limit - x= lower limit
like this :
(l/2)^3/3- (-l/2)^3/3
got this first term ?

- anonymous

yes

- hartnn

can you try for other 2 terms ?

- hartnn

\[\int\limits_{-l/2}^{+l/2}(x^2-2rx+r^2)dx\\= \int\limits_{-l/2}^{+l/2}x^2dx-2r\int\limits_{-l/2}^{+l/2}x dx+r^2\int\limits_{-l/2}^{+l/2}1dx = \\ =[x^3/3]_{-l/2}^{+l/2} -2r [x^2/2]_{-l/2}^{+l/2}+r^2[x]_{-l/2}^{+l/2}= \\ = [(l/2)^3/(3)-(-l/2)^3/(3)]-2r[(l/2)^2/(2)-(-l/2)^2/(2)]\\ +r^2[l/2-(-l/2)] \\ = 2*l^3/24 -2r(2*l^2/8)+r^2(2l/2) \\ =l^3/12 - rl^2/2+r^2l\]

- anonymous

hey this one is nt d answer.

- anonymous

- anonymous

- anonymous

plz can u provide me conversion chats

- anonymous

example : how to convert kg into mass. like this can u provide me the chat in which about all unit conversion given.

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