anonymous
  • anonymous
please tell to how to solve question related to calclus and where &how to use integration &differentiation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
rather a broad question, isn't it?
hartnn
  • 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
  • hartnn
finding area*

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anonymous
  • anonymous
\[gm/l \int\limits_{-l/2}^{+l/2}(r-x)^2dx\]
hartnn
  • hartnn
you need to first expand (r-x)^2 =.... ?
anonymous
  • anonymous
don't know
hartnn
  • hartnn
\[(a-b)^2=a^2-2ab+b^2 \\ (r-x)^2 =... ?\]
anonymous
  • anonymous
\[r^2-2rx+x^2\]
hartnn
  • hartnn
thats correct, now \[\int x^n dx=x^{n+1}/(n+1) \\ \int x^2dx=... ?\]
anonymous
  • anonymous
x^3/4
hartnn
  • 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
  • anonymous
why u hav taken 2r &r out?
hartnn
  • hartnn
because when we are integrating with respect to x, 'r' will be constant and can be taken out of integration and differentiation..
anonymous
  • anonymous
how u will know that x is a constant? or any digit or letter is a constant?
hartnn
  • hartnn
do you see 'dx' in integration, ? means only 'x' is variable, all other letters are treated to be constant.
anonymous
  • anonymous
the letter attach to d like dx always only that letter r variable?
anonymous
  • anonymous
@hartnn
hartnn
  • hartnn
sorry, yes. only that letter is variable. so here r is constant, x is variable.
anonymous
  • anonymous
okay after that
hartnn
  • 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
  • anonymous
yes
hartnn
  • hartnn
can you try for other 2 terms ?
hartnn
  • 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
  • anonymous
hey this one is nt d answer.
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
@Hero
anonymous
  • anonymous
plz can u provide me conversion chats
anonymous
  • 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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