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Ruchi.

  • 3 years ago

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

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  1. anonymous
    • 3 years ago
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    rather a broad question, isn't it?

  2. hartnn
    • 3 years ago
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    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 :)

  3. hartnn
    • 3 years ago
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    finding area*

  4. Ruchi.
    • 3 years ago
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    \[gm/l \int\limits_{-l/2}^{+l/2}(r-x)^2dx\]

  5. hartnn
    • 3 years ago
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    you need to first expand (r-x)^2 =.... ?

  6. Ruchi.
    • 3 years ago
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    don't know

  7. hartnn
    • 3 years ago
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    \[(a-b)^2=a^2-2ab+b^2 \\ (r-x)^2 =... ?\]

  8. Ruchi.
    • 3 years ago
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    \[r^2-2rx+x^2\]

  9. hartnn
    • 3 years ago
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    thats correct, now \[\int x^n dx=x^{n+1}/(n+1) \\ \int x^2dx=... ?\]

  10. Ruchi.
    • 3 years ago
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    x^3/4

  11. hartnn
    • 3 years ago
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    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 ?

  12. Ruchi.
    • 3 years ago
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    why u hav taken 2r &r out?

  13. hartnn
    • 3 years ago
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    because when we are integrating with respect to x, 'r' will be constant and can be taken out of integration and differentiation..

  14. Ruchi.
    • 3 years ago
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    how u will know that x is a constant? or any digit or letter is a constant?

  15. hartnn
    • 3 years ago
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    do you see 'dx' in integration, ? means only 'x' is variable, all other letters are treated to be constant.

  16. Ruchi.
    • 3 years ago
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    the letter attach to d like dx always only that letter r variable?

  17. Ruchi.
    • 3 years ago
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    @hartnn

  18. hartnn
    • 3 years ago
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    sorry, yes. only that letter is variable. so here r is constant, x is variable.

  19. Ruchi.
    • 3 years ago
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    okay after that

  20. hartnn
    • 3 years ago
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    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 ?

  21. Ruchi.
    • 3 years ago
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    yes

  22. hartnn
    • 3 years ago
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    can you try for other 2 terms ?

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

  24. Ruchi.
    • 3 years ago
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    hey this one is nt d answer.

  25. Ruchi.
    • 3 years ago
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    @satellite73

  26. Ruchi.
    • 3 years ago
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    @Hero

  27. Ruchi.
    • 3 years ago
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    plz can u provide me conversion chats

  28. Ruchi.
    • 3 years ago
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    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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