## Ruchi. Group Title please tell to how to solve question related to calclus and where &how to use integration &differentiation? one year ago one year ago

1. satellite73

rather a broad question, isn't it?

2. 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 :)

3. hartnn

finding area*

4. Ruchi.

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

5. hartnn

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

6. Ruchi.

don't know

7. hartnn

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

8. Ruchi.

$r^2-2rx+x^2$

9. hartnn

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

10. Ruchi.

x^3/4

11. 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 ?

12. Ruchi.

why u hav taken 2r &r out?

13. hartnn

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

14. Ruchi.

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

15. hartnn

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

16. Ruchi.

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

17. Ruchi.

@hartnn

18. hartnn

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

19. Ruchi.

okay after that

20. 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 ?

21. Ruchi.

yes

22. hartnn

can you try for other 2 terms ?

23. hartnn

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

24. Ruchi.

hey this one is nt d answer.

25. Ruchi.

@satellite73

26. Ruchi.

@Hero

27. Ruchi.

plz can u provide me conversion chats

28. Ruchi.

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