Antidifferentiation Problem....?

Question

Can anyone tell me how to anti-differentiate a function with an absolute value in a definite range?
i.e. : Antidifferentiation of | x-3 | where the range is between -2 and 5....

Answer 1

when u are antidifferentiating | f(x) |, u should follow this:

U know that | m| = m, if m >= 0
= -m, if m < 0, where m is any real number.

So |f(x)| = f(x) , if f(x) >=0,
= -f(x) , if f(x) < 0, ............... (*)

Hence always look for values of x for which, f(x) >=0 and f(x) <0 and partition the range of integration accordingly and then it's like any other integration.

Now come to ur example.

f(x) = x-3 here.
f(x) >= 0 => x-3 >=0 => x>=3
f(x) < 0 => x-3 < 0 => x< 3

Let me use S[-2,5] for denoting anti-differntiation in the range [2,5].

so , S[-2,5] (Ix-3|) = S[-2,3](|x-3|) + S[3,5](|x-3|)
=S[-2,3] (-(x-3)) + S[3,5](x-3) ...(by (*))

This u must have done in ur first class of anti-differentiation!!

Cheers!

Answer 2

If x>3 x-3>0 and your function y=x+3

ifx<3 x-3<0 and y= 3-x

So between -2 and 3 The integral is 3x-x^2/2 +C and

between 3 and 5 it is x^2/2 +3x +C

Answer 3

essential: x^-7 dx=kx^-6 + C x^(-7+a million)/(-7+a million) = ok x^(-6) + C x^(-6) / (-6) = ok x^(-6) + C multiply for the time of with the help of x^(-6) -a million/6 = ok + Cx^(-6) ok = -a million/6 -C/x^6 to locate a numerical fee for ok, you opt for some preliminary situations.