Integrate (3x - 2) dx?

Question

The (3x -2) is within modulous.. This is what is annoying me.. Please explain how to integrate with the modulous.. thanks

Answer 1

you can try to rewrite as sqrt( (3x - 2) (3x - 2) ) and integrate.

or split for x : 3x-2>=0 and < 0, you will get two primitives then.

Answer 2

I'm not sure what you mean by the word "modulous" since
I never saw the word before. However, if you treat this like
an ordinary indefinite integral; that is,
the integral of (3x-2) dx, we integrate and get 3x²/2 - 2x + c
where c is any constant number.
Hope this helps

Answer 3

whoops! took the derivative the 1st time I spoke back. enable's define a function u(x) to help us u = a million-2x^2 du = -4x dx so vital of 3x*sqrt(a million-2x^2)dx could properly be written in terms of u as vital of -3/4 sqrt(u) du enable { stand for taking the vital provided that i won't be able to make the actual sign for it { (-3/4) sqrt(u) du (-3/4) { sqrt(u) du (-a million/2) u^(3/2) + c plugging decrease back in for u = a million-2x^2 (-a million/2)(a million-2x^2)^(3/2) + c enable's verify via taking the derivative (-a million/2)(a million-2x^2)^(3/2) + c (a million/2)(-3/2)(a million-2x^2)^(a million/2)(-4x) 3x * sqrt(a million-2x^2) yep!

Answer 4

♠ y’=|3x-2| =sign(3x-2)*(3x-2),
where sign(3x-2) is ‘almost’ const,
♦ hence y= sign(3x-2)*(3x^2 /2 –2x) +C, where
sign(3x-2) =-1, if x < 2/3;
sign(3x-2) =0, if x = 2/3;
sign(3x-2) =+1, if x > 2/3;

Answer 5

(1.5x^2)-2x