calculus antiderivative?

Question

evaluate the following

INT [ from 1 to e] lnx dx

Answer 1

e
∫ln x dx =
1

u = ln x ..................... dv = dx
du = 1/x dx ............... v = x

∫u dv = uv - ∫v du =>
∫ln x dx = x*ln x -∫x(1/x) dx
∫ln x dx = x*ln x - x + c

=>
e
∫ln x dx = [x*ln x - x] (from 1 to e)
1

e
∫ln x dx = [e*ln e - e - 1*ln 1 + 1]
1

e
∫ln x dx = e - e - 1*0 + 1
1

e
∫ln x dx = 1
1

₢

Answer 2

The antiderivative of ln x is equal to x lnx - x.

So now let's apply the fundamental theorem of calculus.

F(1) = 1 ln 1 -1
F(e) = e ln e - e

1(0) -1 = 0-1= -1
e (1) - e = e -e = 0

0- (-1) = 1

1

Answer 3

the intregral of the natural log of x is -x + x ln(x)

So use the fundamental theorem of calculus to evaluate it with respect to the endpoints 1 and e,
INT [ from 1 to e] lnx dx = (-e + e ln(e)) - (-1 + 1 ln(1))
since ln(1) = 0, ln(e) = 1,
INT [ from 1 to e] lnx dx = (-e + e) +1 - 0 = 1

So the answer is 1

Answer 4

∫ln x dc = x ln x - ∫1/x (x) dx(you can to use integration by parts)
= x ln x -∫ dx
= x ln x - x + c

∫[ from 1 to e]= (1ln1-1)-(eln(e)-e)
= (0-1)- ((e*1)-e)
= -1-0
= -1