∫1/(4x - 1) [0,1] u = 4x - 1
du = 4 dx
1/4∫1/(u) (leaving limits until the end)
1/4 ln abs(u)
1/4 ln abs(4x - 1) [0,1]
1/4 ln(3) - ln(1)
1/4 ln 3
Says it diverges, so I did it wrong. Can someone explain why?
du = 4 dx
1/4∫1/(u) (leaving limits until the end)
1/4 ln abs(u)
1/4 ln abs(4x - 1) [0,1]
1/4 ln(3) - ln(1)
1/4 ln 3
Says it diverges, so I did it wrong. Can someone explain why?
-
You're treating it as though it were proper. That's leading you to an erroneous conclusion. The integrand is not defined at x = 1/4. This point is inside your interval of integration. To handle this integral correctly, you have to split into
1/4 . . .1
∫ .... + ∫ ....
0 . . . 1/4
Each of these terms have an infinite discontinuity at an end point of the interval.
Consider the fist term. You'd get (using the work you've already done)
1/4
∫ 1/(4x - 1) dx = ¼lim ln|4t - 1| - ¼ln|-1| = -∞
0 . . . . . . . . . . . t->1/4-
Since each integral in the sum diverges (you can check the other one), your integral is divergent.
Remember that if the integrand isn't defined on the interval a ≤ x ≤ b, the integral is improper and has to be treated accordingly.
1/4 . . .1
∫ .... + ∫ ....
0 . . . 1/4
Each of these terms have an infinite discontinuity at an end point of the interval.
Consider the fist term. You'd get (using the work you've already done)
1/4
∫ 1/(4x - 1) dx = ¼lim ln|4t - 1| - ¼ln|-1| = -∞
0 . . . . . . . . . . . t->1/4-
Since each integral in the sum diverges (you can check the other one), your integral is divergent.
Remember that if the integrand isn't defined on the interval a ≤ x ≤ b, the integral is improper and has to be treated accordingly.