How to solve this improper integral

Question

$\int _{ 0 }^{ 4 }{ \frac { 4 }{ x-4 } } $

My attempt

$\lim _{ b\rightarrow 4^- }$$\int _{ 0 }^{ b }{ \frac { 4 }{ x-4 } } $

My doubt is why did we need to take Limit as $b\rightarrow 4^- $ but not just $b\rightarrow 4 $

Answer 1

This is because the bottom bound of your improper integral is less than the $x$-coordinate of the singularity and your top bound is the $x$-coordinate of the singularity. Thus, you approach the singularity from the right side. Realize that $\lim_{x\to4} {4\over x-4}$ doesn't exist.

Also - your integral is not finite, it diverges. If we approach the antiderivative at $4$ from the right side, you get $-\infty$.

Answer 2

Because the integral is between $0$ and $4$. If it was between, say $4$ and $6$, we would take the limit as $b\to4^+$.

Answer 3

Note that by definition we have that

$$\int _{ 0 }^{ 4 }{ \frac { 4 }{ x-4 } }dx=\lim _{ b\rightarrow 4^- } \int _{ 0 }^{ b }{ \frac { 4 }{ x-4 } }dx$$

because we are approching the "problematic" point by the left in the interval of integration [0,b].