# Homework Help: Converge or diverge

1. Mar 24, 2014

### cathy

1. The problem statement, all variables and given/known data
∫from 1 to 8 of (-1/(x-3)^2 dx
Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, state your answer as INF. If it diverges to negative infinity, state your answer as MINF. If it diverges without being infinity or negative infinity, state your answer as DIV.

ANS: DIVERGES TO -INFINITY

MY PROBLEM: I KEEP GETTING THAT IT DIVERGES TO +INFINITY

3. The attempt at a solution

solved for antiderv and got 1/(x-3)
so lim c--> c- [(1/(x-3)) from 1 to c]+ c--> c+ (1/(x-3))from c to 8.
Working this out, I get
infinity - 3/10 + infinity.
How is the answer diverge to -infinity?

2. Mar 24, 2014

### vela

Staff Emeritus
I think what you actually meant was the integral becomes
$$\lim_{c \to 3^-} \left.\frac{1}{x-3}\right|_1^c + \lim_{c \to 3^+} \left. \frac{1}{x-3}\right|_c^8.$$ Please show us how you evaluated this to get $+\infty$.

3. Mar 24, 2014

### cathy

[1/c-3 approaching from the left gives infinity + 1/2] + [1/5 - 1/c-3 from the positive side is -infinity]
so I get infinity + (-infinity) = infinity
I know I did something wrong there. I just don't know what.

4. Mar 24, 2014

### vela

Staff Emeritus
Check the sign of the first infinity.

5. Mar 24, 2014

### cathy

isn't it positive? when i graphed it, that graph approaching from the left went straight up

6. Mar 24, 2014

### vela

Staff Emeritus
Not if you graphed 1/(x-3). What's the sign of x-3 when you approach from the left?

7. Mar 26, 2014

### cathy

Ohh I see what you mean. I was not distributing my negative after I took the antiderivative and I was graphing -1/(x-3). Thank you very much.
So it would be (-inf + 1/2) + (-1/5 -infinity)= -infinity.

8. Mar 26, 2014

### vela

Staff Emeritus
Right!