Solve ∫from 1 to 8 of (-1/(x-3)^2 dx: Diverge to -INF

[cathy]

Homework Statement

∫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

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?

 

[vela]

Homework Statement

∫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

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?

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 $+\infty$.

 

[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.

 

[vela]

Check the sign of the first infinity.

 

[cathy]

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

 

[vela]

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

 

[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.

 

[vela]

Right!