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

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In summary, the integral ∫from 1 to 8 of (-1/(x-3)^2 dx is divergent and diverges to negative infinity. This is because the limit as x approaches 3 from the left is positive infinity and the limit as x approaches 3 from the right is negative infinity, resulting in an overall limit of negative infinity. Therefore, the integral cannot be evaluated and is divergent to negative infinity.
  • #1
cathy
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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?
 
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  • #2
cathy said:

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##.
 
  • #3
[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
Check the sign of the first infinity.
 
  • #5
isn't it positive? when i graphed it, that graph approaching from the left went straight up
 
  • #6
Not if you graphed 1/(x-3). What's the sign of x-3 when you approach from the left?
 
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  • #7
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
Right!
 
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1. What does the notation ∫ from 1 to 8 mean?

The notation ∫ from 1 to 8 indicates that the integral is being taken over the interval from 1 to 8 on the x-axis.

2. What does the expression (-1/(x-3)^2) represent in the integral?

The expression (-1/(x-3)^2) represents the integrand, which is the function being integrated over the interval from 1 to 8.

3. What does the notation dx at the end of the integral mean?

The notation dx indicates the variable of integration, in this case x. It is used to specify which variable the integral is being taken with respect to.

4. What does it mean for an integral to diverge to -∞?

An integral that diverges to -∞ means that the value of the integral becomes increasingly negative as the limits of integration approach the given interval. In other words, the area under the curve becomes infinitely negative.

5. How can you determine if an integral will diverge to -∞?

An integral will diverge to -∞ if the function being integrated has a vertical asymptote within the interval of integration. In this case, the function (-1/(x-3)^2) has a vertical asymptote at x=3, which falls within the interval from 1 to 8, leading to a divergent integral.

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