# Help Improper Integrals Type 2 argg

• cmab
In summary, the Math homework asks for the integral from 1 to 3 of 1/(sqrt[abs(x-2)])dx. The first integral is from 1 to 2 and the second is from 2 to 3. The limit for this problem has to approch 2 from the right for the x-2, and from the left for -(x-2).
cmab
I don't understand my Math Homework, here's the number that I don't get.

[int a=1 b=3] 1/(sqrt[abs(x-2)]) dx

sqrt = square root
abs = absolute value
Integral from 1 to 3

Can anyone explain this to me clearly, I'll really appreciate

Thanks

I tried and it keeps giving me 0.

2[sqrt(abs(x-2))] from 1 to t and 2[sqrt(x-2)] from t to 3

(2[sqrt(abs(1-2))])-(2[sqrt(abs(t-2))])
and
(2[sqrt(abs(t-2))])-(2[sqrt(abs(3-2))])

(lim t->2+ (2[sqrt(abs(t-2))])-(2[sqrt(abs(1-2))]) )
+
(lim t->2- (2[sqrt(abs(3-2))])-(2[sqrt(abs(t-2))]) )
=4

in the book it says 3/4

Last edited:
Try and remember whenever you have a problem dealing with absolute values, think, "split in two". That is, split it into two parts: one part is when the value between absolute values is positive, and another part when the value is negative. So:

$$|x-2|=x-2 \quad\text{ when }\quad x-2\geq 0$$

$$|x-2|=-(x-2) \quad\text{ when }\quad x-2<0$$

so, split the integral into two, you know, at 2, and then solve it. Remember to replace the absolute value expression in the radical with the appropriate form (from above) for each interval. Can you do this?

saltydog said:
Try and remember whenever you have a problem dealing with absolute values, think, "split in two". That is, split it into two parts: one part is when the value between absolute values is positive, and another part when the value is negative. So:

$$|x-2|=x-2 \quad\text{ when }\quad x-2\geq 0$$

$$|x-2|=-(x-2) \quad\text{ when }\quad x-2<0$$

so, split the integral into two, you know, at 2, and then solve it. Remember to replace the absolute value expression in the radical with the appropriate form (from above) for each interval. Can you do this?

Nice, i check it out on gragphmatica, and the graph of both seems to match the original function. Thanks dude, if i have a question i<ll comme back here :tongue:

But the limit for that problem have to approch 2 from the right for the x-2, and from the left for -(x-2). Right ?

No limit considerations are involved. Just solve two integrals. I'll set up one for you:

$$\int_1^2 \frac{dx}{\sqrt{2-x}}$$

What's the other one?

I did the thing, and it still gives me 4

## 1. What is an improper integral type 2?

An improper integral type 2 is an integral where either the upper or lower limit of integration is infinite or the function being integrated has an infinite discontinuity within the interval of integration. This type of integral requires special techniques to evaluate.

## 2. How do I know if an integral is type 2?

An integral is considered type 2 if it meets the criteria of having an infinite limit of integration or an infinite discontinuity within the interval of integration. This can be determined by looking at the function being integrated and the limits of integration.

## 3. What are some techniques for evaluating improper integrals type 2?

Some techniques for evaluating improper integrals type 2 include splitting the integral into smaller, finite integrals, using substitution or integration by parts, and using the limit comparison test or the comparison test to determine convergence or divergence.

## 4. How do I determine if an improper integral type 2 converges or diverges?

To determine convergence or divergence of an improper integral type 2, you can use the limit comparison test or the comparison test. If the limit of the integral is finite, then the integral is said to converge. If the limit is infinite, then the integral is said to diverge.

## 5. Can improper integrals type 2 have multiple points of discontinuity?

Yes, improper integrals type 2 can have multiple points of discontinuity within the interval of integration. This can make evaluating the integral more challenging, but it can still be done using the techniques mentioned above.

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