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# quick integration question watch

1. Is this integrable? [(-y^2)-(2y)+3]^(1/2)

How would i go about doing this?
2. .

Complete the square on , and then try making a suitable hyperbolic/trigonometric substitution. I've not tried it, but I think the following substitution will work:

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(after obtaining )
3. Hey thanks for the reply, i used the subsitution y+1=2sinu (as u said) and i got and answer of
(1/2)(y+1)[(4-(y+1)^2)] + sin^-1 [(y+1)/2]

do you reckon it looks right?

sorry if ur having trouble with reading my answer (im not good with that math typing software thing)
4. (Original post by dopemann)
Hey thanks for the reply, i used the subsitution y+1=2sinu (as u said) and i got and answer of
(1/2)(y+1)[(4-(y+1)^2)] + sin^-1 [(y+1)/2]

do you reckon it looks right?

sorry if ur having trouble with reading my answer (im not good with that math typing software thing)
I think it should be 2 sin^-1 etc.
5. Another question, if i was to put the limits of integration between y=1 and y=0 , would i get my answer in degrees? because sin^-1 [(y+1)/2] would give degrees if i put in y=1 or y=0?
7. Yeh sorry mr m, i forgot to write that
8. Ok, no problem, as the general suggested, when working with differentiation and integration, you should ALWAYS work in radian measure.
9. are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
10. (Original post by dopemann)
are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
The value of that integral with those limits is 1.23 to 3 sf.

Imagine the area represented by that integral. The value of the function when is and its value when is . If you calculate the area of a rectangle of height and width you will only obtain as the answer and the area of the integral must clearly be less than this.
11. how did you get that? that looks like the right answer but i dont understand how to get that...
12. (Original post by dopemann)
are you sure its radians, because i get the answer 119.1339746 (seems like degrees to me)
Is your calculator in degree mode?
13. o i understand now... i must use radian mode when plugging in y=1 and y=0 because after all these are in radians (obviously - silly me). so yeh i get the answer 1.23.

Thankyou everyone, appreciate the help
14. You are welcome.

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