Question

hi there I have some integration problems. check this please

Answer 1

\[\int\limits_{2}^{3} \frac{ dx }{ \sqrt{3-x} }\]

Answer 2

\[\int\limits_{0}^{e^2}-\frac{ 1 }{ x(\ln x)^2 } dx\]

Answer 3

\[\int\limits_{2}^{3}(3-x)^{\frac{ -1 }{2 }}\]\[(3-x)^{\frac{ -1 }{2 } + 1}\]\[\frac{ (3-x)^{\frac{ 1 }{2 }} }{ \frac{ 1 }{ 2 }}\times (-1)\]
\[-2(3-x)^{\frac{ 1 }{2 }} \] apply the limit here 2to 3 you will get the answer.

Answer 4

for the second one .. substitute ln(x) = u
then du = dx/x ... can you do it now ?

Answer 5

for the first one i did it without limit. Is it wrong?

Answer 6

Wow! you guys keep giving a way answers and hints. No help at all!!!

Answer 7

I got 2, for the first one

Answer 8

the second, I got 1/2

Answer 9

Which one are you doing right now?

Answer 10

lets check the firs please

Answer 11

OK! So let's do this again but I'm going to teach you how to solve it yourself instead of giving you the answers. Is that cool with you?

Answer 12

ok

Answer 13

thanks

Answer 14

ya it would be good i cna also learn how to explain.. bcos i am not good at explaining integrals.. @calculusfunctions thanks.

Answer 15

\[\int\limits_{2}^{3}\frac{ dx }{ \sqrt{3-x} }\]First of all, what type of integral do we have here? Definite integral or indefinite integral?

Answer 16

@nubeer thank you! Your answer was right so now I'm going to teach you guys how to arrive at the answer using a different approach. Cool? First I'm waiting for your reply to the question I asked. C'mon guys!

Answer 17

defined.

Answer 18

@ahmed84 yes but I think you mean definite, not defined. lol

Answer 19

no I do not know.

Answer 20

hmm because i missed dx , formula or integral sign?

Answer 21

loool

Answer 22

A definite integral is one which is evaluated over a given interval. Do you understand?

Answer 23

yep I understood

Answer 24

@nubeer no not because of that. See if you can figure out why. If not then you will know at the end of this lesson.

Answer 25

Now have you ever learned the substitution method?

Answer 26

yep I know it

Answer 27

yes..

Answer 28

I will right what I have done and you mark it. Deal?

Answer 29

u = 3-x
du/dx = -1

Answer 30

then x = 2 so u = 1
x = 3 so u = 0

Answer 31

Alright then let's view the integral again\[\int\limits_{2}^{3}\frac{ dx }{ \sqrt{3-x} }\]What do you think u should equal?

Answer 32

u = 3-x

Answer 33

Please do not type while I am. Deal?

Answer 34

ok

Answer 35

Yes! u = 3 - x Very good. Now what?

Answer 36

now we will take du/dx?

Answer 37

OK so what is du in terms of dx?

Answer 38

du = -dx

Answer 39

Very good! Now do we leave it like that or do we manipulate it?

Answer 40

I have to manipulate it

Answer 41

I mean in the integrand are we replacing dx or -dx?

Answer 42

Yes so then what?

Answer 43

\[\int\limits_{1}^{0} \frac{ du }{ \sqrt{u} }\]

Answer 44

is it right?

Answer 45

sorry I miss (-) in front of the integral

Answer 46

i think a minus sign should also be there.

Answer 47

Yes! now before we evaluate this integral\[-\int\limits\limits_{1}^{0}\frac{ du }{ \sqrt{u} }\]there is one thing else we can do first. Do you know what that is?

Answer 48

take u in numerator?

Answer 49

I don't mean write the radical in exponent form. Not yet.

Answer 50

Do you see the the interval of the integrand? The lower limit is larger than the upper limit. Does that bother anyone else besides me?

Answer 51

Do you know how to rectify this?

Answer 52

no .. i dont know.

Answer 53

The interval should be from a to b, where a is less than b. Hence \[\int\limits_{b}^{a}f(x)dx =-\int\limits_{a}^{b}f(x)dx\]

Do you understand?

Answer 54

ya i get it..

Answer 55

In other words, whenever you switch the lower and upper limit, you multiply the integral by negative one.

Answer 56

Good so then can you guys tell me what are new integral looks like now? Also @ahmed84 this was your question so why are you not participating more.

Answer 57

I mean ... our new integral ...

Answer 58

lol i think here should be a tab of rules in which few rules should be defined like if you post u have to participate and always say please and thank you... lol that would be good. :P

Answer 59

One step at a time so can you delete that answer my question please?

Answer 60

@nubeer thanks but I don't need a thank you. I just need you to learn. That's my reward.

Answer 61

lol i am jsut saying.. its just a way of respect for the one from whom u learn.. thats all..

Answer 62

@nubeer I know. I wish more students were like you. It's sad that as a teacher I really don't see more.

Answer 63

yeah true.. :(
well lets not get side tracked and continue the question..

Answer 64

I know I'm still waiting to receive your answers to my last question.

Answer 65

\[\int\limits_{0}^{1}\frac{ du }{ \sqrt{u} }\]

Answer 66

@ahmed84 do you agree?

Answer 67

yep I agree with you. I actually went to my teacher today and she said you have to do it like this it is correct . I will write my answer and you see it

Answer 68

No I mean do you agree with @nubeer 's step.

Answer 69

\[=-\int\limits_{1}^{0} \frac{ du }{ \sqrt{u}}= -2[u^.5] = 2 \]

Answer 70

do I need to right limit or not?

Answer 71

So now we have\[\int\limits_{0}^{1}u ^{-\frac{ 1 }{ 2 }}du\]Right?

Answer 72

@ahmed84 I'm trying to teach the logical steps you need to arrive at the final answer you you keep posting the final answer. Do you not want my help?

Answer 73

of course , I want your help.

Answer 74

Anyway your answer @ahmed84 is correct so we'll move on.

Answer 75

From now on I'd like to see the complete solution from you rather than just the answer if you want me to continue teaching you. Deal?

Answer 76

Now what was the next integral you needed help with?

Answer 77

the second please

Answer 78

I need this \[\int\limits_{0}^{e^2}- \frac{ 1 }{ x(linx)^2 }dx\]

Answer 79

I came up with this answer, 1/2

Answer 80

Next we have
\[- \int\limits_{0}^{e ^{2}}\frac{ 1 }{ x (\ln x)^{2} }dx \]

Answer 81

I wrote u = lnx
du = 1/x dx

Answer 82

Are you certain that the lower limit is 0?

Answer 83

yep this what i got

Answer 84

OK! so then we have to approach this integral from the positive side of 0 because the domain of y = ln x is x > 0. Understand? Have you ever done these types of problems before?

Answer 85

yep I understood

Answer 86

In other words,\[\lim_{t \rightarrow 0^{+}}\int\limits_{t}^{e ^{2}}-\frac{ 1 }{ x(\ln x)^{2} }dx\]Do you understand?

Answer 87

no, here is my problem in fact. I do not know about the limit??/?

Answer 88

So are you telling me that you have never had a lesson on these types of problems or are you telling me that you have learned them but do not understand them? Which one is it?

Answer 89

I did integration but I do not understand limit

Answer 90

My question was not whether you have done integration because I can see that you have, rather my question was have you ever learned integrals with limits?

Answer 91

It was long time ago

Answer 92

OK well it's simple. First solve\[\int\limits_{t}^{e ^{2}}-\frac{ 1 }{ x(\ln x)^{2} }dx\]Solve this integral first. Once we have done this then we need to worry about the limit part. So go ahead and show me all your steps. Alright?

Answer 93

ok

Answer 94

\[-\int\limits_{0}^{e^2} \frac{ 1 }{ u^2 }du = -\int\limits_{0}^{e^2} u^-2= [\frac{ u^-1 }{-1 }] = [\frac{ 1 }{ u }]\]

Answer 95

Please take a closer look at the integral I wrote. I replaced the lower limit with a t so why do you still have 0 ?

Answer 96

no I think

Answer 97

Also shouldn't we have changed the lower and upper limits in terms of u now? Please do it again.