Question

integral 1 to 9 (x-1)/ sqrt x dx

Answer 1

rmalik, this is the 5th or 6th integration problem you've posted.

This one is solved in the same way as all the others.

Answer 2

write as two fractions

Answer 3

So I suggest you go back first and make sure you understand the solutions to all those other problems. If you do, you'll have no problem with one.

Answer 4

but its difficult...how..i know sqrt x is x^1/2

Answer 5

Yes, so
\[\frac{x-1}{\sqrt{x}} = x^{1/2} - x^{-1/2}\]

Answer 6

So now, if you understand the other problems, you should know how to integrate this.

Answer 7

but what happens with the numerator?

Answer 8

\[\frac{x-1}{\sqrt{x}} = \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}}\]

Answer 9

Now,
\[\sqrt{x} = x^{1/2}\]
and
\frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = x^{-1/2}

Answer 10

\[\frac{1}{\sqrt{x}} = \frac{1}{x^{1/2}} = x^{-1/2}\]

Answer 11

So that is how I changed the function you are integrating into the form a few posts above

Answer 12

Now ... let me ask you this. In general, what is
\[\int\limits x^p \ dx\]
where p is not -1 ?

Answer 13

what is this indefinite integral?

Answer 14

all of your exercises have depended on this integral

Answer 15

x^2/2

Answer 16

That's right for p = 1. But what about p in general

Answer 17

i dont know

Answer 18

Right, so to get to that answer, can you tell me what is the derivative of x^n?
\[\frac{d \ }{dx} x^n\]

Answer 19

nx

Answer 20

Not quite.

Answer 21

or would it be x^n+1

Answer 22

not quite

Answer 23

if you dont want to help me dont bother

Answer 24

\[\frac{d \ }{dx} x^n = nx^{n-1}\]

Answer 25

I am helping you. Be patient. Do you recognize this formula?

Answer 26

yes

Answer 27

Right. Now, integration is "anti-differentiation", so what is the integral of x^{n-1}
\[\int\limits x^{n-1} dx\]

Answer 28

i.e., what function when differentiated gives x^(n-1) ?

Answer 29

but im not understanding the problem i asked...since its x^1/2-x^-1/2...then i solve the integral which would be 1^1/2-1^-1/2-9^1/2-9^-1/2

Answer 30

In general, for p not equal to -1
\[\int\limits x^p \ dx = \frac{1}{p+1} x^{p+1} + C\]

Answer 31

And that is because
\[\frac{d \ }{dx} x^{p+1} = (p+1) x^p\]

Answer 32

So now, with that formula you can calculate
\[\int\limits (x^{1/2} - x^{-1/2}) \ dx \]

Answer 33

yes like i showed above is that how i would do it?

Answer 34

No, what you wrote above was wrong. That's why I took back into the basics here so you would understand why.

Answer 35

Using the integration formula above for x^p, what is
\[\int\limits x^{1/2} \ dx\]
?

Answer 36

3x^2/3/2

Answer 37

p = 1/2 so the integral is
\[\frac{1}{1 + 1/2} x^{1 + 1/2} = \frac{1}{3/2} x^{3/2} = \frac{2}{3} x^{3/2}\]

Answer 38

o i had it the wrong way

Answer 39

What now is
\[\int\limits x^{-1/2} \ dx\]

Answer 40

hold on i will write on paper first

Answer 41

it would be 2x^1/2?

Answer 42

Yes.

Answer 43

ok cool now please help with next steps i think i get it now

Answer 44

So now you know how to do
\[\int\limits_1^9 \frac{x -1}{\sqrt{x}} \ dx\]

Answer 45

wait wait wait....so heres what i am doing....

Answer 46

im sorry i dont

Answer 47

what is the indefinite integral?
\[\int\limits \frac{x-1}{\sqrt{x}} dx\]
?

Answer 48

2x^1/2?

Answer 49

so i would just plug in the intrgral og 0 to 9?

Answer 50

If that were right, then the derivative of (2x^(1/2)) would be the function we're trying to integrate.

But it's not. The derivative of (2x^(1/2)) is just x^(-1/2)

Answer 51

What is the function we're trying to integrate?

Answer 52

i really dont understand and im serious

Answer 53

We are trying to integrate the function
\[\frac{x-1}{\sqrt{x}}\]

Answer 54

Now, as it stands, that's not a function we can deal with. So we figure out how to write in terms of things we understand.

Answer 55

you said its = to x^1/2=x^-1/2

Answer 56

\[\frac{x-1}{\sqrt{x}} = \frac{x-1}{x^{1/2}} = \frac{x}{x^{1/2}} - \frac{1}{x^{1/2}}\]

Answer 57

And we can simplify that further
\[\frac{x}{x^{1/2}} - \frac{1}{x^{1/2}} = x^{1-1/2} - x^{-1/2} = x^{1/2} - x^{-1/2}\]

Answer 58

Do understand everything I just did?

Answer 59

so it becomes x^1/2-x^-1/2..dx which result to become (2x^3/2)/3-2x^1/2

Answer 60

Yes

Answer 61

then i compute the intergral??

Answer 62

Yes

Answer 63

FINALLY!!!!!! haha

Answer 64

Tell me when you have the answer.

Answer 65

so it would be...well no answer yet...im up to (2^3/2)/3-(2^1/2)-(54/3)-6..im not sure if that's right

Answer 66

Ok. Let's make sure agree on the indefinite integral
\[\int\limits x^{1/2} - x^{-1/2} \ dx = \frac{2}{3} x^{3/2} - 2 x^{1/2} \ ( \ + C \ )\]

Answer 67

yes

Answer 68

So now if you want the definite integral from x = 1 to 9, substitute x=9 of this expression minus x = 1 of this expression.

What do you get when you substitute x = 1?

Answer 69

(2(1)^3/2)/3-2(1)^1/2= 2/3 -2?

Answer 70

yes ... which is ....?

Answer 71

-4/3?

Answer 72

of course.

Answer 73

Now, what is the integral evaluated at x = 9?

Answer 74

well i got 54/3-6

Answer 75

(54/3)-6?

Answer 76

which simplifies to ...

Answer 77

35/3?

Answer 78

which is 40/3?

Answer 79

No ... 54 = 9 x 6 = 18 x 3
Hence 54/3 = 18

Answer 80

Right. The definite integral is equal to 40/3.

Answer 81

That is the final answer.

Answer 82

ok wonderfull thanks sooooooooooooooooooooooo much!!

Answer 83

So now, do yourself a favor. Take a blank piece of paper and work the entire problem again from the beginning.

Then take another piece of blank piece of paper and work the entire problem without looking at the first piece or our discussion here.

When you can do that, then you really understand this problem.

Answer 84

ok great help thanks agian

Answer 85

Also, go back and look at the other questions you've asked and make sure you understand the solutions.

Answer 86

ok ... good luck.