Question

how to solve this:
\[\int\limits_{4}^{9} \frac{ x+1 }{ \sqrt{x} }\]

Answer 1

You can split the fraction.

Answer 2

Then you get two integrals that you can evaluate easily.

Answer 3

Apply a change of variable, it's more simple.

Answer 4

sqrtx=u would by my substitution here.

Answer 5

\[\frac{ x }{ \sqrt{x} } (\frac{ 1 }{ \sqrt{x} })\]

Answer 6

in case you want to split it, remember the denominator affects all the terms:
\[\frac{ x+1 }{ \sqrt{x} }=\frac{ x }{ \sqrt{x} }+\frac{ 1 }{ \sqrt{x} }\]

Answer 7

\[\sqrt{x}=u\]\[\frac{dx}{\sqrt{x}}=2du\]
rewrite our original substitution \[x=u^2\]
we can now plug it in \[\int\limits_2^3 (u^2+1)2du\] notice the bounds change nicely since 9 and 4 are perfect squares.

Answer 8

\[\frac{ x }{x ^{\frac{ 1 }{ 2 }} } \]

Answer 9

@Owlcoffee : oh i get it..

Answer 10

Kainui: must i change it to u ?

Answer 11

Kainui is doing a very nice change of variable.
When you change it to u, the whole integral changes.

Answer 12

No, it's just one way of many to do it. All are good, but I think this is easiest for me. You might like something else, I'm just showing you my way. You're allowed to have your own path, so find out what it is and take it. =)

Answer 13

can i know what is the other alternative way? i want to compare which one might be the easiest for me.

Answer 14

Sure, that's what all the other people here are trying to say. I can help explain those better if you'd like.

Answer 15

There are many ways of finding a primitive of a integral.
Now a "convenient" way is a different thing.

Answer 16

yes, i'd like you explain it. 'convenient' way ?

Answer 17

Change of variable, like kainui is saying, or you could split the fraction inside the integral to make it into a sum of integrals that are more simple:
\[\int\limits_{4}^{9}\frac{ x+1 }{ \sqrt{x} }dx\]
\[\int\limits_{4}^{9}\frac{ x }{ \sqrt{x} }dx + \int\limits_{4}^{9}\frac{ 1 }{ \sqrt{x} }\]
\[\int\limits_{4}^{9}x ^{\frac{ 1 }{ 2 }}dx + \int\limits_{4}^{9}x ^{-\frac{ 1 }{ 2 }}dx\]
but the "way" is up to you, and when you analyze and decide the smplest way of solving it.

Answer 18

why this one, \[\int\limits_{4}^{9} x^{\frac{ 1 }{ 2 }} dx\] not negative 1/2 like the other one.

Answer 19

property of exponents ;)

Answer 20

hmm ? what is tht

Answer 21

Oh, I can't explain it now, I have to leave for anathomy class. But it's a sequence of properties that bounds the exponents. type in google images "laws of exponent". And you'll see.

Answer 22

okay, thank you .

Answer 23

\( \dfrac{x^a}{x^b} = x^a x^{-b} = x^{a-b} \qquad x^a x^b = x^{a+b} \)

\( \dfrac{x}{\sqrt{x}} = \dfrac{x^1}{x^{1/2}} = x^1 x^{-1/2} = x^{1-1/2} = x^{1/2} \)
This is all good Algebra to know for manipulating your expressions.

Answer 24

Or if you like, \( \dfrac{x}{\sqrt{x}} = \dfrac{\sqrt{x}^2}{\sqrt{x}} =\dfrac{\sqrt{x} \sqrt{x}}{\sqrt{x}} \)

Answer 25

@AccessDenied thank you! i understand it totally...