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is the value of dy similar to the value of dx?

why it become a^2?

then we need to re-substitute back after integration, a^2= 1+x^2 before integrating w.r.t x

*after integration w.r.t y,

in this we are integrating y and we assume as x as constant?

what do you mean by w.r.t y, ?

yes, true, so we'll get answer in 'x' after first integrating w.r.t y (with respect to y)

i'm sorry don't know how to plot this..

i'm sorry i forgot our previous lessons about that..

i know this x=0,y=0,x=2 but i don't know y=x^2/2..i'm sorry

so, plot these x=0,y=0,x=2 using 'Draw' tool here.

|dw:1362400369683:dw|

is it correct?

where is the line x=2 ?

|dw:1362400538076:dw|

sorry i forgot x = 2..

why it becomes parabola?

|dw:1362400889186:dw|
to
|dw:1362400935932:dw|

i'm sorry i forgot how to find the equation of the line

by the way, did you get that final diagram ?

not yet..

which part you have doubt with ?

\[x =2 and x = \sqrt{2y}\] i don't know how to fnd the equation of this line..

x= 2, x=\(\sqrt{2y}\)
so, 2= \(\sqrt{2y}\)
can u find 'y' from here ?

no..

\[y=\sqrt{2}/2 ?\]

yes that's true i almost forgot the topics about integration that's why i am like this..

du = 2x + 2y

yes..

so, your inner integral will become,
\(\int \dfrac{du}{2\sqrt u}\)
can you integarte this ?

1/2..is it correct?

as long as weve gotten this u substitution correct:
\[\int \frac{1}{2\sqrt{u}}du=\sqrt{u}\]

y^2+2y+1 = (y+1)^2 sooo
\[\int\sqrt{5+y^2}-(y+1)~dy\]

why it becomes subtraction?

why it become y^2+2y+1 = (y+1)^2 ?

because: \(\Large \int_{a}^{b} f(x)~dx=F(b)-F(a)\)

ah ok tnx..

in this time we are integrating y?

what's next?

the next integration is next ...
\[\int\sqrt{5+y^2}-(y+1)~dy\]

\[\sqrt{5 + y} - y\]

is it correct?i'm sorry have a problem in integration..

the square root part seems a little off to me

and it looks like you tried a derivative onthe y+1 parts instead of integeration

\[-\int y+1~dy=-(\frac{y^2}{2}+y)\]

yes.. i'm sorry a have a problem in integration..

why is that u not integrate \[\sqrt{5 + y^2}\]?

ah.. wat is this \[\sqrt{a^2 + x^2}\]? how will i use this?

i looked it up in a table, just to be sure. Look at #30

a^2 = 5, and x = y

\[\int \sqrt{5+y^2}dy=\frac12y\sqrt{5+y^2}+\frac12 5^2ln(y+\sqrt{5+y^2})\]from y=0 to y=2

the 5^2 ln part is a typo; just 5 ln ...

\[y/2 \sqrt{5 + y^2} + 5/2 \ln (y + \sqrt{5 + y^2}\]

yes

what's next?

plug in the values of your limit to determine the values

\[(3+\frac52 ln(5))-(\frac54ln(5))-4\]
the -4 is from the simpler part we did earlier

with any luck this simplifies to:\[\frac54ln(5)-1\]

is this the answer? why -4 ?

is that the answer?

is what the answer?

i'm sorry..i don't understand it. it's a little bit difficult... :(

i must study first and review about integral in order for me to understand this..

good luck :)

please teach me again next time..thanks for your time.. :)