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Can you do it ?

find antiderivative twice for a = 3-t

Therefore, the answer is 18.
Something which appeared in the (NUST) test.

REALLY? O_O

Yeah

An O'levels student who have Add math can solve this too! D: D: D:

Hmm,good :)

What about hyperbolic functions ?
You've done that ?

give me the problem, then i can tlel.

\[\int\limits_{}^{}e^x[\sinh^{-1} x+\frac{ 1 }{ \sqrt{1+x^2} }]dx=\]

INtegration by parts?

idk how to deal with sinh^-1

So you don't know hyperbolic functions ?

use e^x(f(x) + f'(x)) rule
differentiation of arcsinh(x) = 1/sqrt(...)

I guess no, then. :S

We have to learn that parallelogram thing. But i can find the area/vol of a curve and stuff

Lets see.
find the Area A+B
|dw:1356722864399:dw|

NOOO! Give me more difficult one.

Ok.

find the area A
|dw:1356723204131:dw|

|dw:1356723438919:dw|

Let me grab a pen and paper.

Hmm :)

Is calc allowed?

No not allowed+You changed the question lol :P

Sinx graph starts from zero. lol.

Oh okay then :/

Haha.

Now solve it .

isk what's sin(pi/4). i can only solve with the help of calc.

180/4

area under cos is 1/sqrt2

Sin45=0.707

0.707-0.293 ?

0.414

Don't do it in decimals.

dang.

1/sqrt2 - (-1+sqrt2)/sqrt2

Making:
(2-sqrt2)/sqrt2

Making:
\[\frac{2\sqrt2-2}{2}\]

sqrt2-1 final answer?

\[\sqrt{2}-1 \ sq.unit\]
Yeah right :D

Wheo! B)

One more.
Find the area enclosed by x=y^2 and y=x-2

umm okay. solving.

k.

1,4

area under line = 2 squnits

trying to figure out about x=y^2

1/2 final answer?

Let me give you options lol
A)9/2 sq.units
B)9/4 //
C)9/7 //
D)7/9 //
E)4/9 //

Can't fig out.

I also couldn't figure this out lol :P

Find the area enclosed by x=y^2 and y=x-2

this is double integration problem.. right ?

i don't think so.

plot both curves..

|dw:1356725537575:dw|

sorry, got disconnected.

\(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy\)

find y2 and solve the integral

sorry, y1 and y2 limits should be
|dw:1356727654740:dw|