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Integral Calculus

Blazing goIITian

Joined:	10 Jun 2007
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1 Apr 2009 21:14:12 IST

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247

Definite Integration

None

The whole area contained between the curve y²(a-x) = x²(a+x) and the line x=a (a>0) is

(a)

(b)

(c)

(d)

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Comments (7)

kabi

Blazing goIITian

Joined: 11 Jan 2009

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2 Apr 2009 00:35:32 IST

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Sahil Gupta

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2 Apr 2009 01:09:29 IST

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I am unable to understand how you got the graph.

pranav agrawal

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2 Apr 2009 01:11:44 IST

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SHE HAS DRAWN IT FROM SOME SOFTWARE~~RATHER THIS KINDA GRAPHS ARE NOT IN THE IIT SYLLABUS~~DNT PAY MUCH ATTENTION TO THEM, RATHER, LEARN THOSE BASIC ONE AND WAYS OF GRAPHICAL TRANSFORMATION FROM THE KNOWN BASIC ONES~~~~~~~~~~~~RATE IF LIKE~~

kabi

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2 Apr 2009 01:17:40 IST

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So area required

=2 lim_{x ---> a}integration ( 0 to x )x Sqrt [ (a+x) / (a -x)] dx

=2 lim_{x ---> a}integration ( 0 to x ) x (a+x)/ Sqrt (a^2-x^2) dx

=2 lim_x--->a [ integration ( 0 to x ) xa/ Sqrt (a²-x² )dx + integration (0 to x ) x² / Sqrt ( a² - x² ) dx

for first integration put x² = t

for second just write x² = x² - a²+a²

so it will become Sqrt ( a² - x²)+ a² / Sqrt ( a² - x² )

so u will get this answer after integration

- 2 aSqrt ( a² - x²)+ a² sin^-1( x/a) - x Sqrt (a² - x² )

now putting the limit from 0 to x with x tending a

=a² pi / 2 + 2 a²

= 2 a² ( pi/ 4 +1)

Thank u

kabi

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Joined: 11 Jan 2009

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2 Apr 2009 01:26:03 IST

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Well i used my graphical calulator but pranav i don't think its not a basic graph . It will a min to draw it without any calculator .

pranav agrawal

Blazing goIITian

Joined: 24 Sep 2008

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2 Apr 2009 01:27:56 IST

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how to draw without that calculator??? pls tell~

kabi

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Joined: 11 Jan 2009

Posts: 575

2 Apr 2009 01:41:52 IST

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y = + - x Sqrt ( a+x / (a-x))

Let us do for positive part

domain x >= -a excluding x = a

f( - a ) = 0

f(0)= 0

differentiate f(x) there is a point of max b/w 0 to -a

for x >0

f(x) will increase and as x tends to a f(x) will tends to infinity

now just reflect it for negative part

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