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Integral Calculus

Blazing goIITian

Joined:	27 Jan 2007
Post:	559

20 Nov 2007 18:33:21 IST

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560

Find the Value

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The tangent represented by the graph of the function y=f(x) at point with abscissa x=1 forms an angle pi/6 and at the point x=2 an angle of pi/3 and at x=3 an angle of pi/4. Then find the value of ₁³ f'(x)f''(x)dx + ₂³ f''(x)dx.

Ans 4/3 - root(3)

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Comments (2)

sowjanya gudipati

Cool goIITian

Joined: 13 Oct 2007

Posts: 75

20 Nov 2007 19:41:31 IST

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given the slope of the curve at the points x=1,2,3

slope of the curve f(x)=f¹(x)

f¹(1)=1/root3, f¹(2)=root3, f¹(3)=1

let f¹(x)=t

then f¹¹(x)dx=dt

_{[ ]}

^{[ ]} f¹(x)f¹¹(x)dx=_{[ ]}

^{[ ]} tdt within the limits 1to3

=t²/2 with in the limits 1 to 3

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

_{[ ]}

^{[ ]} dt=t with in the limits 2to 3

=1-root3

1/3+1-root3=4/3-root3

f¹(x) is derivative of f(x)

if there is any doubt in any part ask me

Bipin Dubey

Forum Expert

Joined: 23 Jan 2007

Posts: 7942

20 Nov 2007 20:27:44 IST

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Given
f'(1) = tan

/6 = 1/

3
f'(2) = tan

/3 =

3
f'(3) = tan

/4 = 1

₁

³ f'(x)f''(x)dx + ₂

³ f''(x)dx

= ₁

³ f'(x)d(f'(x)) + ₂

³ f''(x)dx

= {(f'(x))²/2}₁³ + {f'(x}₂³

= (f'(3))²/2 - (f'(1))²/2 + f'(3) - f'(2)

= 1²/2 - (1/

3)²/2 + 1 -

3

= 4/3 -

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