IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

johri_anshuman

Given a function y=f(x) such that the angle of the tangent with the horizontal at x=1 is pi/6, at x=2 is pi/3 and at x=3 is pi/4. Then evaluate

₁

³ f''(x).f'(x)dx + ₂

³ f''(x) dx

Rates Assured

rooney

Q> ₁

³ f''(x).f'(x)dx + ₂

³ f''(x) dx
Using integration by parts

f''(x).f'(x)dx = f'(x)

f''(x)dx -

f''(x)(

f''(x)dx)dx

f''(x).f'(x)dx = f'(x)

f''(x)dx -

f''(x).f'(x)dx
2

f''(x).f'(x)dx = (f'(x))²

f''(x).f'(x)dx = ((f'(x))²)/2

Use this in initial question

[ ((f'(x))²)/2 ]₁³ + [ f'(x) ] ³₂
= ( 1 - 1/3 )/2 + 1 - 3^1/2= 4/3 - (3)^1/2