3neone there
If f:R→R is a monotonic,differentiable real valued function a,b are two real numbers and
∫f(x)+f(a))(f(x)-f(a))dx=I1
upper limit→ b and lower limit→a
∫x(b-f-1(x))dx=I2
upper limit→f(b)
lower limit→f(a)
I2/I1=____
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UP 0 DOWN 0 0 2
2 Answers
1Applying uv rule in first integral
I1=
[f(b)2-f(a)2]b - a∫b2xf(x)f'(x)dx
for I2
as f(x) is monotonic we can take x=f(k) and put the limits a and b and further
f-1(f(k)) then = k
so we get it as
a∫bf(k)f'(k)(b-k)dk
we can safely put k as x
so
a∫bf(x)f'(x)(b-x)dx
=>
ba∫bf(x)f'(x)dx-a∫bxf(x)f'(x)dx
putting f(x)=t in the first expression we get finally
=>
b[f(b)2-f(a)2]/2 - a∫bxf(x)f'(x)dx
thus clearly I1/I2=2