integration

It may be d({x})

where {x} = fractional part.

i dont know. answer given is 17

i found this thing in a book.
∫_a^b f(x)dg(x) + ∫_a^b g(x)df(x)= g(b)f(b)-g(a)f(a)
substitute f(x)=(x²+1) & g(x)=[x].
thus, ∫₀ ³ (x² +1)d[x] + ∫₀ ³ 2x[x]dx = 30-0
or, ∫₀ ³ (x² +1)d[x] = 30- ∫₀ ³ 2x[x]dx
or, ∫₀ ³ (x² +1)d[x] = 30-13=17

@aditya- how will the answer vary? if we use the identity in the second line it is correct.
could someone please help to prove the identity?

d[f(x)g(x)] = f(x).d{g(x)} + g(x).{f(x)}
then integrate with limits on both sides.

i think the answer will vary due to - g(b)f(b)-g(a)f(a)
putting b=3 , g(b)=3 but putting b=3^- , g(b)=2

finally got through !

I = ₀∫¹ (x²+1) d[x] = ₀∫^1-h (x²+1)d[x] + _1-h∫¹ (x²+1)d[x]
I₁ I₂

here h→0⁺

now for I₁, d[x] =0 thus I₁=0

for I₂ , in the range 1-h to 1 , x²+1 being constant comes outside the integral

thus I₂ = (1²+1) _1-h∫¹ d[x]

now d[x] =1 since value of [x] changes abruptly at x=1

thus I₂ = 2 * 1 =2

similarly calculating the required integrals from 1to 2 and 2 to 3 ,

we get I (from 0 to 3) = (1²+1) + (2²+1) +(3²+1) = 17

integral (0,3) (x^2+1)d[x] = 17

[x]=x - {x}

d[x]= dx - d{x}

integral (0,3) (x^2+1)d[x]= integral (0,3) (x^2+1)(dx-d{x})
integral (0,3) (x^2+1)d[x]=integral (0,3) (x^2+1)dx - integral (0,3) (x^2+1)d{x}

17=12 - integral (0,3) (x^2+1)d{x}

integral (0,3) (x^2+1)d{x}=- 5

but since (x^2+1) and {x} is positive how is dis continuous summation -ve???

(Note: Ans is 17 and correct...but y dis anomaly?)

yeah you are right. however this was a comprehension given in arihant. and that proof was given.