Suppose f,f' and f" are continuous functions is [0,e] and that f'(e)=f(e)=f(1)=1 and the intergral of f(x)/x2 from o to e is 1/2, then the value of the integral f"(x)lnxdx from o to e is??????
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2 Answers
1let us consider it as a indefinite one after that we will place the values
let ∫f''(x)lnxdx=I
let f1=first function anf f2=second function in all cases
applying by parts with lnx=f1 and f''(x)=f2
we get
I= lnx∫f''(x)dx - ∫d(lnx)/dx(∫f''(x)dx)dx
=lnxf'(x) - ∫(1/x)f'(x)dx
applying byparts again for the second term with f1=1/x and f2=f'(x)
I= lnxf'(x) - [ 1/x∫f'(x)dx - ∫d(1/x)/dx(∫f'(x)dx)dx]
= lnxf'(x) - [ (1/x)f(x) + ∫f(x)/x2dx]
now applying the limits from 0 to e
we see that ln0 is undefined hence this definite integral is not defined!!
