If `int_0^x {int_0^u f(t) dt}du` is equal to
(a) `int_0^x (x-u) f(u)`
(b) `int_0^xuf(x-u)du`
(c) `x int_0^x f(u) du`
(d) `x int_0^x uf(u-x)du`

Options

A. `int_(0)^(x)(x-u)f(u)du`
B. `int_(0)^(x) uf(x-u)du`
C. `x int_(0)^(x)f(u)du`
D. `x int_(0)^(x)uf(u-x)du`

Answer

A, B

Explanation

L.H.S `=int_(0)^(x) {int_(0)^(u)f(t)dt}du`
Integrating by parts choose 1 as the second function. Then,
L.H.S`={uint_(0)^(u)f(t)dt}_(0)^(x)-int_(0)^(x)f(u)u du`
`=x int_(0)^(x)f(t)dt-int_(0)^(x)f(u)u du`
`=x int_(0)^(x)f(u)du-int_(0)^(x)f(u) udu-int_(0)^(x)f(u)(x-u)du`
`=R.H.S`.