MCQ
$\int_{}^{} {[f(x)\,g''(x) - f''(x)\,g(x)]\,dx} $=
- A$\frac{{f(x)}}{{g'(x)}}$
- B$f'(x)g(x) - f(x)g'(x)$
- ✓$f(x)g'(x) - f'(x)g(x)$
- D$f(x)g'(x) + f'(x)g(x)$
✓
Answer
Correct option: C.
$f(x)g'(x) - f'(x)g(x)$
c
(c)$\int_{}^{} {[f(x)\,g''(x) - f''(x)\,g(x)]\,dx} $
$ = \int_{}^{} {f(x)\,g''(x)\,dx} - \int_{}^{} {f''(x)\,g(x)\,dx} $
$ = \left( {f(x)\,g'(x) - \int_{}^{} {f'(x)g'(x)\,dx} } \right) - \left( {g(x)\,f'(x) - \int_{}^{} {g'(x)\,f'(x)\,dx} } \right)$
$ = f(x)\,g'(x) - f'(x)\,g(x).$
(c)$\int_{}^{} {[f(x)\,g''(x) - f''(x)\,g(x)]\,dx} $
$ = \int_{}^{} {f(x)\,g''(x)\,dx} - \int_{}^{} {f''(x)\,g(x)\,dx} $
$ = \left( {f(x)\,g'(x) - \int_{}^{} {f'(x)g'(x)\,dx} } \right) - \left( {g(x)\,f'(x) - \int_{}^{} {g'(x)\,f'(x)\,dx} } \right)$
$ = f(x)\,g'(x) - f'(x)\,g(x).$
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