int [f(x)g''(x) - f''(x)g(x)] dx = | vedclass

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(C) We need to evaluate the integral $I = \int [f(x)g''(x) - f''(x)g(x)] dx$. Using the linearity property of integrals,we can write: $I = \int f(x)g'

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$f(x)g'(x) - f'(x)g(x)$

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(C) We need to evaluate the integral $I = \int [f(x)g''(x) - f''(x)g(x)] dx$. Using the linearity property of integrals,we can write: $I = \int f(x)g''(x) dx - \int f''(x)g(x) dx$. Applying integration by parts to the first term $\int f(x)g''(x) dx$ (taking $u = f(x)$ and $dv = g''(x) dx$): $\int f(x)g''(x) dx = f(x)g'(x) - \int f'(x)g'(x) dx$. Applying integration by parts to the second term $\int f''(x)g(x) dx$ (taking $u = g(x)$ and $dv = f''(x) dx$): $\int f''(x)g(x) dx = g(x)f'(x) - \int g'(x)f'(x) dx$. Substituting these back into the expression for $I$: $I = [f(x)g'(x) - \int f'(x)g'(x) dx] - [g(x)f'(x) - \int g'(x)f'(x) dx]$. The integral terms $\int f'(x)g'(x) dx$ cancel out: $I = f(x)g'(x) - f'(x)g(x) + C$. Thus,the correct option is $C$.

$\int [f(x)g''(x) - f''(x)g(x)] dx =$

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