If f(x) = g(x),then the value of int {f'(x) c | vedclass

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(D) Given that $f(x) = g(x)$.We need to evaluate the integral $I = \int f'(x) \cdot g(x) \, dx$.Since $f(x) = g(x)$,we can substitute $f(x)$ with $g(x

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$A$ or $C$ both

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(D) Given that $f(x) = g(x)$.We need to evaluate the integral $I = \int f'(x) \cdot g(x) \, dx$.Since $f(x) = g(x)$,we can substitute $f(x)$ with $g(x)$ or vice versa.Method $1$: Using substitution.Let $f(x) = t$,then $f'(x) \, dx = dt$.Since $f(x) = g(x)$,we have $I = \int f(x) \cdot f'(x) \, dx = \int t \, dt = \frac{t^2}{2} + c = \frac{{\{f(x)\}}^2}{2} + c$.Since $f(x) = g(x)$,this is also equal to $\frac{{\{g(x)\}}^2}{2} + c$.Method $2$: Using integration by parts.$I = \int g(x) \cdot f'(x) \, dx = g(x) \cdot f(x) - \int g'(x) \cdot f(x) \, dx + c$.Since $f(x) = g(x)$,we have $f'(x) = g'(x)$.$I = g(x) \cdot g(x) - \int f'(x) \cdot g(x) \, dx + c$.$I = {\{g(x)\}}^2 - I + c$.$2I = {\{g(x)\}}^2 + c$.$I = \frac{1}{2}{\{g(x)\}}^2 + c$.Both options $A$ and $C$ represent the same value because $f(x) = g(x)$.Therefore,the correct option is $D$.

If $f(x) = g(x)$,then the value of $\int {f'(x) \cdot g(x)} \, dx$ is

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