If f(x) and g(x) are inverse functions of eac | vedclass

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(A) Given that $f(x)$ and $g(x)$ are inverse functions,we have $g(x) = f^{-1}(x)$,which implies $f(g(x)) = x$.By differentiating both sides with respe

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(A) Given that $f(x)$ and $g(x)$ are inverse functions,we have $g(x) = f^{-1}(x)$,which implies $f(g(x)) = x$.By differentiating both sides with respect to $x$,we get $f'(g(x)) \cdot g'(x) = 1$,which implies $\frac{1}{f'(g(x))} = g'(x)$.Substituting this into the integral,we get $\int_{1}^{3} (g(x) + x \cdot g'(x)) dx$.This expression is the derivative of the product $x \cdot g(x)$,i.e.,$\frac{d}{dx} (x \cdot g(x)) = g(x) + x \cdot g'(x)$.Therefore,the integral becomes $\int_{1}^{3} \frac{d}{dx} (x \cdot g(x)) dx = [x \cdot g(x)]_{1}^{3}$.Since $f(1) = 3$,we have $g(3) = 1$. Since $f(3) = 1$,we have $g(1) = 3$.Evaluating the definite integral: $[3 \cdot g(3) - 1 \cdot g(1)] = [3(1) - 1(3)] = 3 - 3 = 0$.

If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

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