If int f(x) , dx = g(x),then int f^{-1}(x) , | vedclass

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(B) Given that $\int f(x) \, dx = g(x)$.We need to evaluate $\int f^{-1}(x) \, dx$.Using integration by parts,let $u = f^{-1}(x)$ and $dv = dx$. Then

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$x f^{-1}(x) - g(f^{-1}(x))$

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(B) Given that $\int f(x) \, dx = g(x)$.We need to evaluate $\int f^{-1}(x) \, dx$.Using integration by parts,let $u = f^{-1}(x)$ and $dv = dx$. Then $du = \frac{d}{dx} f^{-1}(x) \, dx$ and $v = x$.$\int f^{-1}(x) \, dx = x f^{-1}(x) - \int x \frac{d}{dx} f^{-1}(x) \, dx$.Let $f^{-1}(x) = t$,then $x = f(t)$ and $dx = f'(t) \, dt$.Also,$\frac{d}{dx} f^{-1}(x) \, dx = dt$.Substituting these into the integral:$\int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(t) \, dt$.Since $\int f(t) \, dt = g(t)$,we have:$\int f^{-1}(x) \, dx = x f^{-1}(x) - g(t) = x f^{-1}(x) - g(f^{-1}(x))$.Thus,the correct option is $B$.

If $\int f(x) \, dx = g(x)$,then $\int f^{-1}(x) \, dx$ is equal to

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