If f is defined on R such that f(x) f(-x) = 9 | vedclass

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(D) Let $I = \int_{-23}^{23} \frac{dx}{3+f(x)} \dots (i)$ Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,where $a = -23$ and $b

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$\frac{46}{6}$

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(D) Let $I = \int_{-23}^{23} \frac{dx}{3+f(x)} \dots (i)$ Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,where $a = -23$ and $b = 23$,we have $a+b = 0$. So,$I = \int_{-23}^{23} \frac{dx}{3+f(-x)}$. Given $f(x)f(-x) = 9$,we have $f(-x) = \frac{9}{f(x)}$. Substituting this into the integral: $I = \int_{-23}^{23} \frac{dx}{3 + \frac{9}{f(x)}} = \int_{-23}^{23} \frac{f(x) dx}{3f(x) + 9} = \int_{-23}^{23} \frac{f(x) dx}{3(f(x) + 3)} \dots (ii)$ Adding $(i)$ and $(ii)$: $2I = \int_{-23}^{23} \left( \frac{1}{3+f(x)} + \frac{f(x)}{3(f(x)+3)} \right) dx$ $2I = \int_{-23}^{23} \frac{3 + f(x)}{3(3+f(x))} dx = \int_{-23}^{23} \frac{1}{3} dx$ $2I = \frac{1}{3} [x]_{-23}^{23} = \frac{1}{3} (23 - (-23)) = \frac{46}{3}$ $I = \frac{46}{6}$

If $f$ is defined on $R$ such that $f(x) f(-x) = 9$,then find the value of $\int_{-23}^{23} \frac{dx}{3+f(x)}$.

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