Let f: [2, 5] to [2, 5] be a bijective functi | vedclass

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(D) We use the property of definite integrals for inverse functions: $\int_{a}^{b} f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = b f(b) - a f(a)$.Given

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$21$

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(D) We use the property of definite integrals for inverse functions: $\int_{a}^{b} f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = b f(b) - a f(a)$.Given $f: [2, 5] 	o [2, 5]$ is a bijection,we have $f(2) = 2$ and $f(5) = 5$ (or vice versa,but since the derivative of the inverse is positive,$f$ is strictly increasing,so $f(2)=2$ and $f(5)=5$).Thus,the integral becomes $\int_{2}^{5} f(x) dx + \int_{2}^{5} f^{-1}(x) dx$.Using the formula with $a = 2$ and $b = 5$:$\int_{2}^{5} f(x) dx + \int_{f(2)}^{f(5)} f^{-1}(x) dx = 5 \cdot f(5) - 2 \cdot f(2)$.Substituting the values $f(5) = 5$ and $f(2) = 2$:$= 5(5) - 2(2) = 25 - 4 = 21$.

Let $f: [2, 5] \to [2, 5]$ be a bijective function such that $\frac{d}{dx}(f^{-1}(x)) > 0$ for all $x \in [2, 5]$. Then $\int_{2}^{5} (f(x) + f^{-1}(x)) dx$ is

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