The value of the integral int_{4}^{10} frac{[ | vedclass

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(D) Let $I = \int_{4}^{10} \frac{[x^2]}{[(x-14)^2] + [x^2]} dx$ ... $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a

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$3$

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(D) Let $I = \int_{4}^{10} \frac{[x^2]}{[(x-14)^2] + [x^2]} dx$ ... $(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a=4$ and $b=10$,we have $a+b-x = 14-x$.Substituting $x$ with $14-x$ in the integral:$I = \int_{4}^{10} \frac{[(14-x)^2]}{[(14-x-14)^2] + [(14-x)^2]} dx$$I = \int_{4}^{10} \frac{[(14-x)^2]}{[(-x)^2] + [(14-x)^2]} dx$$I = \int_{4}^{10} \frac{[(14-x)^2]}{[x^2] + [(14-x)^2]} dx$ ... $(2)$Adding $(1)$ and $(2)$:$2I = \int_{4}^{10} \frac{[x^2] + [(14-x)^2]}{[x^2] + [(14-x)^2]} dx$$2I = \int_{4}^{10} 1 dx$$2I = [x]_{4}^{10} = 10 - 4 = 6$$I = 3$

The value of the integral $\int_{4}^{10} \frac{[x^2]}{[(x-14)^2] + [x^2]} dx$,where $[x]$ denotes the greatest integer function,is

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