Let the area enclosed by the lines x + y = 2, | vedclass

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(A) The function is defined as $f(x) = \min \left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$.For $0 \leq x < 1$,$[x] = 0$,so $f(x) = \min \left\{x^2 + \frac

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) The function is defined as $f(x) = \min \left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$.For $0 \leq x $x^2 + \frac{3}{4} = 1 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2}$.Thus,$f(x) = x^2 + \frac{3}{4}$ for $0 \leq x The area $A$ is bounded by $x=0, y=0, x+y=2$ and $f(x)$.$A = \int_0^{1/2} (x^2 + \frac{3}{4}) dx + \int_{1/2}^1 (1) dx + \int_1^2 (2-x) dx$.$A = \left[ \frac{x^3}{3} + \frac{3x}{4} \right]_0^{1/2} + [x]_{1/2}^1 + \left[ 2x - \frac{x^2}{2} \right]_1^2$.$A = (\frac{1}{24} + \frac{3}{8}) + (1 - \frac{1}{2}) + ((4 - 2) - (2 - \frac{1}{2}))$.$A = \frac{10}{24} + \frac{1}{2} + (2 - \frac{3}{2}) = \frac{5}{12} + \frac{6}{12} + \frac{6}{12} = \frac{17}{12}$.Therefore,$12A = 17$.

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f(x) = \min \left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$,where $[x]$ denotes the greatest integer $\leq x$,be $A$. Then the value of $12A$ is $............$.

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