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  • Let the area enclosed by the lines,<img alt="y=0 x=0" src="https://entrancecorner.oncodecogs.com/gif.latex?

Let the area enclosed by the linesx+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 x, be A. Then the value of 12A is______.

Option: 1

17


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \int_0^{\frac{1}{2}}\left(\mathrm{x}^2+\frac{3}{4}\right) \mathrm{dx}+\frac{1}{2} \times\left(\frac{3}{2}+\frac{1}{2}\right) \times 1 \\ & =\left[\frac{\mathrm{x}^3}{3}+\frac{3 \mathrm{x}}{4}\right]_0^{\frac{1}{2}}+1 \\ & \mathrm{~A}=\frac{1}{24}+\frac{3}{8}+1 \\ & 12 \mathrm{~A}=\frac{1}{2}+\frac{36}{8}+12 \\ & =\frac{1}{2}+\frac{9}{2}+12 \\ & =5+12 \\ & =17 \end{aligned}

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Deependra Verma

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