Exams Joint Entrance Examination Mathematics Area Under Curves Question 78 Question 78 of 230 Medium NUMERICAL
Let the area enclosed by the lines x + y = 2 , y = 0 , x = 0 x+y=2,y=0,x=0 x + y = 2 , y = 0 , x = 0 and the curve f ( x ) = m i n x 2 + 3 4 , 1 + [ x ] f(x)=min{{x}^{2}+\frac{3}{4},1+[x]} f ( x ) = min x 2 + 4 3 , 1 + [ x ] where [ x ] [x] [ x ]
denotes the greatest integer ≤ x \le x ≤ x , be A A A . Then the value of 12 A 12A 12 A is \_\_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 17
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= [ ∫ 0 1 2 ( x 2 + 3 4 ) d x ] + [ 1 2 ( 3 2 + 1 2 ) × 1 ] =[\int _{0}^{\frac{1}{2}}({x}^{2}+\frac{3}{4})dx]+[\frac{1}{2}(\frac{3}{2}+\frac{1}{2})\times 1] = [ ∫ 0 2 1 ( x 2 + 4 3 ) d x ] + [ 2 1 ( 2 3 + 2 1 ) × 1 ]
= [ x 3 3 + 3 x 4 ] 1 2 0 + 1 = 1 24 + 3 8 − 0 + 1 = 1 + 9 + 24 24 = 34 24 = 17 12 \begin{gathered} ={[\frac{{x}^{3}}{3}+\frac{3x}{4}]}_{\frac{1}{2}}^{0}+1 \\ =\frac{1}{24}+\frac{3}{8}-0+1=\frac{1+9+24}{24}=\frac{34}{24}=\frac{17}{12} \end{gathered} = [ 3 x 3 + 4 3 x ] 2 1 0 + 1 = 24 1 + 8 3 − 0 + 1 = 24 1 + 9 + 24 = 24 34 = 12 17
12 A = 12 × 17 12 = 17 12A=12\times \frac{17}{12}=17 12 A = 12 × 12 17 = 17 Answer the question to unlock