Exams Joint Entrance Examination Mathematics Area Under Curves Question 73 Question 73 of 230 Medium NUMERICAL
Let the area of the region ( x , y ) : x − 2 y + 4 ⩾ 0 , x + 2 y 2 ⩾ 0 , x + 4 y 2 ≤ 8 , y ⩾ 0 {(x,y):x-2y+4\geqslant 0,x+2{y}^{2}\geqslant 0,x+4{y}^{2}\le 8,y\geqslant 0} ( x , y ) : x − 2 y + 4 ⩾ 0 , x + 2 y 2 ⩾ 0 , x + 4 y 2 ≤ 8 , y ⩾ 0 be m n \frac{m}{n} n m , where m m m and n n n are coprime numbers. Then m + n m+n m + n is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 119
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A = ∫ 0 1 [ ( 8 − 4 y 2 ) − ( − 2 y 2 ) ] d y + ∫ 1 3 / 2 [ ( 8 − 4 y 2 ) − ( 2 y − 4 ) ] d y = [ 8 y − 2 y 3 3 ] 1 0 + [ 12 y − y 2 − 4 y 3 3 ] 3 / 2 1 = 107 12 = m n ∴ m + n = 119 \begin{gathered} A=\int _{0}^{1}[(8-4{y}^{2})-(-2{y}^{2})]dy+ \\ \int _{1}^{3/2}[(8-4{y}^{2})-(2y-4)]dy \\ ={[8y-\frac{2{y}^{3}}{3}]}_{1}^{0}+{[12y-{y}^{2}-\frac{4{y}^{3}}{3}]}_{3/2}^{1}=\frac{107}{12}=\frac{m}{n} \\ \therefore m+n=119 \end{gathered} A = ∫ 0 1 [( 8 − 4 y 2 ) − ( − 2 y 2 )] d y + ∫ 1 3/2 [( 8 − 4 y 2 ) − ( 2 y − 4 )] d y = [ 8 y − 3 2 y 3 ] 1 0 + [ 12 y − y 2 − 3 4 y 3 ] 3/2 1 = 12 107 = n m ∴ m + n = 119 Answer the question to unlock