Question

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2. Find the area enclosed by the two curves. You must determine the endpoints of integration. a. y = x4 – 28x, y = -x b. y = (x + 3)2 – 9, y = -f(x+3)2 + 3

Answer 1

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ANSWER – (A)Region bounded by curve y = x+ - 28x & y = - x is shown below. y = -X -80 -60 -40 -20 0 20 40 60 y = x4 – 28x -20 -40 + Now, to find intersection point → x4 – 28x = -x + x4 – 27x = 0 − x(x3 – 27) = 0 = x = 0 & x3 – 27 = 0 + x = 3 then points are (0,0) & (3,0). =So area is given by S. [-~= (x*–28x)] dx – A = S. ° [-x* +27x] dx 3 PA= BA= +*+2) --=[***] PA= 729 -0BA= #ANS.... 10 1 ANSWER – (B)Region enclosed by curve y = (x + 3)2 -9 & y = -3(x+3)2 + 3 is shown below. y = (x + 3)2 - 9 -15 -10 -5 0 10 1 y = - (x + 3)2 + 3 1 Intersection of two curves is given by = (x + 3)2 – 9 = - (x + 3)2 + 3 4 = (x + 3)2 + =(x + 3)2 = 3 +9—=(x + 3)2 = 12 + (x + 3)2 = 9+ (x + 3) = +3 = x = +3 – 3 → x = -6,0 then y = 0,0 i.e. points are (-6,0) & (0,0). Now area enclosed by them is given by PA= (x + 3)2 + + -934 BA= L[+$*+32 + 12] Basic rule(1) | (x + a)" dx = (x + a)+1 n +1 4(x + 3) BA= + 12x 3 3 -6 4 (0+3) 4 (-6+3) BA= +0 + 12(-6) 3 3 3 3 PA = 48 #Ans...