Question

Q6 By sketching the graphs, find the area of the region bounded by, y = x4 - 3x2 – 7 and y = x2 + 5. 07 Using integration, find the area of the triangle joining the points whose coordinates are (4,0), (8,6)and (0,4).

plzzz

Answer 1

value of Evaluating a function Step 3 at some points by v -1 as! as-3 f 1-3) = 47 2 A (-1) = -9 f(1) = -9 & x=3 (3) - 47 Hence the graph is symmetric about y-axis step of finding Minimize / Maximize f(x) = x - 32° -7 f (x)= 42² - 6x f' (a). o 4a²-6x = 0 al 442-6)=0 x =0 x²=6 . X=0 a = t do x=2 1.22 f"(x) = 19x² - 6 t" 220 strike 4"(+ 1.22) >0 Hence function has minimum valus at point 11 2 x = t 1.22

having vertex Y= x²ts since this a upward parabola 10,5) x - 3x² - 7 = xurs QU- 49?-12-0 x - 6x² + 2x² - 12=0 2? la?-6) +212-6)=0 x ²-6=0 x²+2=0 22=6 x = a ſo

Steps. plotting graph. or wie Jeuw dise 207 1st 2.44= 56 11000 А -15- - 20 Area of sout OA B C D E = 2 x area of OABC = ax (10x²ts) - (x - 30²771da = 2x (+ x +49 ²+12) da - 2x [*****+12) • 2x [- 88.18 + 19.596+ 29.39] - 2x [ 17.636 + 48.986] - 2X 31.35 – 62.7 sq unit Area

7. using integration, find the area of the triangle joining the points whose coordinates are (4,0), 18,6) and 10,4). ttt -64-4 -2 -8 204 6 8 Али АОАВ = | Ала одос + Ала АОАС -4 Equation of line OB 0(4,0) 6 (0,4) y-Y, = Yo-Y, (x-x) X-X, Y = 1 10-4) Y = -x+4) Equation of line oA 0(4,0) A (8,6) > Y-Yo - Yo-Yo (x-x.) Y 6 (X-Y) y = 3 (2-4) age,

Equation of line AB 418,6) B(0,4) Y. - Y, = Ya Y, (x-x) Y-6 = = (x-8) Yob & + (x-8) y = -2 +6 Area of A DAB= (n +4 - (-X+Y)] dx + E +4-61-6) Jak - les de 4 fta-o) de - 6891+1027 = 10 + [640) + 80 + 10 - 40] 10 + 10 20 sq. unit | Area of AO AB=