Question

1. Find the area under the graph of the following function over the given interval. y = 6- x2 [-1,2] 2. Evaluate. S(x2 + x – 4)dx 3. Find the area of the region bounded by the graphs of the given equations. y = x2 – 2x y = 2 - x

Answer 1

4 = 6*, [1,2 Area ander a cerve poltolar a=1, biz, f(x) = 6x² § 16-2²1dx The to 16-x²) at 1sxs without absoluter equivalent expression sh (6x) dx 16x²) dx sode- seda xxn+/ 6x- nti 3 2 6x - [6 (2) – (23] - [6c.) – u 10 3 2 3.66666 2 (x2+x-4) de Sade & Seda - Se de + 서 - 4xtc int, ntl a 2 - 4x] 2 +623_40)] - [* +49-443 1.25 y = x - 2x Y = 2 x Area between Ceasles A = 1+ (x-91x)/dx f(x) = x-2x, g(x) = 2x find Intersection points f(x) x²-2x = 2- x

x²-x - 2 = 0 (x + 1)(x-2) =0 Xti = 0 X-2 - 0 x = -1 x = 2 a=-1, b=2 5 1 2 - 2x - (2-x) dx The equivalent expression to (x - 2x - (2-x)) at Kx52 without absolutes 32 (x² + x + 2) dx - 4 Stx +x+2) dx - Sadat Sada & Jade + x 2 texte 2 [- + 2 + 2x] - Boyu 4307 - 200 کے 9 = 4.5 2