Question

Use calculus to find the area bound by the graphs of y = arcsin(2x) - 4x2, y = 0, x = 0, and x = 12.

Answer 1

y sin (22) -42°, 4-0 a=0 제일 12 Area- $"" (sin" (22) – 42°) -o da 12 s 17 'sinlaa) da- $ *42%d2_0 Now solve SS15'(29) da In Integrate by parts Subu u-22 du-ada = sint(u) (du/) esin cujdy Seg!= fg - Sf g fs sinicu), g'-1 fla muz 19=4 (sin(u) = usini(u) - Ju u due solve su Vue dy -- Subu (~42 do= -audy

1 2 3 . ude - (auto) els 19- 1-4² Vir du Vr-u? equ @ Subu in Our foku) - 1585'cu) V-u? 2 eqa la subu in @ Sinu ), à causi5'(u) - VI-42 usinta vi-u 5 g tu=22 22 Soin'caa)da = sin (32) √1-422 4 egi ② subu in 0 2x sini(22) J-422 - (* 0 2 sin (22) V-42² - 432 pusincu) L-4Live - 410-10-40) a 24 sint (24) + +5√23il I Area =