Question

Calculate the finite area enclosed by y = x ^ {^ {3}} and its tangent line at x = \ frac {-1} {2}.

Answer 1

Ansi but n=1 3 so point P(-1/2 - 1/3 at bant P: here and 3(+)= $; grem y=x, and x = -1 x=-1/2 in y = x3 = (-1/2) = -1/8 at which we have to find tangent y = x3, thon (dy) en y = a (x3) = 34" (dy) (3+²)/ (1/2 , /3) (-1/2 , /3) slope of tangent at pon't p: so eq? of tangent atp atp ( 4 ;-) to kluose slope és mt=3 és : 9 - Xo = m+ (x-x1) => 7+5 = 3(x+1) y + 1/3 3x +3 - J J F 2 uthah 11 (mt) al ar 11 -] to the cure 2/ 4 4 8

solving y = 3x11 and y = x3 = 3x+1) 41°= 34+1 x3 g=**) 4 = 23 4 opber x=-1 cuve (lower curve) 4x 3 X-1 (=/ sol -3x-1= 0, but 40 +603-3 (1)-170 5) 0 = I so lule gee to=6 so 4x3_3u-1 = (2x+1)*(x-1)= 0 => x = 1 *== </2; but a= 2) So enclosed Area = X=1 rubber curre-lower curve) dx n= *x+1-*3 ) ale = 3x+1-4x3) dx 1 = 2 ( 34 tratu * ((1+x=8)-(744) *]] *+*+in) = 4(2= #(2-5)= 끌 4 3 2 127 squ unit 64