find the area between the region of the curve y=3-x^2 and the line y=-1

Question

intergrate with respect to a ,x,b andy

Answer 1

y=-x^2+3=parabola, vertikal axis, upside down... go thru y=-1 at x=-2 & x=+2, so integrate y from x=-2 tu x=+2... integral=-(1/3)x^3... at x=-2, this=-8 / (-3)=+8/3... at x=+2, it = +8/(-3)= -8/3... integral from -2 tu +2 is -2*(8/3)=-16/3... tu get area, take absolute value=+16/3