Question

Find the area bounded by the curve y=x2 and the straight line y=2+x

A.4 1/6
B. 4 1/2
C. 5 1/6
D. 5 1/2

Answer 1

`x^2=2+x\\ x^2-x-2=0\\ x^2+x-2x-2=0\\ x(x+1)-2(x+1)=0\\ (x-2)(x+1)=0\\ x=2 \vee =-1\\\\ \displaystyle A=\int \limits_(-1)^22+x-x^2\, dx\\ A=\left[2x+(x^2)/(2)-(x^3)/(3)\right]_(-1)^2\\ A=2\cdot2+(2^2)/(2)-(2^3)/(3)-\left(2\cdot(-1)+((-1)^2)/(2)-((-1)^3)/(3)\right)\\ A=4+2-(8)/(3)-\left(-2+(1)/(2)+(1)/(3)\right)\\ A=6-(8)/(3)+2-(1)/(2)-(1)/(3)\right)\\ A=8-(9)/(3)-(1)/(2)\\ A=8-3-(1)/(2)\\ A=5-(1)/(2)\\ A=4(1)/(2)`