Question

find the area enclosed by y^2=2x+1 and x-y-1=0

Answer 1

this can be solved easily when you draw graph of both curve in one coordinate system. see figure, here it is clear that,

area enclosed by graph is shown in shaded region,

e.g., area enclosed by graph is A =$\int\limits^{3}_{-1}\{(y+1)-\frac{1}{2}(y^2-1)\}dy$

=$\int\limits^3_{-1}\{y+1-\frac{y^2}{2}+\frac{1}{2}\}dy$

=$\int\limits^3_{-1}\{y-\frac{y^2}{2}+\frac{3}{2}\}dy$

=$\{\frac{y^2}{2}-\frac{y^3}{6}+\frac{3y}{2}\}^3_{-1}$

= 16/3 sq unit