Question

Which integral gives the area of the region in the first quadrant bounded by the graphs of y2 = x − 1, x = 0, y = 0, and y = 2?

Answer 1

Interesting that only integrals along the $x$-axis are suggested when integrating along the $y$-axis would be much simpler... Anyway, you have to split the interval of integration into two. The "height" of the region is not uniform over the entire interval.

When $y=0$, we have $0^2=x-1\implies x=1$. When $y=2$, we have $2^2=x-1\implies x=5$. Then the area we want is given by

$\displaystyle\int_0^12\,\mathrm dx+\int_1^52-\sqrt{x-1}\,\mathrm dx$

which seems to agree with the last option.