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?

the integral from 0 to 1 of 2, dx plus the integral from 1 to 5 of the quantity 2 minus square root of x minus 1, dx

the integral from 0 to 5 of 2 minus the square root of x minus 1, dx

the integral from 0 to 1 of 2, dx plus the integral from 1 to 5 of the quantity 3 minus x, dx

the integral from 0 to 1 of 2, dx minus the integral from 1 to 5 of the quantity 2 minus square root of x minus 1, dx

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-√(x-1)\,\mathrm dx`

which seems to agree with the last option.