Sketch first:

5 10 15 20 25 30 1 2 3 4 5 6 x y
x
dy
x = y2 + 1

The curve x = y2 + 1, showing the portion "under" the curve from y = 1 to y = 5.

In this case, we express x as a function of y:

`y=sqrt{x-1}`

`y^2=x-1`

`x=y^2+1`

So the area is given by:

`A=int_1^5(y^2+1) dy=[y^3/3+y]_1^5`

`=45 1/3\ text[sq units]`

Note: For this particular example, we could have also summed it horizontally (integrating `y` and using `dx`), but we would need to break it up into sections first.