Sketch first:

The curve *x* = *y*^{2} + 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.