Question

Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of `y=4-x^2` and `y=1+2sinx`, how do you find the area?

Answer 1

`R=1.764` units squared

Explanation:

In area problems, the first thing to do is graph the functions in question so you have a visual sense of the problem. You can check out the functions in this example graphed here.

From the graph, we see that the region `R` is the curvy-looking triangle. To find the area of `R`, we subtract the area of `1+sinx` from the area of `4-x^2`. The limits of integration are from the `y`-axis `(x=0)` to the point of intersection `(x=1.102)`.

Putting this in math terms, we have:
`R=(int_0^1.102 4-x^2dx)-(int_0^1.102 1+2sinxdx)`

The properties of integrals say we can split these two into mini-integrals like so:
`R=(int_0^1.102 4dx-int_0^1.102 x^2dx)-(int_0^1.102 1dx+2int_0^1.102 sinxdx)`

Now we perform the integration:
`R=(4[x]_0^1.102-[x^3/3]_0^1.102)-([x]_0^1.102-2[cosx]_0^1.102)`

And then evaluate:
`R=(4.408-0.446)-(1.102-2(cos1.102-cos0))`
`R=3.962-(1.102-2(0.452-1))`
`R=3.962-(1.102+1.096)`
`R=3.962-2.198`
`R=1.764` units squared