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?

1 Answer
Feb 21, 2016

Answer:

#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