The area of the region bounded above by y=1+sec^2x , below by y=0 on the left by x=0 and on the right by x=pi/4 is approximately...
a) 1
b) 1.25
c) 1.5
d) 1.75
e)2

Question

The area of the region bounded above by y=1+sec^2x , below by y=0 on the left by x=0 and on the right by x=pi/4 is approximately...
a) 1
b) 1.25
c) 1.5
d) 1.75
e)2

accessdenied · Answer

Have you tried anything so far on this problem that you could not get to work?  Or are you not sure how to begin?

anonymous · Answer

i am not sure how to begin

accessdenied · Answer

Well, what we should be looking for is an integration expression over our region that sums up all the areas from x=0 to x=pi/4  under the y=1+sec^2 x  graph and above the y=0 graph.
A rough sketch of the graph:
|dw:1398567657335:dw|

accessdenied · Answer

|dw:1398567750682:dw|
We should add up all these rectangular areas  y * dx = (1 + sec^2 ) dx,  over the region x=0 to x=pi/4.  That is the definition of our definite integral:
$ \displaystyle \int_{0}^{\pi/4} (1 + \sec^2 x) \ dx $

Is this setup clear?