How do you find the area bounded by #y=4-x^2#, the x and y axis, and x=1?

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Answer 1 · 2016-12-04T00:23:03

First picture what this region would look like by envisioning its graph (or just looking straight at it):

graph{4-x^2 [-9.54, 10.46, -3.92, 6.08]}

Thinking about how this is bounded from side to side, we see it's bounded by the #y#-axis and the line #x=1#.

Since it's also bounded by the #x#-axis, we're looking for the positive area in that "slice" of the graph between #0lt=xlt=1#:

graph{(4-x^2)sqrt(x-x^2)/sqrt(x-x^2) [-2, 3, -1, 4.72]}

This are can be found through integrating the function from #x=0# to #x=1#, or:

#int_0^1(4-x^2)dx#

Integrating (finding the antiderivative) and keeping the bounds gives:

#=[4x-x^3/3]_0^1#

#=[4(1)-1^3/3]-[4(0)-0^3/3]#

#=(4-1/3)-(0-0)#

#=11/3#

The area under the specified curve with the specified bounds is #11/3#.