How do you evaluate the definite integral by the limit definition given #intsqrt(9-x^2)dx# from [-3,3]?

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Answer 1 · 2017-01-22T03:58:31

This answer does not use the limit definition. (I didn't notice that phrase until I had already posted this answer.)

Although it is possible to evaluate #int_-3^3 sqrt(9-x^2) dx# by using a trigonometric substitution, there is another way.

Observe that #y = sqrt(9-x^2)# is an upper semicircle .

#y = sqrt(9-x^2)#

#y^2 = 9-x^2# (for #y >= 0#)

#x^2+y^2=9# (for #y >=0#).

The last (without the restriction on #y#) is the circle of radius #3#, centered at #(0,0)#.

With restricted #y#, we have the semicircle above the #x# axis from #x= -3# to #x=3#.

The integral is the area under the curve and above the #x# axis.

The area of the semicircle is #1/2 (pi(3)^2) = (9pi)/2#.

So

#int_-3^3 sqrt(9-x^2) dx = (9pi)/2#