Calculate the area of the shaded and bounded region by the equation #y= 10/(x^2-10x+29# curve and the line #x=3# (see graphic). ?

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Answer 1 · 2017-11-07T12:16:07

#(15pi)/4#

The shaded area is equivalent to #int_3^oo10/(x^2-10x+29)\ dx#.

First, complete the square for the quadratic in the denominator:
#=int_3^oo10/((x-5)^2+4)\ dx#.

Now, substitute #u=x-5# and #du=dx#:
#=int_-2^oo10/(u^2+4)\ du#.

Since #int\ dx/(x^2+a^2)=arctan(x/a)/a+C#, we know that this definite integral equals
#=[5arctan(u/2)]_-2^oo#

#=lim_(u->oo)(5arctan(u/2))-5arctan(-2/2)#

#=(5pi)/2+(5pi)/4#

#=(15pi)/4#

Calculate the area of ​​the shaded and bounded region by the equation #y= 10/(x^2-10x+29# curve and the line #x=3# (see graphic). ?