Find the area of the region enclosed by #y=x^2-2x# and #y=3#?

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#y=x^2-2x# and #y=3#

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Answer 1 · 2018-02-21T14:46:46

#32/3# unit#"s"^2#

Find the points of intersection #M# and #N#:

#x^2-2x=y=3#

#x^2-2x-3=0#

#(x-3)(x+1)=0#

So #M# is #x=3rarry=3#

And #N# is #x=-1rarry=3#

The area bounded by two curves is given by:

#int_a^b[f(x)-g(x)]dx#, where #f(x) and g(x)# are functions, and #a and b# are the #x# coordinates of the intersection points.

Here, #b=3# and #a=-1#

#f(x)=x^2-2x#

#g(x)=3#

Inputting:

#int_-1^3[(x^2-2x)-(3)]dx#

#int_-1^3(x^2-2x-3)dx#

#int_-1^3(x^2)dx-int_-1^3(x)dx-3int_-1^3(x^0)dx#

#(x^3/3-x^2/2-3x)_-1^3#

#(-9)-(5/3)#

#-32/3#

But since area cannot be negative, we have #32/3# unit#"s"^2# as the area.