graph them
first zeros
(x-4)(x-1) = 0
x = 1 and x = 4
and
x = 1 and x = 1 (vertex on x axis, in lower quadrants(sheds water)
SO
We are looking only at some region between x = 1(where they intersect, and somewhere between x = 1 and x = 4 where they hit again
Where is that?
x^2 -5 x +4 = -(x^2-2x+1) =-x^2+2x -1
2 x^2 - 7 x + 5 = 0
x = [ 7 +/- sqrt (49-40)]/4
x = [ 7+/- 3 ]/4
x = 1 (we know that already)
and x = 10/4 = 5/2 = 2.5
which is about what we thought
SO
do your area between x = 1 and x = 2.5 :)
Calculate the area of the region bounded by the graphs of the given equations
y=x^2-5x+4 and y=-(x-1)^2
The answer is supposed to be 9/8, but I'm getting -111/8. Idk what I'm doing wrong
2 answers
too bad you can't be bothered to show us your work ...
The curves intersect at (1,0) and (5/2,-9/4). A quick look at the graphs shows that vertical strips will be easiest to work with, so the area is
∫[1,5/2] (-(x-1)^2)-(x^2-5x+4) dx
= ∫[1,5/2] -2x^2+7x-5 dx
= -2/3 x^3 + 7/2 x^2 - 5x [1,5/2]
= (-25/24) - (-13/6)
= 27/24
= 9/8
The curves intersect at (1,0) and (5/2,-9/4). A quick look at the graphs shows that vertical strips will be easiest to work with, so the area is
∫[1,5/2] (-(x-1)^2)-(x^2-5x+4) dx
= ∫[1,5/2] -2x^2+7x-5 dx
= -2/3 x^3 + 7/2 x^2 - 5x [1,5/2]
= (-25/24) - (-13/6)
= 27/24
= 9/8
1 answer