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
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 :)
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
To find the area of the region bounded by the given equations, you need to find the points of intersection between the two graphs and then integrate the difference of the two equations over that interval.
First, let's find the points of intersection by equating the two equations:
x^2 - 5x + 4 = -(x - 1)^2
Expanding the right side:
x^2 - 5x + 4 = -x^2 + 2x - 1
Combining like terms:
2x^2 - 7x + 5 = 0
To solve this quadratic equation, we can factor it:
(2x - 5)(x - 1) = 0
Setting each factor equal to zero:
2x - 5 = 0 --> x = 5/2
x - 1 = 0 --> x = 1
So the points of intersection are (5/2, ...) and (1, ...).
To determine the area between the curves, we need to integrate the difference of the two equations over the interval [1, 5/2]:
∫[(x^2 - 5x + 4) - (-(x - 1)^2)] dx
Simplifying the equation:
∫[(x^2 - 5x + 4) - (-(x^2 - 2x + 1))] dx
∫[(x^2 - 5x + 4) + (x^2 - 2x - 1)] dx
∫[2x^2 - 7x + 3] dx
Integrating term by term:
(2/3)x^3 - (7/2)x^2 + 3x + C
Evaluating the integral from x = 1 to x = 5/2:
[(2/3)(5/2)^3 - (7/2)(5/2)^2 + 3(5/2)] - [(2/3)(1)^3 - (7/2)(1)^2 + 3(1)]
[(2/3)(125/8) - (7/2)(25/4) + 15/2] - [(2/3) - (7/2) + 3]
[(250/24) - (175/8) + 15/2] - [(2/3) - (7/2) + 3]
Simplifying further:
[(250/24) - (525/24) + 180/24] - [(4/6) - (21/6) + 18/6]
[-(275/24) + 198/24] - [-(1/6) + 33/6]
[-77/24] - [32/6]
[-77/24] - [16/3]
Finding the common denominator:
[-77/24] - [128/24]
-205/24
Therefore, the area of the region bounded by the graphs of the two equations is -205/24, not -111/8.
Please check your calculations again or provide additional information if necessary.
To find the area of the region bounded by two graphs, you need to find the points of intersection between the two curves. Then you can integrate the difference between the two curves over the interval where they intersect.
Let's first find the points of intersection between the two curves:
Setting the two equations equal to each other:
x^2 - 5x + 4 = -(x-1)^2
Expanding the right side:
x^2 - 5x + 4 = -x^2 + 2x - 1
Combining like terms:
2x^2 - 7x + 5 = 0
Factoring the quadratic equation:
(2x - 5)(x - 1) = 0
Now we have two possible values for x: x = 5/2 and x = 1.
Next, we need to find the y-values for these x-values in both equations:
For y = x^2 - 5x + 4, substituting x = 5/2:
y = (5/2)^2 - 5*(5/2) + 4
y = 25/4 - 25/2 + 4
y = 25/4 - 50/4 + 16/4
y = -9/4
For y = -(x-1)^2, substituting x = 5/2:
y = -((5/2) - 1)^2
y = -(5/2 - 2/2)^2
y = -(3/2)^2
y = -9/4
And for y = -(x-1)^2, substituting x = 1:
y = -(1-1)^2
y = -(0)^2
y = 0
Now we have two points of intersection: (5/2, -9/4) and (1, 0).
To calculate the area, we need to integrate the difference between the two curves over the interval where they intersect. Since the second curve is below the first curve in this interval, we integrate y = (x^2 - 5x + 4) - y = -(x-1)^2 from x = 5/2 to x = 1.
Integrating the difference between the curves:
∫ [ (x^2 - 5x + 4) - (-(x-1)^2) ] dx from x = 5/2 to x = 1
Simplifying the expression:
∫ [ x^2 - 5x + 4 + (x-1)^2 ] dx from x = 5/2 to x = 1
∫ [ x^2 - 5x + 4 + (x^2 - 2x + 1) ] dx from x = 5/2 to x = 1
∫ [ 2x^2 - 7x + 5 ] dx from x = 5/2 to x = 1
Integrating term by term:
[ (2/3)x^3 - (7/2)x^2 + 5x ] evaluated from x = 5/2 to x = 1
Now we can substitute the x-values into the integrated expression and subtract the results:
[ (2/3)(1)^3 - (7/2)(1)^2 + 5(1) ] - [ (2/3)(5/2)^3 - (7/2)(5/2)^2 + 5(5/2) ]
[ (2/3) - 7/2 + 5 ] - [ (2/3)(125/8) - (7/2)((25/4)) + (25/2) ]
Multiplying the fractions:
[ (2/3) - (21/6) + (30/6) ] - [ (250/24)(2/3) - (175/8) + (75/2) ]
Simplifying the expression:
[ (2 - 21 + 30)/6 ] - [ (250/72) - (525/72) + (450/72) ]
[ 11/6 ] - [ 175/72 ]
[ 22/12 ] - [ 175/72 ]
[ 11/6 ] - [ 175/72 ]
Finding a common denominator:
[ 132/72 ] - [ 175/72 ]
Subtracting the fractions:
-43/72
Therefore, the area of the region bounded by the graphs of the two equations is -43/72, not -111/8.
It is possible that either the answer was provided incorrectly or there was a computational error in your calculations. Double-check your calculations to ensure the correct result.