littlemanyu

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Okay, so I need to figure out how to solve this problem where I have to find the area of a region bounded by the curves y = x² - 4x + 3 and y = -x² + 2x + 5. Alright, let's start by recalling that when finding the area between two curves, the general approach is to integrate the top function minus the bottom function between the points where they intersect. So first, I need to find the points of intersection between these two parabolas.


Let me set the two equations equal to each other to find their intersection points. That would be x² - 4x + 3 = -x² + 2x + 5. Let me solve this equation for x. If I move all terms to one side, I get x² -4x +3 +x² -2x -5 = 0. Combining like terms, that's 2x² -6x -2 = 0. Hmm, simplifying that, maybe divide both sides by 2 to make it easier: x² -3x -1 = 0. Wait, is that right? Let me check again. Original equation: x² -4x +3 = -x² +2x +5. Adding x² to both sides: 2x² -4x +3 = 2x +5. Then subtract 2x and 5: 2x² -6x -2 = 0. Yes, that's correct. Dividing by 2: x² -3x -1 = 0.


Now, solving this quadratic equation. The quadratic formula is x = [3 ± sqrt(9 + 4)] / 2, since the discriminant is b² -4ac = (-3)² -4(1)(-1) = 9 +4 =13. So the solutions are x = [3 + sqrt(13)] / 2 and x = [3 - sqrt(13)] / 2. Hmm, sqrt(13) is approximately 3.605, so the x-values are roughly (3 + 3.605)/2 ≈ 3.3025 and (3 - 3.605)/2 ≈ -0.3025. So the two curves intersect at these x-values.


Next, I need to determine which function is on top between these two intersection points. To do that, I can pick a test point between the two x-values and plug it into both equations. Let's choose x = 0, which is between -0.3025 and 3.3025. Plugging into the first function: y = 0 -0 +3 = 3. Plugging into the second function: y = -0 +0 +5 =5. So at x=0, the second function (y = -x² +2x +5) is on top. Therefore, between the two intersection points, the upper function is y = -x² +2x +5 and the lower function is y = x² -4x +3.


Therefore, the area A is the integral from x = [3 - sqrt(13)]/2 to x = [3 + sqrt(13)]/2 of [(-x² +2x +5) - (x² -4x +3)] dx. Let's simplify the integrand first. Subtract the lower function from the upper function: (-x² +2x +5) - x² +4x -3. Combining like terms: -x² -x² = -2x², 2x +4x =6x, 5 -3 =2. So the integrand simplifies to -2x² +6x +2.


So the integral becomes A = ∫ [from a to b] (-2x² +6x +2) dx, where a = [3 - sqrt(13)]/2 and b = [3 + sqrt(13)]/2. Let me compute this integral step by step. The antiderivative of -2x² is (-2/3)x³, the antiderivative of 6x is 3x², and the antiderivative of 2 is 2x. So the antiderivative F(x) = (-2/3)x³ +3x² +2x.


Therefore, the area A = F(b) - F(a). Now, I need to compute F at both limits and subtract. However, since the expressions for a and b might be a bit messy, maybe there's a way to simplify this calculation by expanding F(b) - F(a). Alternatively, since the integral is between two roots of the original equation, perhaps there's symmetry or substitution that can help. But maybe it's straightforward to compute it directly.


Alternatively, maybe I can use substitution. Let me denote the limits as x1 = [3 - sqrt(13)]/2 and x2 = [3 + sqrt(13)]/2. Let me compute F(x2) - F(x1). Let's compute each term:




First, compute F(x) = (-2/3)x³ +3x² +2x.


So F(x2) - F(x1) = (-2/3)(x2³ -x1³) +3(x2² -x1²) +2(x2 -x1).


This might be easier to handle by expanding the differences. Let's recall that x2 -x1 = [3 + sqrt(13)]/2 - [3 - sqrt(13)]/2 = (2 sqrt(13))/2 = sqrt(13).


Also, x2 +x1 = [3 + sqrt(13) +3 - sqrt(13)]/2 = 6/2 =3.


Similarly, x2*x1 = [(3 + sqrt(13))/2] * [(3 - sqrt(13))/2] = (9 - (sqrt(13))²)/4 = (9 -13)/4 = (-4)/4 = -1.


Now, let's compute x2³ -x1³. Using the identity a³ -b³ = (a -b)(a² +ab +b²). We already know a -b = sqrt(13), and a +b =3, ab =-1.


First, compute a² +ab +b². Since a +b =3, ab=-1, then a² +b² = (a +b)^2 -2ab = 9 -2*(-1)=9 +2=11. Therefore, a² +ab +b² =11 + (-1)=10. Therefore, x2³ -x1³ = sqrt(13)*10 =10 sqrt(13).


Similarly, x2² -x1² = (x2 -x1)(x2 +x1) = sqrt(13)*3 =3 sqrt(13).


Therefore, substituting back into F(x2) - F(x1):


= (-2/3)(10 sqrt(13)) +3*(3 sqrt(13)) +2*(sqrt(13))


= (-20/3 sqrt(13)) +9 sqrt(13) +2 sqrt(13)


Combine the terms:


First, convert all coefficients to thirds:


-20/3 sqrt(13) +27/3 sqrt(13) +6/3 sqrt(13) = (-20 +27 +6)/3 sqrt(13) =13/3 sqrt(13).


Therefore, the area A is (13/3) sqrt(13). Wait, that seems a bit complicated. Let me check the calculations again to make sure I didn't make a mistake.


Starting with x2³ -x1³. Using the identity a³ -b³ = (a -b)(a² +ab +b²). We had a -b = sqrt(13), and a +b =3, ab=-1. Then a² +b² = (a +b)^2 -2ab =9 -2*(-1)=11. Therefore, a² +ab +b² =11 + (-1)=10. So x2³ -x1³ = sqrt(13)*10. That's correct.


Then, x2² -x1² = (x2 -x1)(x2 +x1)=sqrt(13)*3. Correct.


So F(x2) - F(x1) = (-2/3)(10 sqrt(13)) +3*(3 sqrt(13)) +2*sqrt(13)


Calculating each term:


-2/3 *10 sqrt(13) = -20/3 sqrt(13)


3*3 sqrt(13)=9 sqrt(13)


2*sqrt(13)=2 sqrt(13)


Now, adding them up:


-20/3 sqrt(13) +9 sqrt(13) +2 sqrt(13) = (-20/3 +9 +2) sqrt(13)


Convert 9 and 2 to thirds: 9 =27/3, 2=6/3


So total coefficient: (-20/3 +27/3 +6/3)= (13/3)


Therefore, A= (13/3) sqrt(13). So the area is (13 sqrt(13))/3. That seems correct. Let me verify by another method.


Alternatively, maybe I can compute the integral directly without using the difference of cubes and squares. Let's compute F(b) - F(a) where F(x) = (-2/3)x³ +3x² +2x.


First, compute F(b):


F(b) = (-2/3)b³ +3b² +2b


Similarly, F(a) = (-2/3)a³ +3a² +2a


So F(b) - F(a) = (-2/3)(b³ -a³) +3(b² -a²) +2(b -a)


Which is the same as before. So the previous calculation holds. Therefore, the area is indeed (13 sqrt(13))/3.


Alternatively, maybe I can check with approximate values. Let's compute the integral numerically.


First, the x-values of intersection are approximately x1 ≈ (3 -3.605)/2 ≈ (-0.605)/2 ≈ -0.3025 and x2 ≈ (3 +3.605)/2 ≈6.605/2≈3.3025.


The integrand is -2x² +6x +2. Let's compute the definite integral from -0.3025 to3.3025.


Alternatively, compute the approximate area using numerical integration. Let's approximate the integral.


But maybe it's easier to compute the exact value. Since we have (13 sqrt(13))/3, let's compute that numerically. sqrt(13)≈3.6055, so 13*3.6055≈46.8715. Then divided by3≈15.6238. So the area is approximately15.62 square units.


Alternatively, let's compute the integral step by step numerically. The integrand is -2x² +6x +2. The antiderivative is (-2/3)x³ +3x² +2x. Let's compute F(3.3025) and F(-0.3025).


First, compute F(3.3025):


(-2/3)(3.3025)^3 +3(3.3025)^2 +2*(3.3025)


First, compute 3.3025^3:


3.3025^3 ≈3.30253.30253.3025. Let's compute step by step:


3.30253.3025 ≈10.909 (since 3^2=9, 0.3025^2≈0.0915, cross terms 23*0.3025≈1.815, so total≈9 +1.815 +0.0915≈10.9065)


Then 10.90653.3025≈10.90653 +10.9065*0.3025≈32.7195 +3.299≈36.0185


So (-2/3)*36.0185≈-24.0123


Next term:3*(3.3025)^2≈3*10.9065≈32.7195


Third term:2*3.3025≈6.605


Adding them up: -24.0123 +32.7195 +6.605≈15.3122


Now compute F(-0.3025):


(-2/3)(-0.3025)^3 +3(-0.3025)^2 +2*(-0.3025)


First, (-0.3025)^3≈-0.0277


(-2/3)*(-0.0277)≈0.0185


Next term:3*(0.3025)^2≈3*0.0915≈0.2745


Third term:2*(-0.3025)≈-0.605


Adding them up:0.0185 +0.2745 -0.605≈-0.312


Therefore, F(b) - F(a)≈15.3122 - (-0.312)=15.3122 +0.312≈15.6242, which matches the approximate value of (13 sqrt(13))/3≈15.6238. So that checks out.


Therefore, the exact area is (13 sqrt(13))/3 square units.




To find the area of the region bounded by the curves ( y = x^2 - 4x + 3 ) and ( y = -x^2 + 2x + 5 ), follow these steps:



Find Intersection Points:
Set the equations equal to solve for ( x ):
[
x^2 - 4x + 3 = -x^2 + 2x + 5 \implies 2x^2 - 6x - 2 = 0 \implies x^2 - 3x - 1 = 0
]
Using the quadratic formula:
[
x = \frac{3 \pm \sqrt{13}}{2}
]
The curves intersect at ( x = \frac{3 - \sqrt{13}}{2} ) and ( x = \frac{3 + \sqrt{13}}{2} ).



Determine Upper and Lower Functions:
Test a point between the intersections (e.g., ( x = 0 )):


Upper function: ( y = -x^2 + 2x + 5 )
Lower function: ( y = x^2 - 4x + 3 )



Set Up the Integral:
The area ( A ) is the integral of the difference of the functions between the intersection points:
[
A = \int_{\frac{3 - \sqrt{13}}{2}}^{\frac{3 + \sqrt{13}}{2}} \left[(-x^2 + 2x + 5) - (x^2 - 4x + 3)\right] , dx
]
Simplify the integrand:
[
-2x^2 + 6x + 2
]



Compute the Integral:
Find the antiderivative ( F(x) ):
[
F(x) = -\frac{2}{3}x^3 + 3x^2 + 2x
]
Evaluate from ( a = \frac{3 - \sqrt{13}}{2} ) to ( b = \frac{3 + \sqrt{13}}{2} ):
[
A = F(b) - F(a)
]
Using algebraic identities and simplifications (difference of cubes and squares), the result is:
[
A = \frac{13\sqrt{13}}{3}
]




Final Answer:
[
\boxed{\dfrac{13\sqrt{13}}{3}}
]