Find the Area Between the Curves y=x^2-5x+4 , y=-(x-1)^2

Problem

y=x2−5*x+4,y=−(x−1)2

Solution

1. Expand the second equation to make it easier to compare with the first.

−(x−1)2=−(x2−2*x+1)

−(x−1)2=−x2+2*x−1

2. Find the intersection points by setting the two equations equal to each other.

x2−5*x+4=−x2+2*x−1

2*x2−7*x+5=0

3. Solve for x using the quadratic formula or factoring.

(2*x−5)*(x−1)=0

x=1,x=2.5

4. Determine the upper curve on the interval [1,2.5] by testing a point, such as x=2

(y_1)=2−5*(2)+4=−2

(y_2)=−(2−1)2=−1

Since −1>−2 the curve y=−(x−1)2 is the upper curve.

5. Set up the definite integral for the area A using the formula A=(∫_a^b)((upper−lower)*d(x))

A=(∫_1^2.5)(((−x2+2*x−1)−(x2−5*x+4))*d(x))

A=(∫_1^2.5)((−2*x2+7*x−5)*d(x))

6. Integrate the expression with respect to x

(∫_^)((−2*x2+7*x−5)*d(x))=−(2*x3)/3+(7*x2)/2−5*x

7. Evaluate the definite integral at the boundaries 2.5 and 1

A=(−(2*(2.5)3)/3+(7*(2.5)2)/2−5*(2.5))−(−(2*(1)3)/3+(7*(1)2)/2−5*(1))

A=(−31.25/3+43.75/2−12.5)−(−2/3+7/2−5)

A=(−62.5/6+131.25/6−75/6)−(−4/6+21/6−30/6)

A=(−6.25/6)−(−13/6)

A=6.75/6

A=1.125

Final Answer

(∫_1^2.5)((−(x−1)2−(x2−5*x+4))*d(x))=9/8

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