Find the Area Between the Curves y=x^2+2x-4 , y=x+4

Problem

y=x2+2*x−4,y=x+4

Solution

1. Find the intersection points by setting the two equations equal to each other to determine the limits of integration.

x2+2*x−4=x+4

x2+x−8=0

2. Solve for x using the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a)

x=(−1±√(,1−4*(1)*(−8)))/(2*(1))

x=(−1±√(,33))/2

3. Identify the upper and lower functions on the interval [(−1−√(,33))/2,(−1+√(,33))/2] Testing a point like x=0 shows 0+4>0+2*(0)−4 so y=x+4 is the upper curve.

4. Set up the integral for the area A=(∫_a^b)(((y_upper)−(y_lower))*d(x))

A=(∫_(−1−√(,33))/2^(−1+√(,33))/2)((x+4−(x2+2*x−4))*d(x))

A=(∫_(−1−√(,33))/2^(−1+√(,33))/2)((−x2−x+8)*d(x))

5. Integrate the expression with respect to x

(∫_^)((−x2−x+8)*d(x))=−(x3)/3−(x2)/2+8*x

6. Evaluate the definite integral at the boundaries. For a quadratic a*x2+b*x+c=0 with roots α and β the area is |a|/6*(β−α)3

β−α=(−1+√(,33))/2−(−1−√(,33))/2=√(,33)

A=|−1|/6*(√(,33))3

A=(33√(,33))/6

A=(11√(,33))/2

Final Answer

(∫_(−1−√(,33))/2^(−1+√(,33))/2)((x+4−(x2+2*x−4))*d(x))=(11√(,33))/2

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