Find the Area Between the Curves y=x^2+9x-4 , y=x+2

Problem

y=x2+9*x−4,y=x+2

Solution

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

x2+9*x−4=x+2

2. Solve the quadratic equation by moving all terms to one side.

x2+8*x−6=0

3. Apply the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a) to find the roots.

x=(−8±√(,8−4*(1)*(−6)))/(2*(1))

x=(−8±√(,64+24))/2

x=(−8±√(,88))/2

x=(−8±2√(,22))/2

x=−4±√(,22)

4. Identify the upper and lower functions on the interval [−4−√(,22),−4+√(,22)] Testing a point like x=−4 shows that y=x+2 is the upper curve and y=x2+9*x−4 is the lower curve.

ƒ(x)=x+2

g(x)=x2+9*x−4

5. Set up the integral for the area A=(∫_a^b)((ƒ(x)−g(x))*d(x))

A=(∫_−4−√(,22)^−4+√(,22))(((x+2)−(x2+9*x−4))*d(x))

A=(∫_−4−√(,22)^−4+√(,22))((−x2−8*x+6)*d(x))

6. Evaluate the definite integral using the power rule.

A=[−(x3)/3−4*x2+6*x](−4+√(,22))(−4−√(,22))

7. Simplify the result using the property that for a quadratic a*x2+b*x+c with roots α and β the area is |a|/6*(β−α)3

β−α=(−4+√(,22))−(−4−√(,22))=2√(,22)

A=1/6*(2√(,22))3

A=1/6*(8⋅22√(,22))

A=(176√(,22))/6

A=(88√(,22))/3

Final Answer

Area=(88√(,22))/3

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