Find the Area Between the Curves y=6-x^2 , y=x+4

Problem

y=6−x2,y=x+4

Solution

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

6−x2=x+4

2. Rearrange the equation into a standard quadratic form to solve for x

x2+x−2=0

3. Factor the quadratic to find the specific xvalues where the curves cross.

(x+2)*(x−1)=0

x=−2,x=1

4. Identify the upper curve by testing a value between the intersections, such as x=0 Since 6−0=6 and 0 + 4 = 4,t*h*e*c*u*r*v*e = 6 - x^2$ is the upper boundary.

5. Set up the definite integral for the area by subtracting the lower curve from the upper curve over the interval [−2,1]

(∫_−2^1)((6−x2)−(x+4)*d(x))

6. Simplify the integrand before performing the integration.

(∫_−2^1)(−x2−x+2*d(x))

7. Find the antiderivative of the simplified expression.

[−(x3)/3−(x2)/2+2*x]1(−2)

8. Evaluate the definite integral at the upper and lower limits.

(−1/3−1/2+2)−(8/3−2−4)

7/6−(−10/3)

7/6+20/6

9. Simplify the result to find the final area.

27/6=9/2

Final Answer

(∫_−2^1)((6−x2)−(x+4)*d(x))=9/2

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