Find the Area Between the Curves x=-5 , x=2 , y=x , y=x^2-2

Problem

(∫_−5^2)(|x−(x2−2)|*d(x))

Solution

1. Identify the functions and the interval of integration. The curves are ƒ(x)=x and g(x)=x2−2 and the area is bounded between x=−5 and x=2

2. Determine the points of intersection to see if the curves cross within the interval. Set the equations equal to each other:

x2−2=x

x2−x−2=0

(x−2)*(x+1)=0

The curves intersect at x=−1 and x=2

3. Analyze which function is greater on the sub-intervals [−5,−1] and [−1,2] For x∈(−5,−1) x2−2>x For x∈(−1,2) x>x2−2

4. Set up the definite integrals for the total area A by splitting the interval at the intersection point x=−1

A=(∫_−5^−1)((x2−2−x)*d(x))+(∫_−1^2)((x−(x2−2))*d(x))

5. Find the antiderivatives for each integral:

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

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

6. Evaluate the first integral from −5 to −1

[((−1)3)/3−((−1)2)/2−2*(−1)]−[((−5)3)/3−((−5)2)/2−2*(−5)]

[−1/3−1/2+2]−[−125/3−25/2+10]

7/6−(−265/6)=272/6=136/3

7. Evaluate the second integral from −1 to 2

[−2/3+2/2+2*(2)]−[−((−1)3)/3+((−1)2)/2+2*(−1)]

[−8/3+2+4]−[1/3+1/2−2]

10/3−(−7/6)=20/6+7/6=27/6=9/2

8. Sum the results of the two integrals to find the total area:

A=136/3+9/2

A=272/6+27/6=299/6

Final Answer

(∫_−5^2)(|x−(x2−2)|*d(x))=299/6

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