Find the Area Between the Curves y=x^2 , y=x+3

Problem

y=x2,y=x+3

Solution

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

x2=x+3

x2−x−3=0

2. Apply the quadratic formula to solve for x since the quadratic is not easily factorable.

x=(−(−1)±√(,(−1)2−4*(1)*(−3)))/(2*(1))

x=(1±√(,1+12))/2

x=(1±√(,13))/2

3. Identify the upper and lower functions on the interval [(1−√(,13))/2,(1+√(,13))/2] Testing a point like x=0 shows that 0+3>0 so y=x+3 is the upper curve.

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

A=(∫_(1−√(,13))/2^(1+√(,13))/2)((x+3−x2)*d(x))

5. Find the antiderivative of the integrand.

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

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

β−α=(1+√(,13))/2−(1−√(,13))/2=√(,13)

A=1/6*(√(,13))3

A=(13√(,13))/6

Final Answer

(∫_(1−√(,13))/2^(1+√(,13))/2)((x+3−x2)*d(x))=(13√(,13))/6

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