10. Area and Average Value
b2. Area Between Two Curves - In Pieces
Sometimes the curves intersect more than twice. So you need to do the integral in pieces:
Find the area between the cubic \(y=x^3-x^2\) and the line \(y=2x\).
The curves intersect when \[\begin{aligned} x^3-x^2&=2x \\ x^3-x^2-2x&=0 \\ x(x+1)(x-2)&=0 \\ x&=-1,0,2 \end{aligned}\] Between \(x=-1\) and \(x=0\), the cubic is above the line. Between \(x=0\) and \(x=2\), the line is above the cubic. So the area is
\[\begin{aligned} A&=\int_{-1}^0 (x^3-x^2-2x)\,dx +\int_0^2 (2x-x^3+x^2)\,dx \\ &=\left[\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2\right]_{-1}^0 +\left[x^2-\dfrac{x^4}{4}+\dfrac{x^3}{3}\right]_0^2 \\ &=(0)-\left(\dfrac{1}{4}-\dfrac{-1}{3}-1\right) +\left(4-4+\dfrac{8}{3}\right)-(0) \\ &=\dfrac{-3-4+12+32}{12}=\dfrac{37}{12} \end{aligned}\]
Find the area between the curve \(y=x^3-4x^2+4x=x(x-2)^2\) and the parabola \(y=x^2\).
To find the intersection points, equate the functions: \[x^3-4x^2+4x=x^2\]
\(\displaystyle A=\int_0^1 (x^3-4x^2+4x)-x^2\,dx+\int_1^4 x^2-(x^3-4x^2+4x)\,dx =\dfrac{71}{6}\)
To plot the cubic, we note that its \(x\)-intercepts are at \(0\) and \(2\), it is negative when \(x\) is negative and positive when \(x\) is positive. So there must be a local minimum at \(x=2\). To find the intersection points, we equate the functions: \[\begin{aligned} x^3-4x^2+4x&=x^2 \\ x^3-5x^2+4x&=0 \\ x(x-1)(x-4)&=0 \\ x=0,1,4& \end{aligned}\] So the area is
\[\begin{aligned} A&=\int_0^1 \text{upper}-\text{lower}\,dx +\int_1^4 \text{upper}-\text{lower}\,dx \\ &=\int_0^1 (x^3-4x^2+4x)-x^2\,dx +\int_1^4 x^2-(x^3-4x^2+4x)\,dx \\ &=\int_0^1 x^3-5x^2+4x\,dx +\int_1^4-x^3+5x^2-4x\,dx \\ &=\left[\dfrac{x^4}{4}-\dfrac{5x^3}{3}+2x^2\right]_0^1 +\left[-\dfrac{x^4}{4}+\dfrac{5x^3}{3}-2x^2\right]_1^4 =\dfrac{71}{6} \end{aligned}\]
You can practice computing Areas between Curves which Cross by using the following Maplets (requires Maple on the computer where this is executed):
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