Area Between Multiple Intersections - Calculus 2

1. The curves $y = x^3 - 3x$ and $y = x$ intersect at three points. If the area between the curves in the middle region (where the cubic is below the line) is $A$, and the total area of the two outer regions is $B$, what is the ratio $\frac{B}{A}$?
   Answer: $1$ Setting $x^3 - 3x = x$, we get $x^3 - 4x = 0$, so $x(x^2 - 4) = 0$, giving $x = 0, \pm 2$. The difference $g(x) = x^3 - 4x$ satisfies $g(-1) = 3 > 0$, $g(1) = -3 < 0$. So the cubic is above the line on $(-2,0)$ and below on $(0,2)$. We have $A = \int_0^2 |x^3 - 4x|dx = -\int_0^2 (x^3 - 4x)dx = 4$ and $B = \int_{-2}^0 |x^3 - 4x|dx = \int_{-2}^0 (x^3 - 4x)dx = 4$ by symmetry of $g(x)$ (odd function). Thus $\frac{B}{A} = 1$. Choice B assumes $B = 2A$, choice C reverses the ratio, choice D uses incorrect region identification.
2. The curves $y = x^3 - 6x^2 + 9x$ and $y = x$ intersect at three points. What is the total area of the regions enclosed between these curves?
   Answer: $8$ First find intersection points by solving $x^3 - 6x^2 + 9x = x$, which gives $x^3 - 6x^2 + 8x = 0$, so $x(x^2 - 6x + 8) = 0$, yielding $x = 0, 2, 4$. The difference function is $f(x) = x^3 - 6x^2 + 8x$. Since $f(1) = 3 > 0$ and $f(3) = -9 < 0$, the cubic is positive on $[0,2]$ and negative on $[2,4]$. The total area is $\int_0^2 |f(x)|dx + \int_2^4 |f(x)|dx = \int_0^2 f(x)dx - \int_2^4 f(x)dx = \left[\frac{x^4}{4} - 2x^3 + 4x^2\right]_0^2 - \left[\frac{x^4}{4} - 2x^3 + 4x^2\right]_2^4 = \frac{16}{3} - \left(-\frac{16}{3}\right) = 8$. Choice B gives the area of just one region, C doubles the total incorrectly, and D assumes no sign change.
3. The function $f(x) = x^4 - 5x^2 + 4$ intersects the $x$-axis at four points. What is the total area between $f(x)$ and the $x$-axis?
   Answer: $\frac{44}{15}$ Setting $x^4 - 5x^2 + 4 = 0$ and substituting $u = x^2$, we get $u^2 - 5u + 4 = 0$, so $(u-1)(u-4) = 0$, giving $u = 1, 4$. Thus $x = \pm 1, \pm 2$. Since $f(0) = 4 > 0$ and $f(1.5) = (1.5)^4 - 5(1.5)^2 + 4 = 5.0625 - 11.25 + 4 = -2.1875 < 0$, we have $f(x) > 0$ on $(-2,-1) \cup (1,2)$ and $f(x) < 0$ on $(-1,1)$. The area is $2\int_1^2 f(x)dx - \int_{-1}^1 f(x)dx = 2\left[\frac{x^5}{5} - \frac{5x^3}{3} + 4x\right]_1^2 - \left[\frac{x^5}{5} - \frac{5x^3}{3} + 4x\right]_{-1}^1 = 2 \cdot \frac{22}{15} - 0 = \frac{44}{15}$. Choice A gives area of one region only, C and D use computational errors.

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