# Find the sum of the areas of the planes

The graphs of the functions $f(x) = x^5 - 5x^3 + 7x$ and $g(x) = 2x-1$ trap 4 planes (I have no idea how to translate this properly into English). Calculate the sum of the areas of the planes.

My book does the following:

$$\displaystyle \int_{-1.96}^{1.83} | f(x) - g(x) | dx \approx 5.48$$

I don't understand this. To me it seems you would only get 2 of the 4 planes, namely the ones where $f(x) \geq g(x)$. But why do you also get the area of the other 2 planes with this integral?

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Oh wait, absolute signs.. –  Ylyk Coitus Mar 26 '13 at 16:19
A plot always helps. wolframalpha.com/input/… –  Ron Gordon Mar 26 '13 at 16:26

Graphing such functions is always a good idea to see the regions bound by two given curves: see, e.g., WA's graph:

There are indeed four such regions:

Here, you need to find the points of intersection by setting $f(x) = g(x)$ and solving:

\begin{align} f(x) & = g(x) &\\ \\ & \iff x^5 - 5x^3 + 7x = 2x-1 \\ \\ & \iff x^5 - 5x^3 + 5x + 1 = 0\tag{1} \\ \\ & \iff (x+1)(x^4 - x^3 - 4x^2 + 4x + 1) = 0 \end{align}

Then you'll need to break the integral into the sum of four integrals each with different bounds of integration, which will range within the interval $[\approx -1.96, \approx 1.83]$, and for each integrand, subtract the function of the "lower curve" from function corresponding to the upper curve, depending upon the interval on which to integrate.

$$\int_{x_1}^{x_2} (f(x) - g(x)) dx + \int_{x_2}^{x_3} (g(x) - f(x)) dx + \int_{x_3}^{x_4} (f(x) - g(x)) dx + \int_{x_4}^{x_5} (g(x) - f(x)) dx$$ where $x_1 < x_2 \lt x_3 \lt x_4 \lt x_5$ are your roots (solutions) to $(1)$:

$$x_1: (-, -),\; x_3: (+, -),\; x_4: (-, +), \;x_5: (+, +):\; \frac 14\left(1 \pm \sqrt 5 \pm \sqrt{6(5 - \sqrt 5)}\right);\quad x_2 = -1$$

\begin{align} x_1 & \approx -1.96 \\ \\ x_2 & =\; -1.00 \\ \\ x_3 & \approx -0.21 \\ \\ x_4 & \approx 1.34 \\ \\ x_5 & \approx 1.83 \\ \end{align}

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Thanks, dear friend! –  amWhy Mar 26 '13 at 20:14
The avatar is so cute :D especially his mouth is like a decreasing function. ;-) –  Babak S. Mar 26 '13 at 20:33