Let a and b respectively be the points of local maximum and local minimum of the function \(f(x) = 2x^3 - 3x^2 - 12x\). If A is the total area of the region bounded by \(y=f(x)\), the x-axis and the lines \(x = a\) and \(x = b\), then 4A is equal to ___________.
Show Hint
When calculating the area between a curve and the x-axis, always find the roots of the function within the interval of integration. You must split the integral at these roots and take the absolute value of the function in each sub-interval, which means negating the integral over regions where the function is negative.
Step 1: Find Points of Local Maxima and Minima
We need to find the critical points of \(f(x)\) by setting its first derivative to zero.
\[ f(x) = 2x^3 - 3x^2 - 12x \]
\[ f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) \]
Set \(f'(x) = 0\):
\[ 6(x-2)(x+1) = 0 \]
The critical points are \(x=2\) and \(x=-1\).
To determine which is a maximum and which is a minimum, we use the second derivative test.
\[ f''(x) = 12x - 6 \]
At \(x=-1\): \(f''(-1) = 12(-1) - 6 = -18<0\). This is a point of local maximum. So, \(a = -1\).
At \(x=2\): \(f''(2) = 12(2) - 6 = 18>0\). This is a point of local minimum. So, \(b = 2\). Step 2: Set up the Area Integral
We need to find the area A bounded by \(y=f(x)\), the x-axis, \(x=a=-1\), and \(x=b=2\).
\[ A = \int_{-1}^{2} |f(x)| dx = \int_{-1}^{2} |2x^3 - 3x^2 - 12x| dx \]
We need to find the sign of \(f(x)\) in the interval \([-1, 2]\). Let's find the roots of \(f(x)=0\).
\[ f(x) = x(2x^2 - 3x - 12) = 0 \]
The roots are \(x=0\) and \(x = \frac{3 \pm \sqrt{9 - 4(2)(-12)}}{4} = \frac{3 \pm \sqrt{105}}{4}\).
\(\sqrt{105}\) is slightly more than 10. Let's say 10.2.
Roots are approx \(\frac{3 \pm 10.2}{4}\), which are \(3.3\) and \(-1.8\).
The roots are \(x \approx -1.8\), \(x=0\), \(x \approx 3.3\).
In the interval \([-1, 0]\), let's test \(x=-0.5\). \(f(-0.5) = (-)(+)(-)=+\). So \(f(x) \ge 0\).
In the interval \([0, 2]\), let's test \(x=1\). \(f(1) = 2-3-12 = -13<0\). So \(f(x) \le 0\).
The integral must be split at \(x=0\).
\[ A = \int_{-1}^{0} (2x^3 - 3x^2 - 12x) dx + \int_{0}^{2} -(2x^3 - 3x^2 - 12x) dx \]
Step 3: Evaluate the Integral
The antiderivative of \(f(x)\) is \(F(x) = \int (2x^3 - 3x^2 - 12x) dx = \frac{2x^4}{4} - \frac{3x^3}{3} - \frac{12x^2}{2} = \frac{x^4}{2} - x^3 - 6x^2\).
\[ A = [F(x)]_{-1}^{0} - [F(x)]_{0}^{2} \]
\[ A = (F(0) - F(-1)) - (F(2) - F(0)) = 2F(0) - F(-1) - F(2) \]
\(F(0) = 0\).
\(F(-1) = \frac{(-1)^4}{2} - (-1)^3 - 6(-1)^2 = \frac{1}{2} + 1 - 6 = -\frac{9}{2}\).
\(F(2) = \frac{(2)^4}{2} - (2)^3 - 6(2)^2 = 8 - 8 - 24 = -24\).
\[ A = 2(0) - (-\frac{9}{2}) - (-24) = \frac{9}{2} + 24 = \frac{9+48}{2} = \frac{57}{2} \]
Step 4: Final Answer
We need to find the value of 4A.
\[ 4A = 4 \times \frac{57}{2} = 2 \times 57 = 114 \]
