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Step-by-step Solution
Step 1: Differentiate the given function to find critical points
The given function is
f(x) = 2x^3 - 3x^2 - 12x.
Differentiate with respect to x :
f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(12x) = 6x^2 - 6x - 12.
Factor out the common term:
f'(x) = 6(x - 2)(x + 1).
Step 2: Solve for critical points and identify local maxima/minima
Set f'(x) = 0 :
6(x - 2)(x + 1) = 0 \quad \Longrightarrow \quad x = 2 \text{ or } x = -1.
• At x = -1 , the function f(x) is:
f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1)
= 2(-1) - 3(1) + 12
= -2 - 3 + 12
= 7.
• At x = 2 , the function f(x) is:
f(2) = 2(2)^3 - 3(2)^2 - 12(2)
= 2 \cdot 8 - 3 \cdot 4 - 24
= 16 - 12 - 24
= -20.
By analyzing the sign of f'(x) around x = -1 and x = 2 , or using the second derivative test, we conclude:
• x = -1 corresponds to a local maximum (since f(-1) = 7 ).
• x = 2 corresponds to a local minimum (since f(2) = -20 ).
Thus, a = -1 (point of local maximum) and b = 2 (point of local minimum).
Step 3: Set up the integral for the total area
We want the total area of the region bounded by y = f(x) , the x -axis, and the lines x = a and x = b . That is the area between x = -1 and x = 2 . However, the function crosses the x -axis within this interval, so we split it at the root x=0 (since f(0) = 0 ):
• On [-1, 0] , the function f(x) is 2x^3 - 3x^2 - 12x .
• On [0, 2] , the function changes sign, so to compute area correctly, we take the positive value of the integrand; effectively,
f(x) = 2x^3 - 3x^2 - 12x becomes -(2x^3 - 3x^2 - 12x) if it is negative. Equivalently, one may write the expression as 12x + 3x^2 - 2x^3 if f(x) is negative in that region and we want the positive area.
Hence, the total area A can be written as:
A = \int_{-1}^{0} \bigl(2x^3 - 3x^2 - 12x\bigr)\, dx \;+\; \int_{0}^{2} \bigl(12x + 3x^2 - 2x^3\bigr)\, dx.
Step 4: Evaluate the definite integrals
1) On [-1, 0] :
\int \bigl(2x^3 - 3x^2 - 12x\bigr)\,dx
= \int 2x^3\,dx - \int 3x^2\,dx - \int 12x\,dx
= \frac{2x^4}{4} - x^3 - 6x^2
= \frac{x^4}{2} - x^3 - 6x^2.
Evaluate from x = -1 to x = 0 :
\left[\frac{x^4}{2} - x^3 - 6x^2\right]_{-1}^{0}
= \Bigl( \frac{0^4}{2} - 0^3 - 6 \cdot 0^2 \Bigr)
- \Bigl( \frac{(-1)^4}{2} - (-1)^3 - 6 \cdot (-1)^2 \Bigr)
= \bigl(0 - 0 - 0 \bigr)
- \bigl( \frac{1}{2} + 1 - 6 \bigr)
= 0 - \left(\frac{1}{2} + 1 - 6\right)
= 0 - \left(\frac{3}{2} - 6\right)
= 0 - \left(-\frac{9}{2}\right)
= \frac{9}{2}.
2) On [0, 2] :
\int \bigl(12x + 3x^2 - 2x^3 \bigr)\,dx
= \int 12x\,dx + \int 3x^2\,dx - \int 2x^3\,dx
= 6x^2 + x^3 - \frac{2x^4}{4}
= 6x^2 + x^3 - \frac{x^4}{2}.
Evaluate from x = 0 to x = 2 :
\left[6x^2 + x^3 - \frac{x^4}{2}\right]_0^2
= \Bigl(6 \cdot 2^2 + 2^3 - \frac{(2)^4}{2}\Bigr)
- \Bigl(6 \cdot 0^2 + 0^3 - \frac{0^4}{2}\Bigr)
= \Bigl(6 \cdot 4 + 8 - \frac{16}{2}\Bigr)
- 0
= (24 + 8 - 8)
= 24.
Therefore, the total area is:
A = \frac{9}{2} + 24 = \frac{9}{2} + 24 = \frac{9}{2} + \frac{48}{2} = \frac{57}{2} = 28.5.
Step 5: Multiply the area by 4 to find the required value
The question asks for 4A :
4A = 4 \times 28.5 = 114.
Final Answer
4A = 114.
