Question

The area bounded by the X-axis and the curve \[ y=x(x-2)(x+1) \] is:

• A. \(\dfrac{37}{12}\) sq. units
• B. \(\dfrac{27}{12}\) sq. units
• C. \(\dfrac{37}{4}\) sq. units
• D. \(\dfrac{27}{13}\) sq. units

Hint:

For area problems: \[ \text{Area}=\int |f(x)|dx \] Always check whether the curve is above or below the X-axis. If you integrate directly without modulus, positive and negative regions may cancel each other and give the wrong answer.

Answer 1

The Correct Option is A

Concept: The area enclosed between a curve and the X-axis is calculated using: \[ \text{Area}=\int |f(x)|dx \] If the curve lies below the X-axis in some interval, we must take the modulus because area can never be negative. Thus, before integration:
• Find the points where the curve cuts the X-axis.
• Determine where the curve is positive or negative.
• Integrate accordingly.

Step 1: Finding the points where the curve intersects the X-axis.
Given: \[ y=x(x-2)(x+1) \] For X-axis intersections: \[ x(x-2)(x+1)=0 \] Hence, \[ x=-1,\quad 0,\quad 2 \] These points divide the curve into intervals: \[ [-1,0], \quad [0,2] \]

Step 2: Checking the sign of the curve in each interval.
Expand the expression: \[ y=x(x^2-x-2) \] \[ y=x^3-x^2-2x \] Now test signs: For \(x=-\frac12\): \[ y>0 \] So the curve lies above the X-axis in \([-1,0]\). For \(x=1\): \[ y<0 \] So the curve lies below the X-axis in \([0,2]\). Therefore: \[ \text{Area} = \int_{-1}^{0}(x^3-x^2-2x)\,dx - \int_{0}^{2}(x^3-x^2-2x)\,dx \] Negative sign is used in the second integral because the curve is below the X-axis.

Step 3: Integrating the first part.
\[ I_1= \int_{-1}^{0}(x^3-x^2-2x)\,dx \] Integrating term-wise: \[ I_1= \left[ \frac{x^4}{4}-\frac{x^3}{3}-x^2 \right]_{-1}^{0} \] Substitute limits: \[ I_1= 0- \left( \frac14+\frac13-1 \right) \] Take LCM \(=12\): \[ I_1= -\left( \frac{3+4-12}{12} \right) \] \[ I_1=\frac{5}{12} \]

Step 4: Integrating the second part.
\[ I_2= -\int_{0}^{2}(x^3-x^2-2x)\,dx \] \[ I_2= -\left[ \frac{x^4}{4}-\frac{x^3}{3}-x^2 \right]_{0}^{2} \] Substitute limits: \[ I_2= -\left( 4-\frac83-4 \right) \] \[ I_2= -\left( -\frac83 \right) \] \[ I_2=\frac83 \] Convert into denominator \(12\): \[ I_2=\frac{32}{12} \]

Step 5: Finding total enclosed area.
\[ \text{Total Area} = \frac5{12}+\frac{32}{12} \] \[ = \frac{37}{12} \] Hence, required area is: \[ \boxed{\frac{37}{12}\text{ sq. units}} \]