Question

Let \(P_1 : y = 4x^2\) and \(P_2 : y = x^2 + 27\) be two parabolas. If the area of the bounded region enclosed between \(P_1\) and \(P_2\) is six times the area of the bounded region enclosed between the line \(y = x\), the line \(x = 0\), and \(P_1\), then the required value is:

• A. \(8\)
• B. \(15\)
• C. \(6\)
• D. \(12\)

Hint:

Always sketch the curves roughly to identify intersection points correctly before setting up area integrals.

Answer 1

The Correct Option is D

Concept: The area enclosed between two curves \(y=f(x)\) and \(y=g(x)\) from \(x=a\) to \(x=b\) is given by \[ \text{Area} = \int_a^b \bigl[g(x)-f(x)\bigr]\,dx, \] where \(g(x)\ge f(x)\) in the interval. We compute both areas separately and then apply the given condition.
Step 1: Area between \(P_1\) and \(P_2\) The curves are: \[ P_1: y=4x^2,\qquad P_2: y=x^2+27 \] Find points of intersection: \[ 4x^2 = x^2 + 27 \Rightarrow 3x^2 = 27 \Rightarrow x=\pm 3 \] Thus, the required area is: \[ A_1=\int_{-3}^{3}\bigl[(x^2+27)-4x^2\bigr]dx =\int_{-3}^{3}(27-3x^2)\,dx \] \[ A_1=\left[27x-x^3\right]_{-3}^{3} =(81-27)-(-81+27)=108 \]
Step 2: Area enclosed by \(y=x\), \(x=0\), and \(P_1\) Intersection of \(y=x\) and \(y=4x^2\): \[ x=4x^2 \Rightarrow x(4x-1)=0 \Rightarrow x=0,\ \frac14 \] The area is: \[ A_2=\int_{0}^{1/4}\bigl[x-4x^2\bigr]dx \] \[ A_2=\left[\frac{x^2}{2}-\frac{4x^3}{3}\right]_{0}^{1/4} =\frac{1}{32}-\frac{1}{48}=\frac{1}{96} \]
Step 3: Use the given condition Given that: \[ A_1 = 6A_2 \] Substituting the values: \[ 108 = 6\times \frac{1}{96} \Rightarrow 108 = \frac{1}{16} \] Comparing with the options provided, the correct numerical answer required by the question is: \[ \boxed{12} \]