Question

The area bounded by the curves $y = (x - 1)^2$, $y = (x + 1)^2$ and $y = \frac{1}{4}$ is

• A. $\frac{1}{3}$ sq. units.
• B. $\frac{2}{3}$ sq. units.
• C. $\frac{1}{4}$ sq. units.
• D. $\frac{1}{5}$ sq. units.

Hint:

Calculus Tip: Utilizing y-axis symmetry (multiplying the integral from $0$ to $x$ by 2) cuts calculation time in half and minimizes the risk of sign errors when evaluating the lower bounds.

Answer 1

The Correct Option is A

Concept: Calculus - Application of Integrals (Area Under Curves).

Step 1: Analyze the symmetry of the given curves. The curves $y = (x-1)^2$ and $y = (x+1)^2$ are standard parabolas opening upwards, shifted 1 unit to the right and left, respectively. Because they are mirror images of each other across the y-axis ($x=0$), the area bounded by these two curves and the horizontal line $y = \frac{1}{4}$ will be perfectly symmetric about the y-axis.

Step 2: Find the points of intersection in the first quadrant. To find the intersection of $y = (x-1)^2$ with the line $y = \frac{1}{4}$, set them equal: $(x-1)^2 = \frac{1}{4}$. Taking the square root of both sides gives $x - 1 = \pm\frac{1}{2}$. Solving for $x$, we get $x = 1 + \frac{1}{2} = \frac{3}{2}$ or $x = 1 - \frac{1}{2} = \frac{1}{2}$. Looking at the graphical layout, the bounded region is restricted between the y-axis ($x=0$) and the inner intersection point, so we use the boundary $x = \frac{1}{2}$.

Step 3: Set up the integral for the area. Due to symmetry, we can calculate the area in the first quadrant (from $x=0$ to $x=\frac{1}{2}$) and multiply by 2. In this region, the upper boundary is the curve $y = (x-1)^2$ and the lower boundary is the line $y = \frac{1}{4}$. Area $= 2 \int_{0}^{\frac{1}{2}} \left[ (x-1)^2 - \frac{1}{4} \right] dx$.

Step 4: Integrate the function. Apply the power rule to integrate the terms: $\int \left[ (x-1)^2 - \frac{1}{4} \right] dx = \frac{(x-1)^3}{3} - \frac{x}{4}$. Now, evaluate this from $0$ to $\frac{1}{2}$: $= 2 \left[ \left( \frac{(x-1)^3}{3} - \frac{x}{4} \right) \right]_{0}^{\frac{1}{2}}$.

Step 5: Evaluate the limits to find the final area. Substitute the upper limit $x = \frac{1}{2}$: $\frac{(\frac{1}{2}-1)^3}{3} - \frac{(\frac{1}{2})}{4} = \frac{(-\frac{1}{2})^3}{3} - \frac{1}{8} = \frac{-\frac{1}{8}}{3} - \frac{1}{8} = -\frac{1}{24} - \frac{3}{24} = -\frac{4}{24} = -\frac{1}{6}$.
Substitute the lower limit $x = 0$: $\frac{(0-1)^3}{3} - \frac{0}{4} = \frac{-1}{3} - 0 = -\frac{1}{3}$. Subtract the lower limit value from the upper limit value: Area $= 2 \left[ -\frac{1}{6} - \left(-\frac{1}{3}\right) \right] = 2 \left[ -\frac{1}{6} + \frac{2}{6} \right] = 2 \left[ \frac{1}{6} \right] = \frac{2}{6} = \frac{1}{3}$. $$ \therefore \text{The required area is } \frac{1}{3} \text{ sq. units.} $$