The area of the region enclosed between the parabolas y² = 2x - 1 and y² = 4x - 3 is (A) 1/3 (B) 1/6

Q: Why can't we just integrate with respect to $x$?

You absolutely can, but it is like trying to cut a steak with a spoon. You would have to solve for $y = \pm \sqrt{2x-1}$ and $y = \pm \sqrt{4x-3}$, then split the integral into different intervals where the curves change. It is a recipe for calculation errors. Integrating with respect to $y$ keeps the functions as simple polynomials, which is the hallmark of a smart JEE strategy.

Q: How do I know which curve is the 'right' one?

Great question! Look at the coefficients. For a fixed $y$, $x = \frac{y^2+3}{4}$ is larger than $x = \frac{y^2+1}{2}$. Let us check: $\frac{y^2+3}{4} - \frac{y^2+1}{2} = \frac{y^2+3 - 2y^2 - 2}{4} = \frac{1-y^2}{4}$. Since this is positive for $y \in (-1, 1)$, the curve $x_2 = \frac{y^2+3}{4}$ is indeed to the right. Always test a point if you are unsure!

Q: Why did we multiply by 2 in the integral?

Because the parabolas are symmetric about the $x$-axis. The area above the $x$-axis is identical to the area below it. By calculating the area from $y=0$ to $y=1$ and doubling it, we avoid dealing with the negative limit and simplify the arithmetic. It is a classic trick to save time and reduce the chance of a sign error.

Question

The area of the region enclosed between the parabolas $y^2 = 2x - 1$ and $y^2 = 4x - 3$ is

Options:

(A) $\frac{1}{3}$

(B) $\frac{1}{6}$

(C) $\frac{2}{3}$

(D) $\frac{3}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a coordinate plane. You see two parabolas, $y^2 = 2x - 1$ and $y^2 = 4x - 3$.

Most students immediately think of integrating with respect to $x$. However, these are not typical vertical parabolas; they open to the right.

If you force yourself to integrate with respect to $x$, you walk into a trap of square roots and complex, split-region integrals. The secret to mastering JEE Advanced problems is choosing the most elegant path, which in this case is integrating with respect to $y$.

### The Strategic Shift

Since our parabolas are defined as $y^2 = f(x)$, it is far more natural to express $x$ as a function of $y$. From our first equation, $y^2 = 2x - 1$, we get:

$$x_1 = \frac{y^2 + 1}{2}$$

From the second, $y^2 = 4x - 3$, we get:

$$x_2 = \frac{y^2 + 3}{4}$$

Now, the problem transforms from a messy $x$-integral into a clean, polynomial $y$-integral. This is the kind of strategic thinking that separates top rankers from the rest.

### Finding the Intersection

Before we can calculate the area, we need to know where these curves meet to define our limits of integration. We set $x_1 = x_2$:

$$\frac{y^2 + 1}{2} = \frac{y^2 + 3}{4}$$

Multiplying by 4, we get $2(y^2 + 1) = y^2 + 3$. Expanding this, we find $2y^2 + 2 = y^2 + 3$, which simplifies beautifully to $y^2 = 1$.

Thus, our intersection points are $y = 1$ and $y = -1$.

### The Power of Symmetry

We are looking for the area between these two curves. The integral is $A = \int_{-1}^{1} (x_{right} - x_{left}) dy$.

However, the region is perfectly mirrored across the $x$-axis. Instead of integrating from $-1$ to $1$, we can integrate from $0$ to $1$ and multiply by 2 to avoid potential sign errors.

Our integral becomes:

$$A = 2 \int_{0}^{1} (x_2 - x_1) dy$$

Substituting our expressions, we have:

$$A = 2 \int_{0}^{1} \left( \frac{y^2 + 3}{4} - \frac{y^2 + 1}{2} \right) dy$$

### The Final Calculation

Let us simplify the integrand. Finding a common denominator of 4, we get:

$$\frac{y^2 + 3 - 2(y^2 + 1)}{4} = \frac{y^2 + 3 - 2y^2 - 2}{4} = \frac{1 - y^2}{4}$$

Now our integral is:

$$A = 2 \int_{0}^{1} \frac{1 - y^2}{4} dy = \frac{1}{2} \int_{0}^{1} (1 - y^2) dy$$

Integrating this is straightforward:

$$\frac{1}{2} \left[ y - \frac{y^3}{3} \right]_{0}^{1}$$

Evaluating at the limits, we get $\frac{1}{2} (1 - \frac{1}{3}) = \frac{1}{2} (\frac{2}{3}) = \frac{1}{3}$.

The final result is **$\frac{1}{3}$**. Remember, in JEE, the most elegant solution is usually the right one.

Animated Solution for Mathematics - Definite Integration: The area of the region enclosed between the parabolas $y^{2} = 2 x - 1$ and $y^{2} = 4 x - 3$ is

Select Answer:

Visualized Solution

Visualize the Parabolas

Express $x$ in terms of $y$

Find Intersection Points

Solve for $y$

Set up the Area Integral

Substitute the Expressions

Simplify the Integrand

Integrate and Evaluate

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Strategic Shift

Finding the Intersection

The Power of Symmetry

The Final Calculation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region enclosed between the parabolas y2=2x−1 and y2=4x−3 is