Find the area bounded by the curves x² = y, x² = -y and y² = 4x - 3.

Q: Why can't I just use vertical strips ($dx$)? It feels more natural to integrate with respect to $x$.

You absolutely can, but in JEE Advanced, time is a currency you cannot afford to waste. If you use vertical strips, the upper boundary changes from the rightward parabola to the downward parabola at $x = 3/4$. This forces you to split the integral into two separate, tedious calculations. By choosing horizontal strips ($dy$), the boundaries remain consistent throughout the entire region, turning a two-part problem into a single, elegant integral.

Q: I'm stuck on the intersection point. How do I solve $x^4 - 4x + 3 = 0$?

Don't panic! Whenever you see a polynomial of degree 4, look for integer roots first. Notice that the sum of the coefficients ($1 - 4 + 3$) is zero? That is a massive hint that $x = 1$ is a root. Once you identify $x = 1$, you can use synthetic division or polynomial long division to factor out $(x-1)$. You will find that $(x-1)$ is actually a repeated root, which simplifies the equation significantly.

Q: Why do we multiply the final integral by $2$?

This comes from the symmetry of the region. The curves $y = x^2$ and $y = -x^2$ are mirror images across the $x$-axis. Because the entire bounded region is symmetric about this axis, the area of the top half is identical to the area of the bottom half. By calculating only the top half, we simplify our work and then double it to get the total area.

Question

Find the area bounded by the curves $x^{2}=y, x^{2}=-y$ and $y^{2}=4 x-3$.

Sigma Percentile · Accepted Answer

### The Geometry of Elegance: Mastering Area Under Curves

Welcome, future engineers! Today, we are not just solving a calculus problem; we are embarking on a journey of geometric intuition.

When you face a problem involving the area bounded by curves, the biggest mistake you can make is to dive straight into the algebra. Stop. Breathe. Visualize.

Let us look at the battlefield: we have $y = x^2$ (an upward-opening parabola), $y = -x^2$ (a downward-opening parabola), and $y^2 = 4x - 3$ (a rightward-opening parabola). These three curves dance together to trap a specific region in the coordinate plane.

### Phase 1: The Strategic Choice

In the world of JEE Advanced, the difference between a topper and a struggler is often the choice of the integration variable.

If we were to use vertical strips ($dx$), we would be forced to break our integral into two distinct pieces because the upper and lower boundaries change their identity at $x = \frac{3}{4}$. That is a recipe for calculation errors.

Instead, look at the horizontal perspective. If we use horizontal strips ($dy$), the right boundary is always the curve $y^2 = 4x - 3$, and the left boundary is always the curve $y = x^2$. This consistency is the hallmark of a well-planned solution. Always look for the path of least resistance!

### Phase 2: The Intersection Hunt

Before we can integrate, we must know our limits. We need to find where the upward parabola $y = x^2$ meets the rightward parabola $y^2 = 4x - 3$.

Substituting $y = x^2$ into the second equation gives us $(x^2)^2 = 4x - 3$, or:

$$x^4 - 4x + 3 = 0$$

Do not let the degree four polynomial intimidate you. As we noted, the sum of the coefficients is zero, which immediately reveals $x = 1$ as a root.

Factoring this, we find $(x-1)^2(x^2+2x+3) = 0$. Since the quadratic term has no real roots, our intersection point is locked at $x = 1$.

At this point, $y = 1^2 = 1$. Thus, our upper intersection point is $(1, 1)$, and by symmetry, the lower one is $(1, -1)$.

### Phase 3: The Calculus of Symmetry

Now, we set up our integral. Because of the symmetry about the $x$-axis, we calculate the area of the upper half and multiply by $2$.

Our integral becomes:

$$A = 2 \int_{0}^{1} (x_{right} - x_{left}) \, dy$$

Substituting our expressions for $x$ in terms of $y$, where $x_{right} = \frac{y^2+3}{4}$ and $x_{left} = \sqrt{y}$, we get:

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

Let us expand this to make the integration trivial:

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

### Phase 4: The Final Calculation

Applying the power rule, $\int y^n \, dy = \frac{y^{n+1}}{n+1}$, we integrate term by term:

$$A = 2 \left[ \frac{y^3}{12} + \frac{3y}{4} - \frac{2y^{3/2}}{3} \right]_{0}^{1}$$

Substituting the upper limit $y = 1$ (the lower limit $y = 0$ vanishes), we are left with:

$$A = 2 \left( \frac{1}{12} + \frac{3}{4} - \frac{2}{3} \right)$$

Finding the common denominator of $12$, we get:

$$A = 2 \left( \frac{1 + 9 - 8}{12} \right) = 2 \left( \frac{2}{12} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3}$$

And there it is: **$\frac{1}{3}$ square units**. It is not just a number; it is the result of a strategic choice, a careful observation of symmetry, and the disciplined application of calculus.

Remember, in JEE, the math is the tool, but your intuition is the master. Keep practicing, keep visualizing, and keep falling in love with the process!

Animated Solution for Mathematics - Definite Integration: Find the area bounded by the curves $x^{2} = y, x^{2} = - y$ and $y^{2} = 4 x - 3$ .

Visualized Solution

Identifying the Curves

The Bounded Region

Finding Intersection Points

Solving $x^{4} - 4 x + 3 = 0$

Coordinates of Intersections

Choosing the Strip: $d x$ vs $d y$

Boundaries for Horizontal Strip

Setting up the Integral

Expanding the Integrand

Performing the Integration

Applying the Limits

Fraction Arithmetic

Final Area Calculation

The Sigma Insight: Area Bounded by Curves

The Geometry of Elegance

Phase 1

Phase 2

Phase 3

Phase 4

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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; } Find the area bounded by the curves x2=y,x2=−y and y2=4x−3.