The area (in sq. units) in the first quadrant bounded by the parabola, y = x² + 1, the tangent to it

Q: Wait, why can't we just integrate the tangent line equation from $0$ to $2$?

That is a fantastic question! If you integrate the tangent line equation from $0$ to $2$, you are calculating the area of a trapezoid (or a combination of a rectangle and a triangle) under the tangent line down to the x-axis. But the region we are interested in is bounded by the *parabola* on top and the *tangent* on the bottom. Integrating the tangent line alone would give you the area *under* the tangent, which is not what the problem asks for. We need the area *between* the parabola and the tangent, bounded by the coordinate axes.

Q: Why is the base of the triangle $2 - 3/4$? I'm confused about the x-intercept.

Think of the tangent line as a straight path. It hits the x-axis at a specific point. To find that point, we set the y-coordinate to zero in our tangent equation, $y = 4x - 3$. Setting $y=0$ gives $4x = 3$, so $x = 3/4$. The triangle's base lies on the x-axis, starting from this intercept ($x = 3/4$) and ending at the point of tangency ($x = 2$). The length of this segment is simply the difference: $2 - 3/4 = 5/4$. Always visualize the horizontal span of your shapes!

Q: Is there a way to solve this by integrating with respect to y?

Yes, absolutely! You could express $x$ in terms of $y$ for both the parabola and the tangent line and integrate with respect to $y$. However, that would require splitting the integral into two parts because the 'left' boundary changes at $y=1$. The subtraction method we used is much more elegant and less prone to calculation errors. In JEE, choosing the path of least resistance—like our subtraction strategy—is a hallmark of a smart student.

Question

The area (in sq. units) in the first quadrant bounded by the parabola, $y = x^2 + 1$, the tangent to it at the point $(2, 5)$ and the coordinate axes is:

Options:

(A) $187/24$

(B) $8/3$

(C) $37/24$

(D) $14/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a calculus problem; we are embarking on a journey of geometric visualization. We have a parabola, $y = x^2 + 1$, a graceful curve that starts at $(0, 1)$ and climbs upwards.

We have a point $P(2, 5)$ resting on this curve, and a tangent line that kisses the parabola at this exact spot. Our mission is to find the area trapped in the first quadrant by these boundaries. This is a classic JEE Advanced challenge—it tests not just your ability to integrate, but your ability to see the 'negative space' in a problem.

### The Tangent as a Linear Approximation

Before we can calculate any area, we must define our boundaries. The parabola is given by $y = x^2 + 1$. To find the tangent line, we need its slope.

The derivative, $\frac{dy}{dx} = 2x$, is the heartbeat of the curve; it tells us the instantaneous rate of change at any point. At our point of interest, $x = 2$, the slope is $m = 2(2) = 4$.

Using the point-slope form, $y - y_1 = m(x - x_1)$, we substitute our point $(2, 5)$ and slope $m = 4$ to get $y - 5 = 4(x - 2)$. Simplifying this, we arrive at the elegant linear equation: 
$$y = 4x - 3$$
This line serves as our primary boundary.

### The Art of Subtraction

Now, look at the region. It is bounded by the y-axis ($x=0$), the x-axis ($y=0$), the parabola, and the tangent line. If you try to integrate this directly, you might find yourself tangled in complex limits.

Instead, let us use the 'Sculptor's Technique'. Imagine a block of marble representing the area under the parabola from $x=0$ to $x=2$. This area is defined by the integral:
$$\int_{0}^{2} (x^2 + 1) dx$$

When we calculate this, we get:
$$\left[ \frac{x^3}{3} + x ight]_{0}^{2} = \frac{8}{3} + 2 = \frac{14}{3}$$
This represents the total area under the curve from the y-axis to the point $P$.

However, our region is not the entire area under the parabola. The tangent line cuts through this space. The area we want is the area under the parabola *minus* the area of the triangle formed by the tangent line and the x-axis.

### The Triangle and the Final Calculation

Let us focus on that triangle. Its height is the y-coordinate of point $P$, which is $5$. Its base lies on the x-axis.

To find the base, we determine where the tangent line $y = 4x - 3$ hits the x-axis by setting $y = 0$:
$$4x - 3 = 0 \Rightarrow x = \frac{3}{4}$$
The base of our triangle stretches from $x = 3/4$ to $x = 2$. The length of this base is $2 - \frac{3}{4} = \frac{5}{4}$.

The area of this right-angled triangle is:
$$	ext{Area}_{	ext{triangle}} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{5}{4} 	imes 5 = \frac{25}{8}$$

Finally, we perform our subtraction to find the required area:
$$	ext{Required Area} = \frac{14}{3} - \frac{25}{8}$$

To subtract these, we find a common denominator of $24$:
$$	ext{Required Area} = \frac{112}{24} - \frac{75}{24} = \frac{37}{24}$$

### Conclusion

The final result is **$\frac{37}{24}$ square units**. It is a beautiful, precise number.

You see, calculus is not just about crunching numbers; it is about decomposing complex shapes into simpler, manageable parts. You have successfully navigated the parabola, the tangent, and the geometry of the coordinate plane. Keep this 'subtraction mindset' with you—it will serve you well in the most difficult problems of your JEE journey.

Animated Solution for Mathematics - Definite Integration: The area (in sq. units) in the first quadrant bounded by the parabola, $y = x^{2} + 1$ , the tangent to it at the point $(2, 5)$ and the coordinate axes is:

Select Answer:

Visualized Solution

Visualizing the Curve

Finding the Slope of the Tangent

Slope at Point $P (2, 5)$

Equation of the Tangent Line

Intersection with the X-axis

Defining the Bounded Region

Strategy for Area Calculation

Integrating the Parabola

Evaluating the Parabola Integral

Setting up the Triangle Area

Atomic Compute: Triangle Area

Final Area Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Tangent as a Linear Approximation

The Art of Subtraction

The Triangle and the Final Calculation

Conclusion

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