Let A₁ be the area of the region bounded by the curves y = sin x, y = cos x and y-axis in the first

Q: Wait, why is $A_1$ defined as $\cos x - \sin x$ and not the other way around?

Great question! In the interval $0 \le x \le \frac{\pi}{4}$, the cosine curve starts at $1$ and decreases, while the sine curve starts at $0$ and increases. Therefore, $\cos x$ is the upper boundary and $\sin x$ is the lower boundary. To find the area between them, we always integrate the upper function minus the lower function: $\int (\text{upper} - \text{lower}) dx$.

Q: Can I calculate $A_2$ by integrating $\sin x$ and $\cos x$ separately?

You absolutely can, and it is a valid method! You would integrate $\sin x$ from $0$ to $\frac{\pi}{4}$ and $\cos x$ from $\frac{\pi}{4}$ to $\frac{\pi}{2}$. However, the geometric shortcut of recognizing $A_1 + A_2$ as the integral of $\cos x$ from $0$ to $\frac{\pi}{2}$ is much faster and less prone to calculation errors. Always look for these elegant shortcuts in JEE problems!

Q: Why does $A_1 + A_2$ simplify to just the integral of $\cos x$?

If you look at the graph, $A_1$ is the area between the curves from $0$ to $\frac{\pi}{4}$. $A_2$ is the area under the lower envelope from $\frac{\pi}{4}$ to $\frac{\pi}{2}$. When you add them, you are essentially filling the space under the cosine curve from $0$ to $\frac{\pi}{2}$. It is a beautiful puzzle-piece fit!

Question

Let $A_1$ be the area of the region bounded by the curves $y = \sin x$, $y = \cos x$ and $y$-axis in the first quadrant. Also, let $A_2$ be the area of the region bounded by the curves $y = \sin x$, $y = \cos x$, $x$-axis and $x = \frac{\pi}{2}$ in the first quadrant. Then,

Options:

(A) $A_1 = A_2$ and $A_1 + A_2 = \sqrt{2}$

(B) $A_1 : A_2 = 1 : 2$ and $A_1 + A_2 = 1$

(C) $2 A_1 = A_2$ and $A_1 + A_2 = 1 + \sqrt{2}$

(D) $A_1 : A_2 = 1 : \sqrt{2}$ and $A_1 + A_2 = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are witnessing a beautiful choreography between two of the most fundamental functions in trigonometry: $y = \sin x$ and $y = \cos x$.

In the first quadrant, these two curves perform a delicate dance, and our goal is to measure the areas they carve out. Imagine standing in the first quadrant, where $x$ ranges from $0$ to $\frac{\pi}{2}$.

The sine curve, $y = \sin x$, begins its journey at the origin $(0,0)$ and climbs gracefully to $(\frac{\pi}{2}, 1)$. The cosine curve, $y = \cos x$, starts at the peak $(0,1)$ and descends to the horizon at $(\frac{\pi}{2}, 0)$.

They are destined to cross paths. To find where this happens, we set them equal: $\sin x = \cos x$. Dividing by $\cos x$, we find $	an x = 1$, which tells us the intersection occurs exactly at $x = \frac{\pi}{4}$. This point is our pivot.

### Defining and Conquering $A_1$

Our first region, $A_1$, is bounded by the $y$-axis, the sine curve, and the cosine curve. Looking at our graph, from $x = 0$ to $x = \frac{\pi}{4}$, the cosine curve sits proudly above the sine curve.

Thus, the area $A_1$ is the integral of the difference between the upper and lower boundaries:

$$A_1 = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx$$

Integrating this is a joy. The integral of $\cos x$ is $\sin x$, and the integral of $-\sin x$ is $\cos x$. Evaluating from $0$ to $\frac{\pi}{4}$:

$$A_1 = [\sin x + \cos x]_{0}^{\frac{\pi}{4}} = \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}ight) - (0 + 1) = \sqrt{2} - 1$$

There we have it! A solid, elegant value for $A_1$.

### The Geometric "Aha!" Moment

Now, we approach $A_2$. The problem defines it as the area bounded by the sine curve, the cosine curve, the $x$-axis, and the vertical line $x = \frac{\pi}{2}$.

If you try to calculate this by splitting it into two integrals, you will succeed, but you might miss the elegance of the geometry. Look at the combined region $A_1 + A_2$.

If you place $A_1$ and $A_2$ together, they perfectly fill the area under the cosine curve from $x = 0$ to $x = \frac{\pi}{2}$. This is the "puzzle-piece" insight that separates the masters from the calculators.

$$A_1 + A_2 = \int_{0}^{\frac{\pi}{2}} \cos x dx$$

Calculating this is straightforward:

$$A_1 + A_2 = [\sin x]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}ight) - \sin(0) = 1 - 0 = 1$$

### The Final Calculation

With $A_1 + A_2 = 1$ and $A_1 = \sqrt{2} - 1$, finding $A_2$ is trivial:

$$A_2 = 1 - A_1 = 1 - (\sqrt{2} - 1) = 2 - \sqrt{2} = \sqrt{2}(\sqrt{2} - 1)$$

Finally, the ratio $A_1 : A_2$ is:

$$\frac{A_1}{A_2} = \frac{\sqrt{2} - 1}{\sqrt{2}(\sqrt{2} - 1)} = \frac{1}{\sqrt{2}}$$

We have arrived at our destination. The sum is $1$, and the ratio is **$1 : \sqrt{2}$**. Remember, in JEE Advanced, the most powerful tool in your arsenal is not just your ability to integrate, but your ability to visualize the geometry behind the equations.

Select Answer:

Visualized Solution

Visualizing the Curves

Finding the Intersection Point

Defining Area $A_{1}$

Calculating Area $A_{1}$

Defining Area $A_{2}$

The Geometric Shortcut

Calculating $A_{1} + A_{2}$

Finding Area $A_{2}$

Calculating the Ratio $A_{1} : A_{2}$

Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Defining and Conquering $A_{1}$

The Geometric "Aha!" Moment

The Final Calculation

Similar Questions

The area, enclosed by the curves $y = sin x + cos x$ and $y = ∣ cos x - sin x ∣$ and the lines $x = 0, x = \frac{π}{2}$ is:

Let $A_{1}$ be the bounded area enclosed by the curves $y = x^{2} + 2, x + y = 8$ and $y$ -axis that lies in the first quadrant. Let $A_{2}$ be the bounded area enclosed by the curves $y = x^{2} + 2, y^{2} = x, x = 2$ , and $y$ -axis that lies in the first quadrant. Then $A_{1} - A_{2}$ is equal to

The area of the region between the curves $y = \frac{1 + s i n x}{c o s x}$ and $y = \frac{1 - s i n x}{c o s x}$ bounded by the lines $x = 0$ and $x = π /4$ is

The area enclosed by the curves $y = sin x + cos x$ and $y = ∣ cos x - sin x ∣$ over the interval $[0, π /2]$ is

The area bounded by the curves $y = cos x$ and $y = sin x$ between the ordinates $x = 0$ and $x = \frac{3 π}{2}$ is

The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is:

Let the area of the region bounded by the curve $y = max {sin x, cos x}$ , lines $x = 0, x = \frac{3 π}{2}$ , and the $x$ -axis be $A$ . Then, $A + A^{2}$ is equal to ......... .

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

The area of the region described by $A = {(x, y) : x^{2} + y^{2} \leq 1 and y^{2} \leq 1 - x}$ is:

The area of the region $A = {(x, y) : ∣ cos x - sin x ∣ \leq y \leq sin x, 0 \leq x \leq \frac{π}{2}}$ is

Select Answer:

Visualized Solution

Visualizing the Curves

Finding the Intersection Point

Defining Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Calculating Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

Defining Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

The Geometric Shortcut

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​+A2​

Finding Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​

Calculating the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​:A2​

Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Defining and Conquering .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​

The Geometric "Aha!" Moment

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area, enclosed by the curves y=sinx+cosx and y=∣cosx−sinx∣ and the lines x=0,x=2π​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region between the curves y=cosx1+sinx​​ and y=cosx1−sinx​​ bounded by the lines x=0 and x=π/4 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area enclosed by the curves y=sinx+cosx and y=∣cosx−sinx∣ over the interval [0,π/2] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area bounded by the curves y=cosx and y=sinx between the ordinates x=0 and x=23π​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region bounded by the curves x(1+y2)=1 and y2=2x is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the area of the region bounded by the curve y=max{sinx,cosx}, lines x=0,x=23π​, and the x-axis be A. Then, A+A2 is equal to ......... .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sketch the region bounded by the curves y=x2 and y=1+x22​. Find the area.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region described by A={(x,y):x2+y2≤1 and y2≤1−x} is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region A={(x,y):∣cosx−sinx∣≤y≤sinx,0≤x≤2π​} is