Let the area of the region bounded by the curve y = max{sin x, cos x}, lines x = 0, x = 3π/2, and th

Q: Why do we split the integral at $\frac{\pi}{4}$ and $\frac{5\pi}{4}$? Can't we just integrate the whole thing at once?

Great question! The function $y = \max\{\sin x, \cos x\}$ is not a single smooth curve; it is a piecewise function. Integration requires a continuous integrand. By splitting at the intersection points, we define the function uniquely in each interval, allowing us to handle the transitions between $\sin x$ and $\cos x$ correctly.

Q: Wait, why did we negate the integral in the regions $[\pi, \frac{5\pi}{4}]$ and $[\frac{5\pi}{4}, \frac{3\pi}{2}]$?

Remember, the problem asks for the area bounded by the curve and the $x$-axis. Area is a geometric quantity—it must be positive. When the curve dips below the $x$-axis, the integral $\int y dx$ yields a negative value. To get the area, we take the absolute value, which is equivalent to $-\int y dx$ for those specific regions.

Q: Is there a shortcut to avoid calculating four separate integrals?

While there isn't a single 'magic' formula, you can simplify the process by recognizing the symmetry of the sine and cosine functions. However, for JEE Advanced, the most robust method is to methodically define the function in each sub-interval. The 'cancellation' of the $1/\sqrt{2}$ terms at the end is the reward for your careful setup!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function $y = \max\{\sin x, \cos x\}$ represents the "upper envelope" of the sine and cosine curves. To find the area bounded by this function from $x = 0$ to $x = \frac{3\pi}{2}$, we must identify the intervals where each function dominates.

The switching points occur where $\sin x = \cos x$, which implies $	an x = 1$. Within the interval $[0, \frac{3\pi}{2}]$, this equality holds at:

$$x = \frac{\pi}{4} \quad 	ext{and} \quad x = \frac{5\pi}{4}$$

These points partition our domain into four distinct regions: $[0, \frac{\pi}{4}]$, $[\frac{\pi}{4}, \pi]$, $[\pi, \frac{5\pi}{4}]$, and $[\frac{5\pi}{4}, \frac{3\pi}{2}]$.

### The Four Acts of Integration

We calculate the area $A$ by integrating the maximum function over these intervals:

1. In $[0, \frac{\pi}{4}]$, $\cos x \ge \sin x$:
$$A_1 = \int_{0}^{\frac{\pi}{4}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{4}} = \frac{1}{\sqrt{2}}$$

2. In $[\frac{\pi}{4}, \pi]$, $\sin x \ge \cos x$:
$$A_2 = \int_{\frac{\pi}{4}}^{\pi} \sin x \, dx = [-\cos x]_{\frac{\pi}{4}}^{\pi} = -(-1) - (-\frac{1}{\sqrt{2}}) = 1 + \frac{1}{\sqrt{2}}$$

3. In $[\pi, \frac{5\pi}{4}]$, $\sin x \ge \cos x$ (note that both are negative, so we take the absolute value):
$$A_3 = \int_{\pi}^{\frac{5\pi}{4}} -\sin x \, dx = [\cos x]_{\pi}^{\frac{5\pi}{4}} = -\frac{1}{\sqrt{2}} - (-1) = 1 - \frac{1}{\sqrt{2}}$$

4. In $[\frac{5\pi}{4}, \frac{3\pi}{2}]$, $\cos x \ge \sin x$ (both are negative):
$$A_4 = \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} -\cos x \, dx = [-\sin x]_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} = -(-1) - (-\frac{1}{\sqrt{2}}) = 1 - \frac{1}{\sqrt{2}}$$

### The Grand Summation

Summing these individual areas gives the total area $A$:

$$A = \frac{1}{\sqrt{2}} + \left(1 + \frac{1}{\sqrt{2}}ight) + \left(1 - \frac{1}{\sqrt{2}}ight) + \left(1 - \frac{1}{\sqrt{2}}ight)$$

Simplifying the expression, the terms involving $\frac{1}{\sqrt{2}}$ cancel out:

$$A = 1 + 1 + 1 = 3$$

### Final Calculation

The problem requires the evaluation of $A + A^2$. Substituting $A = 3$:

$$A + A^2 = 3 + 3^2 = 3 + 9 = \mathbf{12}$$

Animated Solution for Mathematics - Definite Integration: 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 ......... .

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Defining the Upper Envelope

Finding Intersection Points

Region 1: $[0, \frac{π}{4}]$

Region 2: $[\frac{π}{4}, π]$

Region 3: $[π, \frac{5 π}{4}]$

Region 4: $[\frac{5 π}{4}, \frac{3 π}{2}]$

Summing the Areas

Final Calculation: $A + A^{2}$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Four Acts of Integration

The Grand Summation

Final Calculation

Similar Questions

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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 $x = \frac{π}{2}$ in the first quadrant. Then,

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Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Defining the Upper Envelope

Finding Intersection Points

Region 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [0,4π​]

Region 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [4π​,π]

Region 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [π,45π​]

Region 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [45π​,23π​]

Summing the Areas

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+A2

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Four Acts of Integration

The Grand Summation

Final Calculation

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