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

Q: Wait, why do we split the integral at $\pi/4$?

The absolute value function $f(x) = |g(x)|$ is defined as $g(x)$ if $g(x) \ge 0$ and $-g(x)$ if $g(x) < 0$. In our case, $g(x) = \cos x - \sin x$. This expression changes sign exactly when $\cos x = \sin x$, which happens at $x = \pi/4$. To integrate this correctly, we must treat the two intervals separately because the function definition changes at this critical point.

Q: How do I know which function is the 'upper' curve?

You can determine this by testing a value within each interval. For example, in $[0, \pi/4]$, pick $x = 0$. Then $y_1 = \sin 0 + \cos 0 = 1$ and $y_2 = |\cos 0 - \sin 0| = 1$. Since they are equal at the boundary, check a point like $x = \pi/6$. You will find $y_1 > y_2$. Alternatively, you can just subtract the functions and ensure the resulting area is positive by taking the absolute value of the integral if you are unsure.

Q: Can I just integrate the absolute value directly without splitting?

Unfortunately, no. The fundamental theorem of calculus requires the integrand to be continuous and well-defined. While the absolute value function is continuous, its derivative is not defined at the cusp, and more importantly, the expression inside the integral changes its algebraic form. Splitting the integral is the standard and most robust way to handle piecewise functions like this.

Question

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

Options:

(A) $4(\sqrt{2} - 1)$

(B) $2\sqrt{2}(\sqrt{2} - 1)$

(C) $2(\sqrt{2} + 1)$

(D) $2\sqrt{2}(\sqrt{2} + 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the area enclosed by the functions $y_1 = \sin x + \cos x$ and $y_2 = |\cos x - \sin x|$ over the interval $[0, \pi/2]$.

These functions represent a smooth wave and a sharp, angular curve, respectively. Our goal is to calculate the total area trapped between them within the specified bounds.

### The Critical Point: Where the Modulus Breaks

The absolute value function is a conditional statement. The behavior of $y_2$ depends entirely on the sign of the expression $\cos x - \sin x$.

To solve this, we must identify the "hinge" point where the expression inside the modulus changes sign. This occurs when $\cos x = \sin x$.

Within our interval of $[0, \pi/2]$, this equality holds precisely at $x = \pi/4$. This point serves as the boundary that splits our integration into two distinct phases.

### Phase One: The First Region

In the interval $[0, \pi/4]$, the cosine function is dominant, meaning $\cos x \ge \sin x$. Consequently, the expression inside the modulus is non-negative, allowing us to remove the absolute value bars.

The function $y_2$ simplifies to $\cos x - \sin x$. We set up the integral for the area $A_1$ as follows:

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

The $\cos x$ terms cancel out, leaving us with the integral of $2 \sin x$. Integrating $2 \sin x$ yields $-2 \cos x$.

Evaluating this from $0$ to $\pi/4$, we obtain:

$$-2(\cos(\pi/4) - \cos(0)) = -2\left(\frac{1}{\sqrt{2}} - 1\right) = 2 - \sqrt{2}$$

### Phase Two: The Second Region

Crossing the threshold at $x = \pi/4$ into the interval $[\pi/4, \pi/2]$, the relationship shifts such that $\sin x \ge \cos x$. Because the expression inside the modulus is now negative, we must negate it to maintain positivity.

Thus, $y_2$ becomes $\sin x - \cos x$. Our second area, $A_2$, is defined as:

$$A_2 = \int_{\pi/4}^{\pi/2} ((\sin x + \cos x) - (\sin x - \cos x)) \, dx$$

Here, the $\sin x$ terms vanish, leaving us with the integral of $2 \cos x$. Integrating $2 \cos x$ gives $2 \sin x$.

Evaluating from $\pi/4$ to $\pi/2$, we calculate:

$$2(\sin(\pi/2) - \sin(\pi/4)) = 2\left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \sqrt{2}$$

### The Grand Finale

The total area $A$ is the sum of the two regions: $A = A_1 + A_2$.

Substituting our calculated values:

$$A = (2 - \sqrt{2}) + (2 - \sqrt{2}) = 4 - 2\sqrt{2}$$

By factoring, we can express the final result as **$A = 2\sqrt{2}(\sqrt{2} - 1)$**. This symmetrical result confirms that when dealing with modulus functions, identifying the critical point is the key to unlocking the solution.

Animated Solution for Mathematics - Definite Integration: The area enclosed by the curves $y = sin x + cos x$ and $y = ∣ cos x - sin x ∣$ over the interval $[0, π /2]$ is

Select Answer:

Visualized Solution

Visualizing the Curves

Handling the Absolute Value

Setting up the First Integral

Simplifying the First Integrand

Computing the First Area

Setting up the Second Integral

Simplifying the Second Integrand

Computing the Second Area

Total Area Calculation

Final Conclusion

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Critical Point

Phase One

Phase Two

The Grand Finale

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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 enclosed by the curves y=sinx+cosx and y=∣cosx−sinx∣ over the interval [0,π/2] is