The area under the curve $$y = |\cos x - \sin x|$$, $$0 \leq x \leq \frac{\pi}{2}$$, and above x-axis is :

The area under the curve $$ y = |\cos x - \sin x| $$ from $$ x = 0 $$ to $$ x = \frac{\pi}{2} $$ and above the x-axis needs to be found. Since the function involves an absolute value, we first analyze the expression inside. Rewrite $$ \cos x - \sin x $$ using trigonometric identities. Express it as $$ \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) $$, because:

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

Thus, $$ y = \left| \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) \right| = \sqrt{2} \left| \cos\left(x + \frac{\pi}{4}\right) \right| $$.

Now, determine where $$ \cos\left(x + \frac{\pi}{4}\right) $$ is positive or negative in the interval $$ [0, \frac{\pi}{2}] $$. The argument $$ x + \frac{\pi}{4} $$ ranges from $$ \frac{\pi}{4} $$ (when $$ x = 0 $$) to $$ \frac{3\pi}{4} $$ (when $$ x = \frac{\pi}{2} $$). Cosine is positive in $$ [\frac{\pi}{4}, \frac{\pi}{2}) $$ and negative in $$ (\frac{\pi}{2}, \frac{3\pi}{4}] $$. The point where it changes sign is at $$ x + \frac{\pi}{4} = \frac{\pi}{2} $$, so $$ x = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} $$. Therefore:

• For $$ 0 \leq x \leq \frac{\pi}{4} $$, $$ \cos\left(x + \frac{\pi}{4}\right) \geq 0 $$, so $$ y = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) $$.
• For $$ \frac{\pi}{4} < x \leq \frac{\pi}{2} $$, $$ \cos\left(x + \frac{\pi}{4}\right) < 0 $$, so $$ y = -\sqrt{2} \cos\left(x + \frac{\pi}{4}\right) $$.

The area is the integral of $$ y $$ from 0 to $$ \frac{\pi}{2} $$, split at $$ x = \frac{\pi}{4} $$:

$$ \text{Area} = \int_{0}^{\frac{\pi}{4}} \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left( -\sqrt{2} \cos\left(x + \frac{\pi}{4}\right) \right) dx $$

Factor out $$ \sqrt{2} $$:

$$ \text{Area} = \sqrt{2} \int_{0}^{\frac{\pi}{4}} \cos\left(x + \frac{\pi}{4}\right) dx - \sqrt{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos\left(x + \frac{\pi}{4}\right) dx $$

Find the antiderivative of $$ \cos\left(x + \frac{\pi}{4}\right) $$. Let $$ u = x + \frac{\pi}{4} $$, so $$ du = dx $$. The antiderivative is $$ \sin\left(x + \frac{\pi}{4}\right) $$.

Evaluate the first integral:

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

Multiply by $$ \sqrt{2} $$:

$$ \sqrt{2} \left(1 - \frac{\sqrt{2}}{2}\right) = \sqrt{2} \cdot 1 - \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \sqrt{2} - \frac{2}{2} = \sqrt{2} - 1 $$

Now, evaluate the second integral:

$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos\left(x + \frac{\pi}{4}\right) dx = \left[ \sin\left(x + \frac{\pi}{4}\right) \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2} + \frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \sin\left(\frac{\pi}{2}\right) = \frac{\sqrt{2}}{2} - 1 $$

Multiply by $$ -\sqrt{2} $$:

$$ -\sqrt{2} \left( \frac{\sqrt{2}}{2} - 1 \right) = -\sqrt{2} \cdot \frac{\sqrt{2}}{2} + \sqrt{2} \cdot 1 = -\frac{2}{2} + \sqrt{2} = -1 + \sqrt{2} $$

Add both parts together:

$$ \text{Area} = (\sqrt{2} - 1) + (-1 + \sqrt{2}) = \sqrt{2} - 1 - 1 + \sqrt{2} = 2\sqrt{2} - 2 $$

Hence, the area is $$ 2\sqrt{2} - 2 $$. Comparing with the options:

• A. $$ 2\sqrt{2} $$
• B. $$ 2\sqrt{2} - 2 $$
• C. $$ 2\sqrt{2} + 2 $$
• D. 0

The correct answer is Option B.