Step 1. Evaluate the integral:
\(\int_{\frac{\pi}{8}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx\)
Step 2. Rewrite \( \sqrt{1 - \sin 2x} \) using trigonometric identities:
\(\int_{\frac{\pi}{8}}^{\frac{\pi}{3}} |\sin x - \cos x| \, dx\)
Step 3. Split the integral based on the intervals where \( \sin x - \cos x \) changes sign:
\(= \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx\)
Step 4. Solve each integral:
\(= -1 + 2\sqrt{2} - \sqrt{3}\)
Thus, we have:
\(\alpha + \beta \sqrt{2} + \gamma \sqrt{3} = -1 + 2\sqrt{2} - \sqrt{3}\)
where \( \alpha = -1 \), \( \beta = 2 \), and \( \gamma = -1 \).
Step 5. Calculate \( 3\alpha + 4\beta - \gamma \):
\(3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6\)
The Correct Answer is: \( 3\alpha + 4\beta - \gamma = 6 \).
The integral can be rewritten as: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left| \sin x - \cos x \right| \, dx \] We split the absolute value function into two parts, as \( \sin x - \cos x \) changes its sign at \( x = \frac{\pi}{4} \).
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left| \sin x - \cos x \right| \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx \]
We compute the two integrals separately. For the first part: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = -1 + 2\sqrt{2} - \sqrt{3} \] For the second part: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3} \] where we are given: \[ \alpha = -1, \quad \beta = 2, \quad \gamma = -1 \]
Now, we compute: \[ 3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6 \]
The final value is: \[ \boxed{6} \]
The area of the region enclosed by the parabolas \( y = x^2 - 5x \) and \( y = 7x - x^2 \) is _________.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,