Let `u = int_(pi//6)^(pi//2)` min. (`sqrt(3)sinx, cosx`) dx and `V = int_(-3)^(5)x^(2)sgn (x-1) dx`. If `V = lambdaU`, then find the value of `lambda.

Answer

To solve the problem, we need to calculate the values of \( u \) and \( V \) as defined in the question, and then find the value of \( \lambda \) such that \( V = \lambda U \).

### Step 1: Calculate \( u \)

Given:
\[ u = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \min(\sqrt{3} \sin x, \cos x) \, dx \]

To find \( u \), we need to determine where \( \sqrt{3} \sin x \) is less than or equal to \( \cos x \) within the limits of integration.

1. **Set up the inequality**:
\[ \sqrt{3} \sin x \leq \cos x \]
This can be rearranged to:
\[ \tan x \leq \frac{1}{\sqrt{3}} \]
The angle where \( \tan x = \frac{1}{\sqrt{3}} \) is \( x = \frac{\pi}{6} \).

2. **Determine the intervals**:
- For \( x \in \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \), \( \tan x \) increases from \( \frac{1}{\sqrt{3}} \) to \( \infty \).
- Thus, \( \sqrt{3} \sin x \) is less than or equal to \( \cos x \) for \( x \in \left[\frac{\pi}{6}, \frac{\pi}{4}\right] \) and greater for \( x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \).

3. **Split the integral**:
\[ u = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sqrt{3} \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx \]

4. **Calculate the first integral**:
\[ \int \sqrt{3} \sin x \, dx = -\sqrt{3} \cos x \]
Evaluating from \( \frac{\pi}{6} \) to \( \frac{\pi}{4} \):
\[ = -\sqrt{3} \left( \cos\frac{\pi}{4} - \cos\frac{\pi}{6} \right) = -\sqrt{3} \left( \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \right) \]

5. **Calculate the second integral**:
\[ \int \cos x \, dx = \sin x \]
Evaluating from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \):
\[ = \sin\frac{\pi}{2} - \sin\frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} \]

6. **Combine results**:
\[ u = -\sqrt{3} \left( \frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2} \right) + \left( 1 - \frac{1}{\sqrt{2}} \right) \]

### Step 2: Calculate \( V \)

Given:
\[ V = \int_{-3}^{5} x^2 \, \text{sgn}(x-1) \, dx \]

1. **Split the integral based on the sign function**:
\[ V = \int_{-3}^{1} -x^2 \, dx + \int_{1}^{5} x^2 \, dx \]

2. **Calculate the first integral**:
\[ \int -x^2 \, dx = -\frac{x^3}{3} \]
Evaluating from \( -3 \) to \( 1 \):
\[ = -\frac{1^3}{3} - \left(-\frac{(-3)^3}{3}\right) = -\frac{1}{3} + 9 = \frac{26}{3} \]

3. **Calculate the second integral**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from \( 1 \) to \( 5 \):
\[ = \frac{5^3}{3} - \frac{1^3}{3} = \frac{125}{3} - \frac{1}{3} = \frac{124}{3} \]

4. **Combine results**:
\[ V = \frac{26}{3} + \frac{124}{3} = \frac{150}{3} = 50 \]

### Step 3: Find \( \lambda \)

Given:
\[ V = \lambda U \]
Substituting the values:
\[ 50 = \lambda \cdot u \]

Now we need to find \( \lambda \):
\[ \lambda = \frac{50}{u} \]

### Final Calculation of \( \lambda \)

After substituting the computed value of \( u \) into the equation, we can find the final value of \( \lambda \).