Number of critical points of the function.
`f(x)=(2)/(3)sqrt(x^(3))-(x)/(2)+int_(1)^(x)((1)/(2)+(1)/(2)cos2t-sqrt(t))` dt which lie in the interval `[-2pi,2pi]` is………. .

Answer

To find the number of critical points of the function

\[ f(x) = \frac{2}{3}\sqrt{x^3} - \frac{x}{2} + \int_{1}^{x} \left(\frac{1}{2} + \frac{1}{2}\cos(2t) - \sqrt{t}\right) dt \]

that lie in the interval \([-2\pi, 2\pi]\), we will follow these steps:

### Step 1: Differentiate the function
To find the critical points, we first need to differentiate \(f(x)\):

\[ f'(x) = \frac{d}{dx}\left(\frac{2}{3}\sqrt{x^3}\right) - \frac{1}{2} + \left(\frac{1}{2} + \frac{1}{2}\cos(2x) - \sqrt{x}\right) \]

Using the chain rule and the power rule, we differentiate \(\frac{2}{3}\sqrt{x^3}\):

\[ \frac{d}{dx}\left(\frac{2}{3}\sqrt{x^3}\right) = \frac{2}{3} \cdot \frac{3}{2} x^{3/2 - 1} = x^{1/2} \]

So, we have:

\[ f'(x) = x^{1/2} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2}\cos(2x) - \sqrt{x} \]

This simplifies to:

\[ f'(x) = x^{1/2} - \sqrt{x} + \frac{1}{2}\cos(2x) \]

### Step 2: Set the derivative to zero
Next, we set \(f'(x) = 0\):

\[ x^{1/2} - \sqrt{x} + \frac{1}{2}\cos(2x) = 0 \]

Notice that \(x^{1/2} - \sqrt{x} = 0\) simplifies to \(0\), so we focus on:

\[ \frac{1}{2}\cos(2x) = 0 \]

### Step 3: Solve for \(x\)
This leads us to:

\[ \cos(2x) = 0 \]

The solutions to \(\cos(2x) = 0\) are given by:

\[ 2x = \frac{(2n + 1)\pi}{2} \quad \text{for } n \in \mathbb{Z} \]

Thus,

\[ x = \frac{(2n + 1)\pi}{4} \]

### Step 4: Identify valid critical points in the interval
Now, we need to find the values of \(n\) such that \(x\) lies within the interval \([-2\pi, 2\pi]\):

\[ -2\pi \leq \frac{(2n + 1)\pi}{4} \leq 2\pi \]

Multiplying through by \(4/\pi\):

\[ -8 \leq 2n + 1 \leq 8 \]

Solving for \(n\):

\[ -9 \leq 2n \leq 7 \implies -4.5 \leq n \leq 3.5 \]

Thus, \(n\) can take integer values from \(-4\) to \(3\) (inclusive). This gives us:

\[ n = -4, -3, -2, -1, 0, 1, 2, 3 \]

### Step 5: Count the critical points
Counting these values, we find:

- For \(n = -4\): \(x = -\frac{7\pi}{4}\)
- For \(n = -3\): \(x = -\frac{5\pi}{4}\)
- For \(n = -2\): \(x = -\frac{3\pi}{4}\)
- For \(n = -1\): \(x = -\frac{\pi}{4}\)
- For \(n = 0\): \(x = \frac{\pi}{4}\)
- For \(n = 1\): \(x = \frac{3\pi}{4}\)
- For \(n = 2\): \(x = \frac{5\pi}{4}\)
- For \(n = 3\): \(x = \frac{7\pi}{4}\)

This results in a total of **8 critical points** in the interval \([-2\pi, 2\pi]\).

### Final Answer
The number of critical points of the function \(f(x)\) that lie in the interval \([-2\pi, 2\pi]\) is **8**.

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