The function f whose graph passes through the point `(0,7//3)` and whose derivative is `f' (x) = x sqrt(1 - x^2)` is given by

Options

A. `f(x) = -1/3 [(1 - x^2)^(3//2) - 8]`
B. `f(x) = 1/3 [(1 - x^2)^(3//2) + 8]`
C. `f(x) = -1/3 [sin^(-1) x + 7]`
D. none of these

Answer

To find the function \( f(x) \) whose derivative is given by \( f'(x) = x \sqrt{1 - x^2} \) and which passes through the point \( (0, \frac{7}{3}) \), we will follow these steps:

### Step 1: Integrate the derivative
Given:
\[ f'(x) = x \sqrt{1 - x^2} \]
We need to find \( f(x) \) by integrating \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = \int x \sqrt{1 - x^2} \, dx \]

### Step 2: Use substitution
Let:
\[ t = 1 - x^2 \quad \Rightarrow \quad dt = -2x \, dx \quad \Rightarrow \quad dx = -\frac{1}{2x} dt \]
Now, we need to express \( x \sqrt{1 - x^2} \) in terms of \( t \):
\[ \sqrt{1 - x^2} = \sqrt{t} \]
Thus, we can rewrite the integral:
\[ f(x) = \int x \sqrt{1 - x^2} \, dx = \int x \sqrt{t} \left(-\frac{1}{2x} dt\right) = -\frac{1}{2} \int \sqrt{t} \, dt \]

### Step 3: Integrate \( \sqrt{t} \)
Now we integrate:
\[ -\frac{1}{2} \int \sqrt{t} \, dt = -\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + C = -\frac{1}{3} t^{3/2} + C \]

### Step 4: Substitute back for \( t \)
Substituting back \( t = 1 - x^2 \):
\[ f(x) = -\frac{1}{3} (1 - x^2)^{3/2} + C \]

### Step 5: Use the point to find \( C \)
We know that \( f(0) = \frac{7}{3} \):
\[ f(0) = -\frac{1}{3} (1 - 0^2)^{3/2} + C = -\frac{1}{3}(1) + C = -\frac{1}{3} + C \]
Setting this equal to \( \frac{7}{3} \):
\[ -\frac{1}{3} + C = \frac{7}{3} \]
Solving for \( C \):
\[ C = \frac{7}{3} + \frac{1}{3} = \frac{8}{3} \]

### Step 6: Write the final function
Substituting \( C \) back into the equation for \( f(x) \):
\[ f(x) = -\frac{1}{3} (1 - x^2)^{3/2} + \frac{8}{3} \]

### Final Answer
\[ f(x) = -\frac{1}{3} (1 - x^2)^{3/2} + \frac{8}{3} \]

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