If `f(x) = int_(0)^(x)t sin t dt`, then f'(x) is

Options

A. `cos x + x sin x`
B. `x sin x`
C. `x cos x`
D. `sin x + x cos x`

Answer

To find \( f'(x) \) for the function defined as

\[ f(x) = \int_{0}^{x} t \sin t \, dt, \]

we will use the Fundamental Theorem of Calculus, specifically Leibniz's rule for differentiation under the integral sign.

### Step-by-Step Solution:

1. **Identify the function and its limits**:
The function \( f(x) \) is defined as the integral of \( t \sin t \) from 0 to \( x \).

2. **Apply the Fundamental Theorem of Calculus**:
According to the theorem, if \( F(t) \) is an antiderivative of \( f(t) \), then:
\[ \frac{d}{dx} \left( \int_{a}^{g(x)} f(t) \, dt \right) = f(g(x)) \cdot g'(x). \]
Here, \( g(x) = x \) and \( a = 0 \).

3. **Differentiate \( f(x) \)**:
We can express \( f'(x) \) as:
\[ f'(x) = t \sin t \bigg|_{t=x} \cdot \frac{d}{dx}(x) - t \sin t \bigg|_{t=0} \cdot \frac{d}{dx}(0). \]

4. **Evaluate the first term**:
For the first term, substitute \( t = x \):
\[ f'(x) = x \sin x \cdot 1. \]

5. **Evaluate the second term**:
For the second term, substitute \( t = 0 \):
\[ f'(x) = 0 \cdot \sin(0) \cdot 0 = 0. \]

6. **Combine the results**:
Thus, we have:
\[ f'(x) = x \sin x - 0 = x \sin x. \]

Therefore, the derivative \( f'(x) \) is

\[ \boxed{x \sin x}. \]