To solve the problem, we need to find the value of the expression:
\[
\int_2^3 f'(x) f''(x) \, dx + \int_1^3 f''(x) \, dx
\]
### Step 1: Determine the slopes of the tangents
We know that the slopes of the tangents at the points (1, f(1)), (2, f(2)), and (3, f(3)) are given by the angles they make with the positive x-axis.
- For the point (1, f(1)), the angle is \(30^\circ\):
\[
f'(1) = \tan(30^\circ) = \frac{1}{\sqrt{3}}
\]
- For the point (2, f(2)), the angle is \(60^\circ\):
\[
f'(2) = \tan(60^\circ) = \sqrt{3}
\]
- For the point (3, f(3)), the angle is \(45^\circ\):
\[
f'(3) = \tan(45^\circ) = 1
\]
### Step 2: Evaluate the first integral
We can use the substitution \(t = f'(x)\), which gives us \(dt = f''(x) \, dx\). Thus, we can rewrite the first integral:
\[
\int_2^3 f'(x) f''(x) \, dx = \int_2^3 t \, dt
\]
The limits of integration will remain the same, and we can evaluate the integral:
\[
\int t \, dt = \frac{t^2}{2}
\]
Now, we need to evaluate this from \(x = 2\) to \(x = 3\):
\[
\left[ \frac{f'(x)^2}{2} \right]_2^3 = \frac{f'(3)^2}{2} - \frac{f'(2)^2}{2}
\]
Substituting the values we found earlier:
\[
= \frac{1^2}{2} - \frac{(\sqrt{3})^2}{2} = \frac{1}{2} - \frac{3}{2} = -1
\]
### Step 3: Evaluate the second integral
The second integral is straightforward:
\[
\int_1^3 f''(x) \, dx = [f'(x)]_1^3 = f'(3) - f'(1)
\]
Substituting the values:
\[
= 1 - \frac{1}{\sqrt{3}}
\]
### Step 4: Combine the results
Now we can combine both parts:
\[
\int_2^3 f'(x) f''(x) \, dx + \int_1^3 f''(x) \, dx = -1 + \left(1 - \frac{1}{\sqrt{3}}\right)
\]
Simplifying this:
\[
= -1 + 1 - \frac{1}{\sqrt{3}} = -\frac{1}{\sqrt{3}}
\]
### Final Answer
Thus, the value of the expression is:
\[
-\frac{1}{\sqrt{3}}
\]
