If `f(x) = 2x^(3)-15x^(2)+24x` and `g(x) = int_(0)^(x)f(t) dt + int_(0)^(5-x) f(t) dt (0 lt x lt 5)`. Find the number of integers for which `g(x)` is increasing.

Answer

To solve the problem, we need to analyze the function \( g(x) \) defined as:

\[ g(x) = \int_{0}^{x} f(t) \, dt + \int_{0}^{5-x} f(t) \, dt \]

where \( f(x) = 2x^3 - 15x^2 + 24x \).

### Step 1: Differentiate \( g(x) \)

Using the Leibniz rule for differentiation under the integral sign, we find \( g'(x) \):

\[ g'(x) = \frac{d}{dx} \left( \int_{0}^{x} f(t) \, dt \right) + \frac{d}{dx} \left( \int_{0}^{5-x} f(t) \, dt \right) \]

The first term differentiates to \( f(x) \). For the second term, we apply the chain rule:

\[ \frac{d}{dx} \left( \int_{0}^{5-x} f(t) \, dt \right) = -f(5-x) \]

Thus, we have:

\[ g'(x) = f(x) - f(5-x) \]

### Step 2: Calculate \( f(5-x) \)

Now we need to compute \( f(5-x) \):

\[ f(5-x) = 2(5-x)^3 - 15(5-x)^2 + 24(5-x) \]

Calculating each term:

- \( (5-x)^3 = 125 - 75x + 15x^2 - x^3 \)
- \( (5-x)^2 = 25 - 10x + x^2 \)

Substituting these into \( f(5-x) \):

\[ f(5-x) = 2(125 - 75x + 15x^2 - x^3) - 15(25 - 10x + x^2) + 24(5-x) \]

Expanding this gives:

\[ = 250 - 150x + 30x^2 - 2x^3 - 375 + 150x - 15x^2 + 120 - 24x \]

Combining like terms:

\[ = -2x^3 + (30x^2 - 15x^2) + (250 - 375 + 120 - 24x) \]
\[ = -2x^3 + 15x^2 - 24x - 5 \]

### Step 3: Set up \( g'(x) \)

Now we can write:

\[ g'(x) = f(x) - f(5-x) = (2x^3 - 15x^2 + 24x) - (-2x^3 + 15x^2 - 24x - 5) \]

Simplifying:

\[ g'(x) = 2x^3 - 15x^2 + 24x + 2x^3 - 15x^2 + 24x + 5 \]
\[ = 4x^3 - 30x^2 + 48x + 5 \]

### Step 4: Find critical points

To find where \( g'(x) = 0 \):

\[ 4x^3 - 30x^2 + 48x + 5 = 0 \]

Using numerical methods or graphing, we find the roots. The roots are approximately:

- \( x \approx -0.098 \) (not in the interval [0, 5])
- \( x = 2.5 \)
- \( x = 5 \)

### Step 5: Determine intervals of increase

We need to analyze the sign of \( g'(x) \) in the intervals:

- \( (0, 2.5) \)
- \( (2.5, 5) \)

Using test points:

1. For \( x = 1 \) (in \( (0, 2.5) \)):
\[ g'(1) = 4(1)^3 - 30(1)^2 + 48(1) + 5 = 4 - 30 + 48 + 5 = 27 > 0 \]
So \( g(x) \) is increasing in \( (0, 2.5) \).

2. For \( x = 3 \) (in \( (2.5, 5) \)):
\[ g'(3) = 4(3)^3 - 30(3)^2 + 48(3) + 5 = 108 - 270 + 144 + 5 = -13 < 0 \]
So \( g(x) \) is decreasing in \( (2.5, 5) \).

### Step 6: Count integers in the increasing interval

The integers in the interval \( (0, 2.5) \) are \( 1 \) and \( 2 \). Thus, there are **2 integers** for which \( g(x) \) is increasing.

### Final Answer

The number of integers for which \( g(x) \) is increasing is **2**.

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