The acute angle between `F_(1)(x)=int_(2)^(x) (2t-5)dt and F_(2)(x)=int_(0)^(x)2t dt," is , if "tantheta=a/b` where a, b are co-prime numbers then b – a is:

Explanation

To find the acute angle between the functions \( F_1(x) \) and \( F_2(x) \), we will follow these steps:

### Step 1: Define the Functions
We have:
\[ F_1(x) = \int_{2}^{x} (2t - 5) \, dt \]
\[ F_2(x) = \int_{0}^{x} 2t \, dt \]

### Step 2: Calculate \( F_1(x) \)
To compute \( F_1(x) \):
\[ F_1(x) = \left[ t^2 - 5t \right]_{2}^{x} = \left( x^2 - 5x \right) - \left( 2^2 - 5 \cdot 2 \right) \]
Calculating the lower limit:
\[ 2^2 - 5 \cdot 2 = 4 - 10 = -6 \]
Thus,
\[ F_1(x) = x^2 - 5x + 6 \]

### Step 3: Calculate \( F_2(x) \)
To compute \( F_2(x) \):
\[ F_2(x) = \left[ t^2 \right]_{0}^{x} = x^2 - 0 = x^2 \]

### Step 4: Find the Intersection Point
Set \( F_1(x) = F_2(x) \):
\[ x^2 - 5x + 6 = x^2 \]
This simplifies to:
\[ -5x + 6 = 0 \implies x = \frac{6}{5} \]

### Step 5: Calculate the y-coordinate at the Intersection
Substituting \( x = \frac{6}{5} \) into \( F_2(x) \):
\[ y = F_2\left(\frac{6}{5}\right) = \left(\frac{6}{5}\right)^2 = \frac{36}{25} \]
Thus, the intersection point is \( P\left(\frac{6}{5}, \frac{36}{25}\right) \).

### Step 6: Find the Slopes \( m_1 \) and \( m_2 \)
Calculate the derivatives:
\[ F_1'(x) = 2x - 5 \quad \text{and} \quad F_2'(x) = 2x \]
Now, evaluate at \( x = \frac{6}{5} \):
\[ m_1 = F_1'\left(\frac{6}{5}\right) = 2\left(\frac{6}{5}\right) - 5 = \frac{12}{5} - 5 = \frac{12}{5} - \frac{25}{5} = -\frac{13}{5} \]
\[ m_2 = F_2'\left(\frac{6}{5}\right) = 2\left(\frac{6}{5}\right) = \frac{12}{5} \]

### Step 7: Calculate \( \tan \theta \)
Using the formula for the tangent of the angle between two curves:
\[ \tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} \]
Substituting \( m_1 \) and \( m_2 \):
\[ \tan \theta = \frac{\left| -\frac{13}{5} - \frac{12}{5} \right|}{1 + \left(-\frac{13}{5}\right)\left(\frac{12}{5}\right)} \]
Calculating the numerator:
\[ -\frac{13}{5} - \frac{12}{5} = -\frac{25}{5} = -5 \implies | -5 | = 5 \]
Calculating the denominator:
\[ 1 - \frac{156}{25} = \frac{25 - 156}{25} = -\frac{131}{25} \]
Thus,
\[ \tan \theta = \frac{5}{-\frac{131}{25}} = \frac{5 \cdot 25}{-131} = -\frac{125}{131} \]
Taking the absolute value:
\[ \tan \theta = \frac{125}{131} \]

### Step 8: Identify \( a \) and \( b \)
From \( \tan \theta = \frac{a}{b} \):
- \( a = 125 \)
- \( b = 131 \)

### Step 9: Calculate \( b - a \)
\[ b - a = 131 - 125 = 6 \]

### Final Answer
Thus, the value of \( b - a \) is:
\[ \boxed{6} \]