If `int_(1)^(a) (a-4x)dx ge 6-5a, a gt 1`, then a equals

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the inequality given in the problem, we will follow these steps:

### Step 1: Set up the integral
We need to evaluate the integral:
\[ I = \int_{1}^{a} (a - 4x) \, dx \]

### Step 2: Perform the integration
To integrate \(a - 4x\), we can separate the integral:
\[ I = \int_{1}^{a} a \, dx - \int_{1}^{a} 4x \, dx \]
Calculating each integral separately:
1. \(\int_{1}^{a} a \, dx = a[x]_{1}^{a} = a(a - 1)\)
2. \(\int_{1}^{a} 4x \, dx = 4\left[\frac{x^2}{2}\right]_{1}^{a} = 2[a^2 - 1]\)

Thus, we have:
\[ I = a(a - 1) - 2(a^2 - 1) \]

### Step 3: Simplify the expression
Now simplifying \(I\):
\[ I = a^2 - a - 2a^2 + 2 = -a^2 - a + 2 \]

### Step 4: Set up the inequality
According to the problem, we have:
\[ -a^2 - a + 2 \geq 6 - 5a \]

### Step 5: Rearrange the inequality
Rearranging gives:
\[ -a^2 - a + 5a + 2 - 6 \geq 0 \]
This simplifies to:
\[ -a^2 + 4a - 4 \geq 0 \]

### Step 6: Multiply through by -1
Multiplying through by -1 (and reversing the inequality):
\[ a^2 - 4a + 4 \leq 0 \]

### Step 7: Factor the quadratic
Factoring the quadratic:
\[ (a - 2)^2 \leq 0 \]

### Step 8: Solve the inequality
The expression \((a - 2)^2\) is non-negative and equals zero when:
\[ a - 2 = 0 \implies a = 2 \]

### Step 9: Consider the constraint
Given the condition \(a > 1\), the only solution that satisfies both the inequality and the condition is:
\[ a = 2 \]

### Conclusion
Thus, the value of \(a\) is:
\[ \boxed{2} \]