The values of 'a' for which `int_0^(a) (3x^(2)+4x-5)dx lt a^(3)-2` are

Options

A. `(1)/(2) lt a lt 2`
B. `(1)/(2) le a le 2`
C. `a le (1)/(2)`
D. `a ge 2`

Answer

To solve the inequality \( \int_0^{a} (3x^2 + 4x - 5) \, dx < a^3 - 2 \), we will follow these steps:

### Step 1: Calculate the integral
We start by calculating the definite integral:

\[ \int_0^{a} (3x^2 + 4x - 5) \, dx \]

Using the power rule for integration, we find:

\[ \int (3x^2) \, dx = x^3, \quad \int (4x) \, dx = 2x^2, \quad \int (-5) \, dx = -5x \]

Thus, the integral becomes:

\[ \int_0^{a} (3x^2 + 4x - 5) \, dx = \left[ x^3 + 2x^2 - 5x \right]_0^{a} \]

### Step 2: Evaluate the integral at the limits
Now we evaluate the integral from 0 to \( a \):

\[ = \left( a^3 + 2a^2 - 5a \right) - \left( 0^3 + 2(0^2) - 5(0) \right) \]

This simplifies to:

\[ = a^3 + 2a^2 - 5a \]

### Step 3: Set up the inequality
We now set up the inequality based on the problem statement:

\[ a^3 + 2a^2 - 5a < a^3 - 2 \]

### Step 4: Simplify the inequality
Subtract \( a^3 \) from both sides:

\[ 2a^2 - 5a < -2 \]

Adding 2 to both sides gives:

\[ 2a^2 - 5a + 2 < 0 \]

### Step 5: Factor the quadratic
Next, we factor the quadratic expression:

\[ 2a^2 - 5a + 2 = 0 \]

To factor, we can use the middle term splitting method. We need two numbers that multiply to \( 2 \times 2 = 4 \) and add to \( -5 \). These numbers are \( -4 \) and \( -1 \):

\[ 2a^2 - 4a - a + 2 < 0 \]

Grouping gives:

\[ 2a(a - 2) - 1(a - 2) < 0 \]

Factoring out \( (a - 2) \):

\[ (2a - 1)(a - 2) < 0 \]

### Step 6: Find critical points
Now we find the critical points by setting each factor to zero:

1. \( 2a - 1 = 0 \) implies \( a = \frac{1}{2} \)
2. \( a - 2 = 0 \) implies \( a = 2 \)

### Step 7: Test intervals
We test the intervals determined by the critical points \( \frac{1}{2} \) and \( 2 \):

- For \( a < \frac{1}{2} \): Choose \( a = 0 \) → \( (2(0) - 1)(0 - 2) = (-1)(-2) = 2 > 0 \)
- For \( \frac{1}{2} < a < 2 \): Choose \( a = 1 \) → \( (2(1) - 1)(1 - 2) = (1)(-1) = -1 < 0 \)
- For \( a > 2 \): Choose \( a = 3 \) → \( (2(3) - 1)(3 - 2) = (5)(1) = 5 > 0 \)

### Step 8: Conclusion
The inequality \( (2a - 1)(a - 2) < 0 \) holds true in the interval:

\[ \frac{1}{2} < a < 2 \]

Thus, the values of \( a \) for which the original inequality holds are:

\[ \boxed{\left( \frac{1}{2}, 2 \right)} \]