The value of `int_0^1e^(x^2-x)`dx` is (a) `<1` (b) `>1` (c) `> `e^(-1/4)` (d)` < e^(-1/4)`

Options

A. `lt1`
B. `gt1`
C. `gte^(-1/4)`
D. `lte^(-1/4)`

Explanation

To solve the integral \( \int_0^1 e^{x^2 - x} \, dx \) and determine its value in relation to the given options, we can follow these steps:

### Step 1: Analyze the function \( x^2 - x \)

We start by analyzing the expression \( x^2 - x \) over the interval \( [0, 1] \).

- At \( x = 0 \):
\[ x^2 - x = 0^2 - 0 = 0 \]

- At \( x = 1 \):
\[ x^2 - x = 1^2 - 1 = 0 \]

- To find the minimum value, we can differentiate:
\[ f(x) = x^2 - x \]
\[ f'(x) = 2x - 1 \]
Setting \( f'(x) = 0 \) gives \( x = \frac{1}{2} \).

- Evaluating at \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \]

Thus, the minimum value of \( x^2 - x \) on the interval \( [0, 1] \) is \( -\frac{1}{4} \) and the maximum value is \( 0 \).

### Step 2: Determine the bounds for \( e^{x^2 - x} \)

Since \( x^2 - x \) varies from \( -\frac{1}{4} \) to \( 0 \), we can find the corresponding values of \( e^{x^2 - x} \):

- At the minimum \( -\frac{1}{4} \):
\[ e^{-\frac{1}{4}} \]

- At the maximum \( 0 \):
\[ e^0 = 1 \]

Thus, \( e^{x^2 - x} \) varies from \( e^{-\frac{1}{4}} \) to \( 1 \).

### Step 3: Set up the integral

Now we can set up the integral:
\[ \int_0^1 e^{x^2 - x} \, dx \]

Using the bounds we found:
\[ e^{-\frac{1}{4}} \leq e^{x^2 - x} \leq 1 \]

### Step 4: Integrate and apply the bounds

Integrating over the interval \( [0, 1] \):
\[ \int_0^1 e^{-\frac{1}{4}} \, dx \leq \int_0^1 e^{x^2 - x} \, dx \leq \int_0^1 1 \, dx \]

Calculating the integrals:
- The left side:
\[ \int_0^1 e^{-\frac{1}{4}} \, dx = e^{-\frac{1}{4}} \cdot (1 - 0) = e^{-\frac{1}{4}} \]

- The right side:
\[ \int_0^1 1 \, dx = 1 \]

Thus, we have:
\[ e^{-\frac{1}{4}} \leq \int_0^1 e^{x^2 - x} \, dx < 1 \]

### Conclusion

From the inequalities we derived:
- The integral \( \int_0^1 e^{x^2 - x} \, dx \) is greater than \( e^{-\frac{1}{4}} \) and less than \( 1 \).

Therefore, the correct options are:
- (c) \( > e^{-\frac{1}{4}} \)
- (a) \( < 1 \)