If `f(x) = int_(x^2)^(x^2 +1) e^(-t^2)`, the interval in which f (x) is increasing is

Options

A. `(0,oo)`
B. `(-oo,0)`
C. `[-2,2]`
D. no where

Answer

To determine the interval in which the function \( f(x) = \int_{x^2}^{x^2 + 1} e^{-t^2} \, dt \) is increasing, we need to follow these steps:

### Step 1: Find the derivative of \( f(x) \)
Using the Leibniz rule for differentiation under the integral sign, we can differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \int_{x^2}^{x^2 + 1} e^{-t^2} \, dt \right) \]

According to the Leibniz rule, we have:

\[ f'(x) = e^{-(x^2 + 1)^2} \cdot \frac{d}{dx}(x^2 + 1) - e^{-x^4} \cdot \frac{d}{dx}(x^2) \]

Calculating the derivatives:

\[ \frac{d}{dx}(x^2 + 1) = 2x \quad \text{and} \quad \frac{d}{dx}(x^2) = 2x \]

Thus, substituting back, we get:

\[ f'(x) = e^{-(x^2 + 1)^2} \cdot 2x - e^{-x^4} \cdot 2x \]

Factoring out \( 2x \):

\[ f'(x) = 2x \left( e^{-(x^2 + 1)^2} - e^{-x^4} \right) \]

### Step 2: Determine when \( f'(x) > 0 \)
For \( f(x) \) to be increasing, we need \( f'(x) > 0 \):

\[ 2x \left( e^{-(x^2 + 1)^2} - e^{-x^4} \right) > 0 \]

This inequality holds when both factors are either positive or negative.

1. **Case 1**: \( 2x > 0 \) implies \( x > 0 \)
2. **Case 2**: \( e^{-(x^2 + 1)^2} - e^{-x^4} > 0 \)

### Step 3: Analyze the second factor
We need to analyze when:

\[ e^{-(x^2 + 1)^2} > e^{-x^4} \]

Taking the natural logarithm (which is a monotonically increasing function), we can simplify this to:

\[ -(x^2 + 1)^2 > -x^4 \]

This simplifies to:

\[ (x^2 + 1)^2 < x^4 \]

Expanding and rearranging gives:

\[ x^4 + 2x^2 + 1 < x^4 \]

This simplifies to:

\[ 2x^2 + 1 < 0 \]

Since \( 2x^2 + 1 \) is always positive for all real \( x \), this condition cannot hold. Therefore, we need to consider the other case.

### Step 4: Analyze when \( x < 0 \)
If \( x < 0 \), then \( 2x < 0 \). In this case, we need:

\[ e^{-(x^2 + 1)^2} - e^{-x^4} < 0 \]

This means:

\[ e^{-(x^2 + 1)^2} < e^{-x^4} \]

This inequality holds true since \( -(x^2 + 1)^2 < -x^4 \) for \( x < 0 \).

### Conclusion
Thus, \( f'(x) > 0 \) when \( x < 0 \). Therefore, the function \( f(x) \) is increasing on the interval:

\[ (-\infty, 0) \]