If `int _(0)^(1) e ^(-x ^(2)) dx =0,` then `int _(0)^(1) x ^(2)e ^(-x ^(2))` dx is equal to

Options

A. `(1)/(2e) (e-1)`
B. `(1)/(2e) (e +1)`
C. `1/e (e -1)`
D. `1/e (e +1)`

Answer

To solve the problem, we need to evaluate the integral

\[ \int_{0}^{1} x^2 e^{-x^2} \, dx. \]

Given that

\[ \int_{0}^{1} e^{-x^2} \, dx = 0, \]

we will use substitution to find the value of the integral.

### Step 1: Substitution
Let \( u = e^{-x^2} \). Then, we differentiate to find \( du \):

\[ du = -2x e^{-x^2} \, dx \quad \Rightarrow \quad dx = -\frac{du}{2x e^{-x^2}}. \]

### Step 2: Express \( x^2 e^{-x^2} \) in terms of \( u \)
From our substitution, we can express \( x^2 e^{-x^2} \) as follows:

\[ x^2 = -\frac{u}{2} \quad \Rightarrow \quad e^{-x^2} = u. \]

### Step 3: Change the limits of integration
When \( x = 0 \), \( u = e^{0} = 1 \).
When \( x = 1 \), \( u = e^{-1} \).

Thus, the limits of integration change from \( 0 \) to \( 1 \) in terms of \( x \) to \( 1 \) to \( e^{-1} \) in terms of \( u \).

### Step 4: Rewrite the integral
Now we rewrite the integral:

\[ \int_{0}^{1} x^2 e^{-x^2} \, dx = \int_{1}^{e^{-1}} -\frac{1}{2} u \, du. \]

### Step 5: Integrate
Now we can integrate:

\[ -\frac{1}{2} \int_{1}^{e^{-1}} u \, du = -\frac{1}{2} \left[ \frac{u^2}{2} \right]_{1}^{e^{-1}} = -\frac{1}{2} \left( \frac{(e^{-1})^2}{2} - \frac{1^2}{2} \right). \]

### Step 6: Evaluate the definite integral
Calculating the above expression gives:

\[ -\frac{1}{2} \left( \frac{e^{-2}}{2} - \frac{1}{2} \right) = -\frac{1}{2} \left( \frac{e^{-2} - 1}{2} \right) = -\frac{1}{4} (e^{-2} - 1). \]

### Step 7: Final expression
Thus, the value of the integral is:

\[ \int_{0}^{1} x^2 e^{-x^2} \, dx = \frac{1 - e^{-2}}{4}. \]

### Summary
The final result is:

\[ \int_{0}^{1} x^2 e^{-x^2} \, dx = \frac{1 - e^{-2}}{4}. \]