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

Options

A. 1: `(1)/(2e) (e-1)`
B. 2: `(1)/(2e) (ea +1)`
C. 3: `1/e (e -1)`
D. 4: `1/e (e +1)`

Answer

To solve the integral \( \int_0^1 x^2 e^{-x^2} \, dx \) given that \( \int_0^1 e^{-x^2} \, dx = a \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( u = -x^2 \). Then, differentiating both sides gives:
\[ du = -2x \, dx \quad \Rightarrow \quad dx = \frac{du}{-2x} \]

### Step 2: Change of Limits
When \( x = 0 \), \( u = 0 \). When \( x = 1 \), \( u = -1 \). Thus, the limits of integration change from \( x: 0 \to 1 \) to \( u: 0 \to -1 \).

### Step 3: Rewrite the Integral
Now, substituting into the integral:
\[ \int_0^1 x^2 e^{-x^2} \, dx = \int_0^{-1} x^2 e^{u} \left( \frac{du}{-2x} \right) \]
Notice that \( x^2 = -u \) (since \( u = -x^2 \)). Therefore, we can rewrite the integral:
\[ = \int_0^{-1} (-u) e^{u} \left( \frac{du}{-2\sqrt{-u}} \right) \]
However, we need to express \( x \) in terms of \( u \). Since \( u = -x^2 \), we have \( x = \sqrt{-u} \):
\[ = \int_0^{-1} \frac{u e^{u}}{-2\sqrt{-u}} \, du \]

### Step 4: Simplifying the Integral
We can factor out the constants:
\[ = -\frac{1}{2} \int_0^{-1} u e^{u} \frac{du}{\sqrt{-u}} \]
This integral is more complex, so we will revert to the original integral and use integration by parts instead.

### Step 5: Integration by Parts
Let \( v = e^{-x^2} \) and \( dv = -2x e^{-x^2} \, dx \). Then, we can write:
\[ \int_0^1 x^2 e^{-x^2} \, dx = -\frac{1}{2} \int_0^1 e^{-x^2} \, d(x^2) \]

### Step 6: Evaluating the Integral
Using the given information \( \int_0^1 e^{-x^2} \, dx = a \):
\[ = -\frac{1}{2} \left[ e^{-x^2} \right]_0^1 = -\frac{1}{2} \left( e^{-1} - e^{0} \right) \]
\[ = -\frac{1}{2} \left( \frac{1}{e} - 1 \right) = \frac{1}{2} \left( 1 - \frac{1}{e} \right) \]

### Final Result
Thus, the value of the integral \( \int_0^1 x^2 e^{-x^2} \, dx \) is:
\[ \frac{1}{2} \left( 1 - \frac{1}{e} \right) \]