Let `I_1=int_0^1e^(x^2)dx and I_2=int_0^(1)2x^(2)e^(x^2)dx` then the value of `I_1 +I_2` is equal to

Options

A. 1
B. 2
C. e
D. `e^2`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and find the sum \( I_1 + I_2 \).

1. **Define the integrals:**
\[ I_1 = \int_0^1 e^{x^2} \, dx \]
\[ I_2 = \int_0^1 2x^2 e^{x^2} \, dx \]

2. **Simplify \( I_2 \):**
We can rewrite \( I_2 \) as:
\[ I_2 = \int_0^1 2x e^{x^2} x \, dx \]
This suggests a substitution. Let \( t = x^2 \), then \( dt = 2x \, dx \) or \( dx = \frac{dt}{2x} \). When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = 1 \).

Thus, we can rewrite \( I_2 \):
\[ I_2 = \int_0^1 e^{t} \, dt \]

3. **Evaluate \( I_2 \):**
The integral \( \int e^t \, dt \) evaluates to \( e^t \). Therefore:
\[ I_2 = \left[ e^t \right]_0^1 = e^1 - e^0 = e - 1 \]

4. **Now we need to evaluate \( I_1 \):**
We will use integration by parts to evaluate \( I_1 \). Let:
\[ u = e^{x^2} \quad \text{and} \quad dv = dx \]
Then:
\[ du = 2x e^{x^2} \, dx \quad \text{and} \quad v = x \]

Applying integration by parts:
\[ I_1 = \left[ x e^{x^2} \right]_0^1 - \int_0^1 x (2x e^{x^2}) \, dx \]
The first term evaluates to:
\[ \left[ x e^{x^2} \right]_0^1 = 1 \cdot e^1 - 0 \cdot e^0 = e \]

The second term is:
\[ -\int_0^1 2x^2 e^{x^2} \, dx = -I_2 \]

Therefore:
\[ I_1 = e - I_2 \]

5. **Substituting \( I_2 \) into \( I_1 \):**
Since we found \( I_2 = e - 1 \):
\[ I_1 = e - (e - 1) = 1 \]

6. **Now calculate \( I_1 + I_2 \):**
\[ I_1 + I_2 = 1 + (e - 1) = e \]

Thus, the final result is:
\[ I_1 + I_2 = e \]