The value of integral $\underset{0}{\overset{2}{\int }}\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\mathrm{dydx}$ is

Double Integral Evaluation Explained

This question asks us to find the value of the given double integral: $$\underset{0}{\overset{2}{\int }}\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\mathrm{dydx}$$ To solve this definite double integral, we need to integrate it step-by-step, starting with the inner integral and then moving to the outer integral.

Step-by-Step Integration Process

We will evaluate the integral by first integrating with respect to $y$ (the inner integral) and then with respect to $x$ (the outer integral).

**1. Integrate the Inner Integral with respect to $y$:**

The inner integral is $\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\mathrm{dy}$. We can rewrite ${\mathrm{e}}^{\mathrm{x}+\mathrm{y}}$ as ${\mathrm{e}}^{\mathrm{x}}\cdot {\mathrm{e}}^{\mathrm{y}}$. Since we are integrating with respect to $y$, ${\mathrm{e}}^{\mathrm{x}}$ is treated as a constant.

So, the inner integral becomes: $$\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{x}}\cdot {\mathrm{e}}^{\mathrm{y}}\mathrm{dy}={\mathrm{e}}^{\mathrm{x}}\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{y}}\mathrm{dy}$$ The integral of ${\mathrm{e}}^{\mathrm{y}}$ with respect to $y$ is simply ${\mathrm{e}}^{\mathrm{y}}$. Now, apply the limits of integration from $0$ to $x$:
$${\mathrm{e}}^{\mathrm{x}}{\left[{\mathrm{e}}^{\mathrm{y}}\right]}_0^{\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}\left({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^0\right)$$
Since ${\mathrm{e}}^0=1$, we have:
$${\mathrm{e}}^{\mathrm{x}}\left({\mathrm{e}}^{\mathrm{x}}-1\right)={\mathrm{e}}^{\mathrm{x}}\cdot {\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\cdot 1$$

$$={\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}$$
This is the result of the inner integral.

**2. Integrate the Outer Integral with respect to $x$:**

Now, substitute the result of the inner integral into the outer integral: $$\underset{0}{\overset{2}{\int }}\left({\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\right)\mathrm{dx}$$ We can integrate each term separately: $$\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}-\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$$

- **For the first term, $\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}$:** Let $u=2x$, then $du=2dx$, which means $dx=\frac{1}{2}du$. So, $\int {\mathrm{e}}^u\frac{1}{2}\mathrm{du}=\frac{1}{2}{\mathrm{e}}^u=\frac{1}{2}{\mathrm{e}}^{2\mathrm{x}}$.
- **For the second term, $\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$:** The integral is simply ${\mathrm{e}}^{\mathrm{x}}$.

Combining these, we get:
$${\left[\frac{1}{2}{\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\right]}_0^2$$
Now, we apply the limits of integration from $0$ to $2$. Substitute the upper limit ($x=2$):
$$\left(\frac{1}{2}{\mathrm{e}}^{2(2)}-{\mathrm{e}}^2\right)=\left(\frac{1}{2}{\mathrm{e}}^4-{\mathrm{e}}^2\right)$$
Substitute the lower limit ($x=0$):
$$\left(\frac{1}{2}{\mathrm{e}}^{2(0)}-{\mathrm{e}}^0\right)=\left(\frac{1}{2}{\mathrm{e}}^0-1\right)=\left(\frac{1}{2}-1\right)=-\frac{1}{2}$$
Now, subtract the value at the lower limit from the value at the upper limit:
$$\left(\frac{1}{2}{\mathrm{e}}^4-{\mathrm{e}}^2\right)-\left(-\frac{1}{2}\right)$$

$$=\frac{1}{2}{\mathrm{e}}^4-{\mathrm{e}}^2+\frac{1}{2}$$

Simplifying the Result

Let's factor out $\frac{1}{2}$ from the expression:
$$\frac{1}{2}\left({\mathrm{e}}^4-2{\mathrm{e}}^2+1\right)$$
Recognize that the term inside the parenthesis, ${\mathrm{e}}^4-2{\mathrm{e}}^2+1$, is a perfect square trinomial of the form $a^2-2ab+b^2=(a-b{)}^2$, where $a={\mathrm{e}}^2$ and $b=1$. So, ${\mathrm{e}}^4-2{\mathrm{e}}^2+1={\left({\mathrm{e}}^2\right)}^2-2\left({\mathrm{e}}^2\right)\left(1\right)+{\left(1\right)}^2={\left({\mathrm{e}}^2-1\right)}^2$.

Therefore, the final value of the double integral is:
$$\frac{1}{2}{\left({\mathrm{e}}^2-1\right)}^2$$

This matches option 2.