The problem 1.e is to solve the integral:

$\int_0^1 \int_0^x y e^x \, dy \, dx$

Step-by-step solution:

Step 1: Break down the double integral

The given double integral can be written as:

$\int_0^1 \left( \int_0^x y e^x \, dy \right) dx$

In this form, the inner integral is with respect to $y$, while the outer integral is with respect to $x$. We will first solve the inner integral.

Step 2: Solve the inner integral

The inner integral is:

$\int_0^x y e^x \, dy$

Note that $e^x$ is constant with respect to $y$, so it can be factored out of the integral:

$e^x \int_0^x y \, dy$

Now solve $\int_0^x y \, dy$. The antiderivative of $y$ is:

$\frac{y^2}{2}$

Now, evaluate this at the limits $0$ and $x$:

$\int_0^x y \, dy = \left[ \frac{y^2}{2} \right]_0^x = \frac{x^2}{2}$

So the inner integral becomes:

$e^x \cdot \frac{x^2}{2} = \frac{x^2 e^x}{2}$

Step 3: Solve the outer integral

Now we substitute this result into the outer integral:

$\int_0^1 \frac{x^2 e^x}{2} \, dx$

Factor out the constant $\frac{1}{2}$:

$\frac{1}{2} \int_0^1 x^2 e^x \, dx$

We now need to compute $\int_0^1 x^2 e^x \, dx$. This can be solved using integration by parts.

Step 4: Integration by parts

We will use integration by parts, where:

$u = x^2 \quad \text{and} \quad dv = e^x \, dx$

First, differentiate and integrate:

$du = 2x \, dx \quad \text{and} \quad v = e^x$

Now apply the integration by parts formula:

$\int u \, dv = uv - \int v \, du$

Substitute $u = x^2$, $du = 2x \, dx$, $v = e^x$, and $dv = e^x \, dx$:

$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx$

Now, we need to compute $\int 2x e^x \, dx$, again using integration by parts.

Let $u = 2x$ and $dv = e^x \, dx$. Then:

$du = 2 \, dx \quad \text{and} \quad v = e^x$

So:

$\int 2x e^x \, dx = 2x e^x - \int 2 e^x \, dx = 2x e^x - 2 e^x$

Thus:

$\int x^2 e^x \, dx = x^2 e^x - (2x e^x - 2 e^x)$

Simplify:

$\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2 e^x = e^x (x^2 - 2x + 2)$

Step 5: Evaluate the definite integral

Now, we evaluate this from 0 to 1:

$\left[ e^x (x^2 - 2x + 2) \right]_0^1$

At $x = 1$:

$e^1 (1^2 - 2(1) + 2) = e (1 - 2 + 2) = e$

At $x = 0$:

$e^0 (0^2 - 2(0) + 2) = 1 \cdot 2 = 2$

So the result of the definite integral is:

$e - 2$

Step 6: Final answer

Now multiply this by $\frac{1}{2}$ to get the final result:

$\frac{1}{2} (e - 2)$

Thus, the solution to the integral is:

$\boxed{\frac{e - 2}{2}}$

---

Would you like further clarification on any step or more details on integration by parts?

Related Questions:

1. What is the importance of integration by parts in solving integrals?
2. Can you show another example where we apply double integration?
3. What happens when the limits of integration are changed?
4. How can we solve integrals involving trigonometric functions?
5. How would we compute integrals that contain multiple variables without using substitution?

Tip:

When solving double integrals, always simplify the inner integral first to reduce complexity for the outer integral.