To compute the definite integral of $\int_0^1 e^{-3x} (3 - 2x^2) \, dx$, we can proceed with the following steps.

Step 1: Break the integral into two parts

We can split the integral as follows:

$\int_0^1 e^{-3x} (3 - 2x^2) \, dx = 3 \int_0^1 e^{-3x} \, dx - 2 \int_0^1 x^2 e^{-3x} \, dx$

Now, we will solve each of these integrals individually.

Step 2: Solve $\int_0^1 e^{-3x} \, dx$

The integral of $e^{-3x}$ is straightforward. Use the substitution $u = -3x$, which gives $du = -3 dx$:

$\int e^{-3x} \, dx = -\frac{1}{3} e^{-3x}$

Now, apply the limits of integration from 0 to 1:

$\int_0^1 e^{-3x} \, dx = \left[ -\frac{1}{3} e^{-3x} \right]_0^1 = -\frac{1}{3} e^{-3} + \frac{1}{3} e^0 = \frac{1}{3} (1 - e^{-3})$

Step 3: Solve $\int_0^1 x^2 e^{-3x} \, dx$

For this integral, we use integration by parts. Let:

• $u = x^2$, so $du = 2x \, dx$
• $dv = e^{-3x} \, dx$, so $v = -\frac{1}{3} e^{-3x}$

Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:

$\int x^2 e^{-3x} \, dx = -\frac{1}{3} x^2 e^{-3x} \Big|_0^1 + \frac{2}{3} \int x e^{-3x} \, dx$

Step 4: Solve $\int_0^1 x e^{-3x} \, dx$

We use integration by parts again. Let:

• $u = x$, so $du = dx$
• $dv = e^{-3x} \, dx$, so $v = -\frac{1}{3} e^{-3x}$

Applying the formula:

$\int x e^{-3x} \, dx = -\frac{1}{3} x e^{-3x} \Big|_0^1 + \frac{1}{3} \int e^{-3x} \, dx$

We already computed $\int e^{-3x} \, dx$ in Step 2. Therefore,

$\int x e^{-3x} \, dx = -\frac{1}{3} \cdot 1 \cdot e^{-3} + \frac{1}{9} (1 - e^{-3}) = \frac{1}{9} (1 - 4e^{-3})$

Step 5: Complete the calculation

Now, substitute this result back into the expression for $\int x^2 e^{-3x} \, dx$:

$\int x^2 e^{-3x} \, dx = -\frac{1}{3} e^{-3} + \frac{2}{9} (1 - 4e^{-3})$

Substituting everything back into the original integral expression, we get:

$\int_0^1 e^{-3x} (3 - 2x^2) \, dx = 3 \cdot \frac{1}{3} (1 - e^{-3}) - 2 \left( -\frac{1}{3} e^{-3} + \frac{2}{9} (1 - 4e^{-3}) \right)$

Simplifying this yields:

$= (1 - e^{-3}) + \frac{2}{9} (8e^{-3} - 2)$

The exact value of the definite integral is:

$= 1 - e^{-3} + \frac{16e^{-3} - 4}{9}$

Final Answer:

$\int_0^1 e^{-3x} (3 - 2x^2) \, dx = \frac{5}{9} + \frac{7}{9} e^{-3}$

Would you like me to break down any steps in more detail or solve similar problems?

Here are five related questions:

1. How would the integral change if the limits were from 0 to infinity?
2. Can you apply integration by parts to other functions involving exponentials and polynomials?
3. How can you evaluate $\int_0^1 x^3 e^{-3x} \, dx$?
4. What happens when you differentiate the result of this integral with respect to $x$?
5. Can this technique be generalized for higher powers of $x$?

Tip: Integration by parts is a powerful technique that often simplifies complex integrals involving products of functions.