This integral problem is a definite integral from 0 to 1. Let me solve it step by step:

Problem:

$\int_0^1 (1 - 2x - 3x^2) \, dx$

Step 1: Break it into simpler terms

The integral can be broken into individual terms: $\int_0^1 (1 - 2x - 3x^2) \, dx = \int_0^1 1 \, dx - \int_0^1 2x \, dx - \int_0^1 3x^2 \, dx$

Step 2: Solve each term

First Term: $\int_0^1 1 \, dx$

$\int 1 \, dx = x \quad \text{(Evaluate from 0 to 1)}.$ $\text{At } x = 1, \, x = 1; \quad \text{at } x = 0, \, x = 0.$ So, $\int_0^1 1 \, dx = 1 - 0 = 1$.

Second Term: $\int_0^1 2x \, dx$

$\int 2x \, dx = x^2 \quad \text{(Evaluate from 0 to 1)}.$ $\text{At } x = 1, \, x^2 = 1^2 = 1; \quad \text{at } x = 0, \, x^2 = 0^2 = 0.$ So, $\int_0^1 2x \, dx = 1 - 0 = 1$.

Third Term: $\int_0^1 3x^2 \, dx$

$\int 3x^2 \, dx = x^3 \quad \text{(Evaluate from 0 to 1)}.$ $\text{At } x = 1, \, x^3 = 1^3 = 1; \quad \text{at } x = 0, \, x^3 = 0^3 = 0.$ So, $\int_0^1 3x^2 \, dx = 1 - 0 = 1$.

Step 3: Combine the results

$\int_0^1 (1 - 2x - 3x^2) \, dx = 1 - 1 - 1 = -1.$

Final Answer:

$\boxed{-1}$

Would you like further clarification or a different approach to solving this problem?

Related Questions:

1. How would the solution change if the limits of integration were $[1, 2]$ instead of $[0, 1]$?
2. How does the power rule for integration apply in this context?
3. What are the graphical interpretations of this integral?
4. How would you compute the integral if the polynomial had higher powers?
5. Can the Fundamental Theorem of Calculus be demonstrated using this example?

Tip:

When solving definite integrals, always remember to apply the limits carefully to avoid sign errors.