The given problem involves solving the integral:

$\int_{-a}^{a} (x^3 + 6x^2 + 8x + a) dx = (2a + 1)^2$

Step 1: Simplify the left-hand side (the integral)

We need to calculate the definite integral from $-a$ to $a$.

Breaking it down term by term:

1. $x^3$: $\int_{-a}^{a} x^3 dx = 0$ This is because $x^3$ is an odd function, and the integral of any odd function over a symmetric interval $[-a, a]$ is zero.

2. $6x^2$: $\int_{-a}^{a} 6x^2 dx = 6 \cdot \int_{-a}^{a} x^2 dx = 6 \cdot \left[ \frac{x^3}{3} \right]_{-a}^{a} = 6 \cdot \frac{a^3}{3} = 2a^3$

3. $8x$: $\int_{-a}^{a} 8x dx = 0$ Like $x^3$, $8x$ is an odd function, so its integral over a symmetric interval is zero.

4. $a$: $\int_{-a}^{a} a dx = a \cdot \int_{-a}^{a} dx = a \cdot (2a) = 2a^2$

Thus, the integral becomes: $2a^3 + 2a^2$

Step 2: Set up the equation

Now, we equate the result of the integral to $(2a + 1)^2$: $2a^3 + 2a^2 = (2a + 1)^2$

Expanding the right-hand side: $(2a + 1)^2 = 4a^2 + 4a + 1$

Now, the equation is: $2a^3 + 2a^2 = 4a^2 + 4a + 1$

Step 3: Simplify the equation

Rearrange everything to one side: $2a^3 + 2a^2 - 4a^2 - 4a - 1 = 0$ $2a^3 - 2a^2 - 4a - 1 = 0$

Step 4: Solve the cubic equation

We now solve the equation: $2a^3 - 2a^2 - 4a - 1 = 0$

Dividing the entire equation by 2 simplifies it: $a^3 - a^2 - 2a - \frac{1}{2} = 0$

Now we need to find the value of $a$ that satisfies this equation. The possible options are:

• $-1$
• $-\frac{1}{2}$
• $0$
• $\frac{1}{2}$
• $1$

Testing $a = -1$: $(-1)^3 - (-1)^2 - 2(-1) - \frac{1}{2} = -1 - 1 + 2 - \frac{1}{2} = 0$

Therefore, $a = -1$ satisfies the equation.

Answer:

The value of $a$ is $\boxed{-1}$.

Would you like further details or explanations on any step? Here are some related questions for practice:

1. What is the general property of odd and even functions in integrals over symmetric limits?
2. How do you expand the square of a binomial expression like $(2a + 1)^2$?
3. What other techniques can be used to solve cubic equations?
4. How can definite integrals be applied in physical problems like calculating areas or volumes?
5. What is the importance of simplifying equations before solving them?

Tip: Always check for symmetry in an integral before evaluating—it can save time!