The problem in the image is:

$\int_{2}^{a} \left( x^3 - 16x \right) dx = -36$

We need to solve for $a$.

Solution

1. Integrate the function $x^3 - 16x$ with respect to $x$: $\int (x^3 - 16x) \, dx = \frac{x^4}{4} - 8x^2 + C$

2. Apply the limits of integration (from 2 to $a$): $\left[ \frac{x^4}{4} - 8x^2 \right]_{2}^{a} = -36$

3. Expand the expression: $\left( \frac{a^4}{4} - 8a^2 \right) - \left( \frac{2^4}{4} - 8 \cdot 2^2 \right) = -36$

4. Calculate the constant part (evaluating the expression at $x = 2$): $\frac{2^4}{4} - 8 \cdot 2^2 = \frac{16}{4} - 8 \cdot 4 = 4 - 32 = -28$

   So, the equation simplifies to: $\frac{a^4}{4} - 8a^2 + 28 = -36$

5. Isolate terms involving $a$: $\frac{a^4}{4} - 8a^2 = -64$

6. Multiply through by 4 to clear the fraction: $a^4 - 32a^2 = -256$

7. Rearrange the equation: $a^4 - 32a^2 + 256 = 0$

8. Substitute $u = a^2$ (a common technique for reducing powers): $u^2 - 32u + 256 = 0$

9. Solve this quadratic equation in $u$: $u = \frac{32 \pm \sqrt{32^2 - 4 \cdot 256}}{2}$ $u = \frac{32 \pm \sqrt{1024 - 1024}}{2} = \frac{32 \pm 0}{2} = 16$

   So, $u = 16$, which means $a^2 = 16$.

10. Take the square root to find $a$: $a = \pm 4$

Since the integral's bounds require $a > 2$, we choose $a = 4$.

Final Answer

$a = 4$

Would you like more details on any step or have any questions?

Here are 5 related questions to further explore this topic:

1. How do we determine the direction of the integral's result in terms of positive or negative values?
2. What if the integral's answer were positive—how would that change the steps?
3. Can we apply the same technique to solve integrals with different limits or integrands?
4. Why is substitution helpful in solving high-degree polynomial equations?
5. How can we verify the solution by plugging $a = 4$ back into the integral?

Tip: When faced with high powers in polynomials, consider substitution to simplify the equation.