1. **State the problem:** Solve the integral equation $$\int_2^x \left(\frac{1}{2}t - 2\right) dt = 3$$ for $x$. 2. **Recall the antiderivative:** The antiderivative of the integrand $\frac{1}{2}t - 2$ is given as $$\frac{1}{4}t^2 - 2t + C$$ where $C$ is a constant. 3. **Evaluate the definite integral:** Using the Fundamental Theorem of Calculus, $$\int_2^x \left(\frac{1}{2}t - 2\right) dt = \left(\frac{1}{4}x^2 - 2x\right) - \left(\frac{1}{4}(2)^2 - 2(2)\right)$$ Calculate the lower limit term: $$\frac{1}{4} \times 4 - 4 = 1 - 4 = -3$$ So the integral becomes: $$\frac{1}{4}x^2 - 2x - (-3) = \frac{1}{4}x^2 - 2x + 3$$ 4. **Set the integral equal to 3 and solve for $x$:** $$\frac{1}{4}x^2 - 2x + 3 = 3$$ Subtract 3 from both sides: $$\frac{1}{4}x^2 - 2x + 3 - 3 = 0$$ $$\frac{1}{4}x^2 - 2x = 0$$ 5. **Factor the equation:** $$x\left(\frac{1}{4}x - 2\right) = 0$$ 6. **Solve each factor:** - $x = 0$ - $$\frac{1}{4}x - 2 = 0 \implies \frac{1}{4}x = 2 \implies x = 8$$ 7. **Final answer:** The solutions are $$x = 0$$ or $$x = 8$$. These values satisfy the integral equation given the limits and the integral value.