1. **State the problem:** Find the area under the curve $y = 2x^{\frac{3}{2}}$ between $x=4$ and $x=8$. 2. **Formula used:** The area under a curve $y=f(x)$ from $x=a$ to $x=b$ is given by the definite integral: $$\text{Area} = \int_a^b f(x)\,dx$$ 3. **Apply the formula:** Here, $f(x) = 2x^{\frac{3}{2}}$, $a=4$, and $b=8$. So, $$\text{Area} = \int_4^8 2x^{\frac{3}{2}}\,dx$$ 4. **Integrate the function:** Recall the power rule for integration: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$ Applying this, $$\int 2x^{\frac{3}{2}} dx = 2 \int x^{\frac{3}{2}} dx = 2 \cdot \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1} = 2 \cdot \frac{x^{\frac{5}{2}}}{\frac{5}{2}} = 2 \cdot \frac{2}{5} x^{\frac{5}{2}} = \frac{4}{5} x^{\frac{5}{2}}$$ 5. **Evaluate the definite integral:** $$\text{Area} = \left[ \frac{4}{5} x^{\frac{5}{2}} \right]_4^8 = \frac{4}{5} \left(8^{\frac{5}{2}} - 4^{\frac{5}{2}}\right)$$ 6. **Calculate powers:** - $8^{\frac{5}{2}} = (8^{\frac{1}{2}})^5 = (\sqrt{8})^5 = (2\sqrt{2})^5$ - $4^{\frac{5}{2}} = (4^{\frac{1}{2}})^5 = (2)^5 = 32$ Calculate $8^{\frac{5}{2}}$: $$ (2\sqrt{2})^5 = 2^5 \cdot (\sqrt{2})^5 = 32 \cdot 2^{\frac{5}{2}} = 32 \cdot 2^{2 + \frac{1}{2}} = 32 \cdot 2^2 \cdot 2^{\frac{1}{2}} = 32 \cdot 4 \cdot \sqrt{2} = 128 \sqrt{2}$$ 7. **Substitute back:** $$\text{Area} = \frac{4}{5} (128 \sqrt{2} - 32) = \frac{4}{5} \cdot 32 (4 \sqrt{2} - 1) = \frac{128}{5} (4 \sqrt{2} - 1)$$ 8. **Final answer:** $$\boxed{\text{Area} = \frac{128}{5} (4 \sqrt{2} - 1)}$$ This is the exact area under the curve between $x=4$ and $x=8$.