1. **Problem statement:** Calculate the volume of the solid formed by rotating the function $f(x) = 2x^2 + 5x + 2$ over the interval $[0,4]$ around the x-axis. 2. **Formula used:** The volume $V$ of a solid of revolution around the x-axis is given by the disk method: $$V = \pi \int_a^b [f(x)]^2 \, dx$$ where $a=0$ and $b=4$. 3. **Calculate $[f(x)]^2$:** $$f(x) = 2x^2 + 5x + 2$$ $$[f(x)]^2 = (2x^2 + 5x + 2)^2$$ Expand: $$= (2x^2)^2 + 2 \cdot 2x^2 \cdot 5x + 2 \cdot 2x^2 \cdot 2 + (5x)^2 + 2 \cdot 5x \cdot 2 + 2^2$$ $$= 4x^4 + 20x^3 + 8x^2 + 25x^2 + 20x + 4$$ Combine like terms: $$= 4x^4 + 20x^3 + (8x^2 + 25x^2) + 20x + 4 = 4x^4 + 20x^3 + 33x^2 + 20x + 4$$ 4. **Set up the integral:** $$V = \pi \int_0^4 (4x^4 + 20x^3 + 33x^2 + 20x + 4) \, dx$$ 5. **Integrate term-by-term:** $$\int 4x^4 \, dx = \frac{4}{5}x^5$$ $$\int 20x^3 \, dx = 5x^4$$ $$\int 33x^2 \, dx = 11x^3$$ $$\int 20x \, dx = 10x^2$$ $$\int 4 \, dx = 4x$$ 6. **Evaluate the definite integral:** $$V = \pi \left[ \frac{4}{5}x^5 + 5x^4 + 11x^3 + 10x^2 + 4x \right]_0^4$$ Calculate each term at $x=4$: $$\frac{4}{5} \cdot 4^5 = \frac{4}{5} \cdot 1024 = \frac{4096}{5} = 819.2$$ $$5 \cdot 4^4 = 5 \cdot 256 = 1280$$ $$11 \cdot 4^3 = 11 \cdot 64 = 704$$ $$10 \cdot 4^2 = 10 \cdot 16 = 160$$ $$4 \cdot 4 = 16$$ Sum: $$819.2 + 1280 + 704 + 160 + 16 = 2979.2$$ At $x=0$, all terms are zero. 7. **Final volume:** $$V = \pi \times 2979.2 \approx 9360.5$$ **Answer:** The volume of the bowl formed by rotating $f(x)$ around the x-axis from 0 to 4 is approximately $9360.5$ cubic units.