1. We are asked to find the area between the curve and the x-axis on the interval $[0,b]$ using definite integrals. 2. The formula for the area under a curve $y=f(x)$ from $x=a$ to $x=b$ is: $$\text{Area} = \int_a^b f(x)\,dx$$ 3. We will apply this formula to each function given. --- **Problem 51: $y=3x^2$** 4. Set up the integral: $$\int_0^b 3x^2\,dx$$ 5. Integrate: $$\int 3x^2\,dx = 3 \cdot \frac{x^3}{3} = x^3$$ 6. Evaluate from 0 to $b$: $$x^3 \Big|_0^b = b^3 - 0 = b^3$$ 7. So, the area is $b^3$. --- **Problem 52: $y=\pi x^2$** 8. Set up the integral: $$\int_0^b \pi x^2\,dx$$ 9. Integrate: $$\int \pi x^2\,dx = \pi \cdot \frac{x^3}{3} = \frac{\pi x^3}{3}$$ 10. Evaluate from 0 to $b$: $$\frac{\pi b^3}{3} - 0 = \frac{\pi b^3}{3}$$ 11. So, the area is $\frac{\pi b^3}{3}$. --- **Problem 53: $y=2x$** 12. Set up the integral: $$\int_0^b 2x\,dx$$ 13. Integrate: $$\int 2x\,dx = 2 \cdot \frac{x^2}{2} = x^2$$ 14. Evaluate from 0 to $b$: $$b^2 - 0 = b^2$$ 15. So, the area is $b^2$. --- **Problem 54: $y=\frac{x}{2} + 1$** 16. Set up the integral: $$\int_0^b \left(\frac{x}{2} + 1\right) dx$$ 17. Integrate term-by-term: $$\int \frac{x}{2} dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}$$ $$\int 1 dx = x$$ 18. So, $$\int_0^b \left(\frac{x}{2} + 1\right) dx = \left(\frac{x^2}{4} + x\right) \Big|_0^b = \frac{b^2}{4} + b - 0 = \frac{b^2}{4} + b$$ 19. So, the area is $\frac{b^2}{4} + b$. --- **Summary:** - Area for $y=3x^2$ is $b^3$ - Area for $y=\pi x^2$ is $\frac{\pi b^3}{3}$ - Area for $y=2x$ is $b^2$ - Area for $y=\frac{x}{2} + 1$ is $\frac{b^2}{4} + b$