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The area $A_1$ is bounded by $y^2 = 4x$, $x=1$, and the x-axis in the first quadrant. We can express $y$ as a function of $x$: $y = \sqrt{4x} = 2\sqrt{x}$. The area $A_1$ is given by the integral: $$A_1 = \int_{0}^{1} 2\sqrt{x} \, dx$$
$$A_1 = 2 \int_{0}^{1} x^{1/2} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} = 2 \cdot \frac{2}{3} \left[ x^{3/2} \right]_{0}^{1} = \frac{4}{3} (1^{3/2} - 0^{3/2}) = \frac{4}{3}$$
The area bounded by $y^2 = 4x$ and $x=4$ is actually the area from $x=0$ to $x=4$. However, the question asks for the area $A_2$ to be bounded by $y^2 = 4x$ and $x=4$. To find $A_2$, we need to integrate from $x=1$ to $x=4$: $$A_2 = \int_{1}^{4} 2\sqrt{x} \, dx$$
$$A_2 = 2 \int_{1}^{4} x^{1/2} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} = \frac{4}{3} \left[ x^{3/2} \right]_{1}^{4} = \frac{4}{3} (4^{3/2} - 1^{3/2}) = \frac{4}{3} (8 - 1) = \frac{4}{3} \cdot 7 = \frac{28}{3}$$
The ratio $A_1 : A_2$ is: $$\frac{A_1}{A_2} = \frac{\frac{4}{3}}{\frac{28}{3}} = \frac{4}{28} = \frac{1}{7}$$ Therefore, $A_1 : A_2 = 1:7$
Final Answer: 1:7
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