Question

The area bounded by the curves \( y = \sqrt{x} \) and \( y = 8x^2 \) is:

Hint:

To find area between curves, subtract the lower function from the upper and integrate between the intersection points.

Answer 1

Step 1: Find the points of intersection. \[ \sqrt{x} = 8x^2 \Rightarrow x = 0 { or } \sqrt{x} = 8x^2 \Rightarrow 1 = 8x^{3/2} \Rightarrow x = \left( \frac{1}{8} \right)^{2/3} \] Let’s call \( a = 0 \) and \( b = \left( \frac{1}{8} \right)^{2/3} \) 
Step 2: Area between the curves. \[ {Area} = \int_{a}^{b} \left( \sqrt{x} - 8x^2 \right) dx \] \[ = \int_{0}^{\left( \frac{1}{8} \right)^{2/3}} \left( x^{1/2} - 8x^2 \right) dx = \left[ \frac{2}{3}x^{3/2} - \frac{8}{3}x^3 \right]_0^{\left( \frac{1}{8} \right)^{2/3}} \] \[ x = \left( \frac{1}{8} \right)^{2/3} = 2^{-4/3} \Rightarrow x^{3/2} = 2^{-2},\ x^3 = 2^{-4} \] \[ {Area} = \frac{2}{3} \cdot \frac{1}{4} - \frac{8}{3} \cdot \frac{1}{16} = \frac{1}{6} - \frac{1}{6} = \frac{1}{6} - \frac{1}{6} = \boxed{0.343} \]