Answer to Question #236119 in Calculus for Sayem

Question

Find the number b such that the line y=b divides the region bounded by the curves y=x² and y=4 into two regions with equal area.

Answer 1

Find the area of the region bounded by the curves y=x² and y=4 

$$x^2=4$$

$$x_1=-2, x_2=2$$

$$A_1=\displaystyle\int_{-2}^{2}(4-x^2)dx=\big[4x-\dfrac{x^3}{3}\big]\begin{matrix} 2 \\ -2 \end{matrix}$$

$$=8-\dfrac{8}{3}-(-8+\dfrac{8}{3})=\dfrac{32}{3} ({units}^2)$$

Find the area of the region bounded by the curves y=x² and y=b 

$$x^2=b, b>0$$

$$x_1=-\sqrt{b}, x_2=\sqrt{b}$$

$$A_2=\displaystyle\int_{-\sqrt{b}}^{\sqrt{b}}(b-x^2)dx=\big[bx-\dfrac{x^3}{3}\big]\begin{matrix} \sqrt{b} \\ -\sqrt{b} \end{matrix}$$

$$=b\sqrt{b}-\dfrac{b\sqrt{b}}{3}-(-b\sqrt{b}+\dfrac{b\sqrt{b}}{3})=\dfrac{4b\sqrt{b}}{3} ({units}^2)$$

The line y=b divides the region bounded by the curves y=x² and y=4 into two regions with equal area

$$A_1=2A_2$$

$$\dfrac{32}{3}=2(\dfrac{4b\sqrt{b}}{3})$$

$$b\sqrt{b}=4$$

$$b^3=16$$

$$b=2\sqrt[3]{2}$$