find the value of b so that the line y=b divides the region bound by the graphs of the two functinos , into two regions of equal area. f(x)=9−x^2 and g(x)=0


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Question Number 173908 by ali009 last updated on 20/Jul/22

$f i n d t h e v a l u e o f b s o t h a t t h e l i n e y = b$ $d i v i d e s t h e r e g i o n b o u n d b y t h e g r a p h s o f$ $t h e t w o f u n c t i n o s, i n t o t w o r e g i o n s o f e q u a l$ $a r e a .$ $f (x) = 9 - x^{2} a n d g (x) = 0$
Commented by mr W last updated on 20/Jul/22

${(\frac{9 - b}{9})}^{\frac{3}{2}} = \frac{1}{2}$ $\Rightarrow b = 9 (1 - \frac{1}{\sqrt[3]{4}}) = \frac{(2 - \sqrt[3]{2}) \times 9}{2} \approx 3.33$
Commented by mr W last updated on 20/Jul/22

Commented by mr W last updated on 20/Jul/22

Commented by mr W last updated on 21/Jul/22

$\frac{b_{2}}{b_{1}} = {(\frac{a_{2}}{a_{1}})}^{2} \Rightarrow \frac{a_{2}}{a_{1}} = {(\frac{b_{2}}{b_{1}})}^{\frac{1}{2}}$ $A_{1} = \frac{2}{3} a_{1} b_{1}$ $A_{2} = \frac{2}{3} a_{2} b_{2}$ $\frac{A_{2}}{A_{1}} = \frac{a_{2} b_{2}}{a_{1} b_{1}} = (\frac{a_{2}}{a_{1}}) (\frac{b_{2}}{b_{1}}) = {(\frac{b_{2}}{b_{1}})}^{\frac{1}{2}} (\frac{b_{2}}{b_{1}}) = {(\frac{b_{2}}{b_{1}})}^{\frac{3}{2}}$
Commented by Erikyatusabes last updated on 21/Jul/22

${(\frac{9 - b}{9})}^{\frac{3}{2}} = \frac{1}{2}$ $\Rightarrow b = 9 (1 - \frac{1}{\sqrt[3]{4}}) = \frac{(2 - \sqrt[3]{2}) \times 9}{2} \approx 3.33$
Commented by Tawa11 last updated on 21/Jul/22

$Great sir$
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