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-

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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