Find-the-area-of-the-region-enclosed-by-the-line-5y-x-6-and-the-curve-y-x-

Question Number 72813 by Learner-123 last updated on 03/Nov/19

F i n d t h e a r e a o f t h e r e g i o n e n c l o s e d

b y t h e l i n e 5 y = x + 6 a n d t h e c u r v e

y = \sqrt{∣ x ∣} .

Commented by ajfour last updated on 03/Nov/19

Commented by ajfour last updated on 03/Nov/19

5 y = x + 6

y = \sqrt{∣ x ∣}

\Rightarrow f o r i n t e r s e c t i o n p o i n t s

{(x + 6)}^{2} = 25 ∣ x ∣

\Rightarrow x^{2} + 12 x - 25 ∣ x ∣ + 36 = 0

f o r x > 0

x^{2} - 13 x + 36 = 0

x = \frac{13}{2} \pm \sqrt{\frac{169}{4} - 36} = \frac{13}{2} \pm \frac{5}{2}

x_{B} = 4, x_{C} = 9

f o r x < 0

x^{2} + 37 x + 36 = 0

(x + 36) (x + 1) = 0

x_{A} = - 1

A r e a A = A_{1} + A_{2} + A_{3}

A_{1} = \int_{- 1}^{0} (\frac{x + 6}{5} - \sqrt{- x}) d x

= {[\frac{x^{2} + 12 x}{10} - \frac{2}{3} {(- x)}^{3 / 2}]}_{- 1}^{0}

= \frac{2}{3} + \frac{11}{10} = \frac{53}{30}

A_{2} = \int_{0}^{4} (\frac{x + 6}{5} - \sqrt{x}) d x

= {[\frac{x^{2} + 12 x}{10} - \frac{2}{3} x^{3 / 2}]}_{0}^{4}

= \frac{64}{10} - \frac{16}{3} = \frac{16}{15}

A_{3} = \int_{4}^{9} (\sqrt{x} - \frac{x + 6}{5}) d x

= \frac{2}{3} x^{3 / 2} - \frac{x^{2} + 12 x}{10}

= \frac{2}{3} (27 - 8) - \frac{1}{10} (81 + 108 - 16 - 48)

= \frac{38}{3} - \frac{25}{2} = \frac{1}{6}

A = \frac{53}{30} + \frac{32}{30} + \frac{5}{30} = 3 s q . u n i t s .

Commented by Learner-123 last updated on 03/Nov/19

t h a n k s s i r .