Please-help-Find-the-area-bounded-by-y-x-2-x-4-x-0-y-0-and-x-3-

Question Number 32756 by NECx last updated on 01/Apr/18

P l e a s e h e l p

F i n d t h e a r e a b o u n d e d b y

y (x + 2) = x^{4}, x = 0, y = 0, a n d x = 3

Answered by MJS last updated on 02/Apr/18

y (x) = {(x - 2)}^{4}

area between x - axis (y = 0) and

y (x) in [0; 3], no change of sign

\underset{0}{\int^{3}} {(x - 2)}^{4} d x = \frac{1}{5} {(x - 2)}^{5} \underset{0}{\overset{3}{∣}} =

= \frac{1}{5} (1 - {(- 2)}^{5}) = \frac{33}{5}

Commented by NECx last updated on 02/Apr/18

p l e a s e I d o n t u n d e r s t a n d h o w y o u

g o t y (x) = {(x - 2)}^{4}

\dots . . T h e r e s n o t h i n g o f s u c h i n t h e

q u e s t i o n .

Commented by MJS last updated on 02/Apr/18

I understood that {it}^{'} s y (x + 2), not

y \times (x + 2) . If this is true,

y (x + 2) = x^{4} \Rightarrow y (x) = {(x - 2)}^{4}

[y (x) = {(x - 2)}^{4} with x = t + 2 =

= y (t + 2) = {(t + 2 - 2)}^{4} = t^{4}]

if {it}^{'} s y \times (x + 2) = x^{4} \Rightarrow

\Rightarrow y = \frac{x^{4}}{x + 2} = x^{3} - 2 x^{2} + 4 x - 8 + \frac{16}{x + 2}

still no change of sign in [0; 3]

\underset{0}{\int^{3}} x^{3} - 2 x^{2} + 4 x - 8 + \frac{16}{x + 2} d x =

= \frac{1}{4} x^{4} - \frac{4}{3} x^{3} + 2 x^{2} - 8 x + 16 \ln ∣ x + 2 ∣ \underset{0}{\overset{3}{∣}} =

= - \frac{15}{4} + 16 \ln n - 16 \ln 2 =

= - \frac{15}{4} + 16 \ln \frac{5}{2} \approx 10.91

Commented by NECx last updated on 02/Apr/18

w o w \dots . n o w I u n d e r s t a n d

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