Please help Find the area bounded by y(x+2)=x^4 ,x=0,y=0,and x=3


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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

	$p l e a s e I d o n t u n d e r s t a n d h o w y o u$ $g o t y (x) = {(x - 2)}^{4}$ $. . . . . T h e r e s n o t h i n g o f s u c h i n t h e$ $q u e s t i o n .$
	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

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