when finding ∫_0 ^2 (2x +4)^5 dx must we change limits?


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 66513 by Rio Michael last updated on 16/Aug/19

$w h e n f i n d i n g \int_{0}^{2} {(2 x + 4)}^{5} d x$ $m u s t w e c h a n g e l i m i t s ?$
Commented by kaivan.ahmadi last updated on 17/Aug/19
$yes.since we must change the parameter t=2x+4⇒dt=2dx { ((x=0⇒t=4)),((x=2⇒t=8)) :} (1/2)∫_4 ^8 t^5 dt=(1/(12))(8^6 −4^6 )=(1/(12))×4^6 (2^6 −1)=((63)/2)×4^6$
$y e s . s i n c e w e m u s t c h a n g e t h e p a r a m e t e r$ $t = 2 x + 4 \Rightarrow d t = 2 d x$ ${\begin{cases} x = 0 \Rightarrow t = 4 \\ x = 2 \Rightarrow t = 8 \end{cases}$ $\frac{1}{2} \int_{4}^{8} t^{5} d t = \frac{1}{12} (8^{6} - 4^{6}) = \frac{1}{12} \times 4^{6} (2^{6} - 1) = \frac{63}{2} \times 4^{6}$
Commented by Rio Michael last updated on 17/Aug/19

$t h a n k y o u s i r .$
Commented by Cmr 237 last updated on 17/Aug/19

$n o ⊛$ $0.5 \int_{4}^{8} t^{5} d t = 0.5 \times \underset{6}{\overset{1}{-}} {[t^{6}]}_{4}^{8}$ $= \underset{12}{\overset{1}{-}} [8^{6} - 4^{6}]$ $= 21504 (\begin{matrix} [\begin{matrix} \end{matrix}] \end{matrix})$
Commented by kaivan.ahmadi last updated on 17/Aug/19

$t h a n k y o u s i r$
	Answered by mr W last updated on 18/Aug/19

	$y o u {m u s t n}^{'} t c h a n g e l i m i t s, i f y o u$ $p r o c e e d l i k e t h i s :$ $\int_{0}^{2} {(2 x + 4)}^{5} d x$ $= \frac{1}{2} \int_{0}^{2} {(2 x + 4)}^{5} d (2 x + 4)$ $= \frac{1}{2 \times 6} {[{(2 x + 4)}^{6}]}_{0}^{2}$ $= \frac{1}{12} [8^{6} - 4^{6}]$ $= 21504$
	Commented by Rio Michael last updated on 19/Aug/19

	$t h a n k s s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum