when-finding-0-2-2x-4-5-dx-must-we-change-limits-

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

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