A AloetheFerret Sep 2019 11 0 Arizona Dec 11, 2019 #1 I would like some help solving these because it is difficult to conceptualize what's even going on. Attachments screenshot (1).png 12.4 KB Views: 30 screenshot (2).png 52.3 KB Views: 7

Cervesa Dec 2014 153 117 USA Dec 11, 2019 #2 (a) use substitution let $u = \dfrac{t}{4} \implies dt = 4 \, du$ lower limit is $t= 0 \implies u = \dfrac{0}{4} = 0$ , upper limit is $t=16 \implies u = \dfrac{16}{4} = 4$ $\displaystyle \int_0^{16} g\left(\dfrac{t}{4}\right) \, dt = \int_0^4 g(u) \cdot 4 \, du = 4 \int_0^4 g(u) \, du = 4 \cdot 3 = 12$ try the same method for (b) Reactions: topsquark

(a) use substitution let $u = \dfrac{t}{4} \implies dt = 4 \, du$ lower limit is $t= 0 \implies u = \dfrac{0}{4} = 0$ , upper limit is $t=16 \implies u = \dfrac{16}{4} = 4$ $\displaystyle \int_0^{16} g\left(\dfrac{t}{4}\right) \, dt = \int_0^4 g(u) \cdot 4 \, du = 4 \int_0^4 g(u) \, du = 4 \cdot 3 = 12$ try the same method for (b)

P Plato MHF Helper Aug 2006 22,490 8,653 Dec 11, 2019 #4 AloetheFerret said: Yeah, I can't figure out b. Little help? Click to expand... Are you really saying that you cannot follow the exact steps for the part a? \(\displaystyle u=4-t\) then \(\displaystyle du=?\) If \(\displaystyle t=0,~u=?~~\&~~t=4,~u=?\)

AloetheFerret said: Yeah, I can't figure out b. Little help? Click to expand... Are you really saying that you cannot follow the exact steps for the part a? \(\displaystyle u=4-t\) then \(\displaystyle du=?\) If \(\displaystyle t=0,~u=?~~\&~~t=4,~u=?\)