Confused at the theory behind these kinds of problems.

Dec 2014
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USA


(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)
 
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Plato

MHF Helper
Aug 2006
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Yeah, I can't figure out b. Little help?
Are you really saying that you cannot follow the exact steps for the part a?
\(\displaystyle u=4-t\) then \(\displaystyle du=?\)
If \(\displaystyle t=0,~u=?~~\&~~t=4,~u=?\)