Definite Integral Question?
Suppose \[\int\limits_{1}^{3} g(t)dt = 12\]
Find \[\int\limits_{5}^{15} g(x/5)\]
I know the answer is 16 but i dont know how to get it
same with \[\int\limits_{-1/3}^{1/3} g(2-3t)dt\]
answer is 60 not 16
Let's keep this in our back pocket, we'll come back to it later: \[\Large\bf\sf \color{#DD4747 }{\int\limits\limits_{1}^{3} g(t)dt \quad=\quad 12}\] So we need to figure this out: \[\Large\bf\sf \int\limits\limits_{5}^{15} g\left(\frac{x}{5}\right)dx\] We'll start by making a substitution:\[\Large\bf\sf u=\frac{x}{5}\]What do you get for your dx?
x^2/10
Hmm no.. maybe I didn't ask the question clearly. We're just taking the derivative of our u.\[\Large\bf\sf du=\frac{1}{5}dx\qquad\to\qquad 5du=dx\]Does that make sense?
ohh okay yeah
Then we'll also want to change the boundaries of our function:\[\Large\bf\sf x=5:\qquad u=\frac{5}{5}=1\]\[\Large\bf\sf x=15:\qquad u=\frac{15}{5}=3\]
So plugging all of these goodies in gives us:\[\Large\bf\sf 5\color{#DD4747 }{\int\limits\limits_{u=1}^3 g(u)\;du}\]
Hmm it's starting to look familiar, yes? :o
Yeah! thanks you really helped me make sense of this
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