Question

calculus

Answer 1

any thoughts?

Answer 2

we can chain it, or just do a change of variable

Answer 3

We did this last semester and I can't find in the book where it showed how to do it haha. I thought you could pull out the 10\[10\int\limits_{0}^{0.1}f(x)dx=15\]Then divide by 10 on both sides and multiply by 0.1. I've tried 150,15,1.5, 0.15...nothing works that I'm doing.

Answer 4

change of variable might help rewrite a function of t, into a function of say u

if u = 10t, then du = 10dt: du/10= dt

t moves from 0 to 1

when t = 0, u = 10(0) = 0
when t = 1, u = 10(1) = 10

\[\int_{0}^{10}\frac{1}{10}f(u)~du\]

Answer 5

\[.1~\int_{0}^{10}f(u)~du=.1\left[~\int_{0}^{1}f(u)~du+\int_{1}^{10}f(u)~du\right]\]
\[.1\left[15+\int_{1}^{10}f(u)~du\right]\]

Answer 6

thats the most sensible process i can come up with.

Answer 7

do your results have to be numerical in value?

Answer 8

yeah they do. That makes sense what you're doing. I wasn't sure how to start that problem. It's been like 3 months. I should be able to figure it out from here. Thanks!

Answer 9

let me see what converting back into t does afterwards

t = u/10; when u=1, t=.1 and when u=10, t=1

looks to me like we are just subtracting that .1(15) from 15

Answer 10

let me rethink that ... sometimes me brain is too quick for my own good

Answer 11

\[.1~\int_{0}^{10}f(u)~du=.1\left[~\int_{0}^{1}f(u)~du+\int_{1}^{10}f(u)~du\right]\]
\[.1(15)~+~.1\int_{u=1}^{u=10}f(u)~du\]
u = 10t, du = 10dt
u=1, t=.1 ; u=10,t=1
\[.1(15)~+~\int_{t=.1}^{t=1}f(10t)~dt\]
.1(15) is what im coming up with

Answer 12

f(10t) dt comes from f(10t)/10

since by the chain rule we can derive it back to f(10t)

given that f(1) - f(0) = 15

f(10(.1))/10 - f(10(0))/10

[f(1) - f(0)]/10 = 15/10 = 1.5 still pans out to me