Changing Limits of Integration given g(u)

[marc017]

Here is a screenshot the question.


By using FTC for the beginning suppose statement i get:
G(10) - G(0) = 21

For evaluating part (a) and using FTC I get:
G(100/10) - G(0/10) = ?
Simplified:
G(10) - G(0) = ?

Since I end up with G(10) - G(0) in both cases I would assume the answer is 21, however that is not the answer and I have started down the wrong path.

Any help would be greatly appreciated.

 

[DrPhil]

Here is a screenshot the question.
View attachment 2540

By using FTC for the beginning suppose statement i get:
G(10) - G(0) = 21

For evaluating part (a) and using FTC I get:
G(100/10) - G(0/10) = ?
Simplified:
G(10) - G(0) = ?

Since I end up with G(10) - G(0) in both cases I would assume the answer is 21, however that is not the answer and I have started down the wrong path.

Any help would be greatly appreciated.

I assume you are using upper-case G(x) for the antiderivative of g(x), that is G'(x) = g(x)

(a) make the substitution u = t/10, t = 10 u
then du = dt/10, dt = 10 du
and the limits t = 0,100 ---> u = 0,10
What did you do with the factor of 10?

(b) substitute u = 10 - t
What is dt in terms of du?
What are the limits of integration for u, corresponding to t from 0 to 10?

 

[pka]

\(\displaystyle \int_0^{100} {f\left( {\dfrac{t}{{10}}} \right)dt} \)

Let \(\displaystyle u=\dfrac{t}{10}\) so \(\displaystyle 10du=dt\) and
\(\displaystyle \[\begin{array}{ccr} t & 0 & {100} \\\hline u &0 & {10} \\ \end{array}
\)

 

[marc017]

Thank you guys that really helped! I got 210 for part (a) and 21 for part (b) which were both correct.