Question

Calculus help please!

Answer 1

Consider a continuous function f such that F
(x) = f(x) for all real values of x. Find 7
2
f(5x)dx

Answer 2

That didn't come out quite right. I'll retype it.

Answer 3

Consider a continuous function f such that F'(x)=f(x) find \[\int\limits_{2}^{7}f(5x)dx\]

Answer 4

F(7)-F(2)

Answer 5

I know its tough. I've been stuck on it for a while. Thanks for tagging those people!

Answer 6

And sorry @Abhishek619 thats not an answer choice. Thanks for trying though.

Answer 7

can you post the choices?

Answer 8

Sure. One sec.

Answer 9

\[\Large \int\limits\limits_{2}^{7}f(5x)dx\]let u=5x,
so
du = 5 dx

Answer 10

5F(7)-5(F(2)
1/5(F(7))-1/5(F(2))
5F(35)-5F(10)
1/5(F(35))-1/5(F(10))

Answer 11

I'm sorry. I'm wrong.
F'(x)=f(x),
F'(5x)=f(5x)
\[\int\limits_{2}^{7}f(5x)dx=\int\limits_{2}^{7}dF(5x)\]
Now, take 5x=t, =>dx=dt/5,
the limits also become 10 and 35.
the answer is
(F(35)-F(10))/5

Answer 12

\[\Large \int\limits\limits\limits_{2}^{7}f(5x)dx \]let u=5x,
so
du = 5 dx\[\large \frac{ du }{ 5 } =dx\]
Change the limits since u = 5x,
so plug in x=2 and then x=7
to get u = 10 and u = 35
\[\Large \frac{ 1 }{ 5 }\int\limits\limits_{10}^{35}f(u)du\]

Answer 13

\[\Large \frac{ 1 }{ 5 }\int\limits\limits\limits_{10}^{35}f(u)du = \frac{ 1 }{ 5 }\left[ F(u) \right]_{10}^{35}\]Then just plug in limits

Answer 14

Thanks @agent0smith and @Abhishek619

Answer 15

That makes perfect sense, thank you!

Answer 16

At first, I was blind to see the 5x. :D