Question

PLEASE HELP I am stuck- I have been waiting for help for a half an hour
Use part I of the Fundamental Theorem of Calculus to find the derivative of
the integral from 4x to 5x of (u+5/u-1)du
I will also show this as an equation below

Answer 1

\[g(x)=\int\limits_{4x}^{5x}\frac{ u+5 }{ u-1 }du\]

Answer 2

I know that this can be separated into two different integrals \[\int\limits_{0}^{5x}\frac{ u+5 }{ ?u-1}du +\int\limits_{4x}^{0}\frac{ u+5 }{ ?u-1}du\]

Answer 3

Substituting u - 1 = k

dk = du

u + 5 = k + 6

Upper bound: 5x - 1
Lower bound: 4x -1

Can you do it from here?

Answer 4

can you step me through it? I want to make sure I don't mess it up :)

Answer 5

Yeah sure.

\[\int\limits_{4x -1 }^{5x -1} {k + 6 \over k} dk = \int\limits_{4x-1}^{5x-1} {1 + {6 \over k}}dk\]

Answer 6

so the anti derivative of that is k + 6ln|k|

Now just evaluate at the points:

5x - 1 + 6ln|5x-1| - (4x - 1 + 6ln|4x-1|)

= x + 6ln|5x-1| - 6ln|4x-1|

\[x + 6\ln({5x-1 \over 4x-1})\]

Answer 7

for some reason my online homework isn't recognizing that as the correct answer.

Answer 8

That's odd, I'm pretty sure that's correct tho.

Answer 9

Try it with the | | instead of the brackets

Answer 10

I think it wants it not with the anti-derivative the first fundamental theorem I think doesn't take the anti-derivate the second theorem does.

Answer 11

Ohh, mybad. I see.

So what you wrote at the start is correct.

Just change it to:

\[\int\limits_{0}^{5x} {u + 5 \over u - 1} du - \int\limits_{0}^{4x}{u + 5 \over u-1} du\]

Answer 12

Now I'll think of how to apply the FTC pt1. Did this a long while ago, so gonna take some time haha

Answer 13

Thank you I appreciate it!

Answer 14

So according to the FTC1, you must find the F(x) function. Where F(x) is the anti derivative of f(x)

Rewrite f(u) as

\[{u + 5 \over u - 1 } = {u - 1 + 6 \over u - 1} = 1 + {6 \over u - 1}\]

So f(u) = 1 + 6/(u-1)

and F(u) = u + 6ln(u-1)

Now F(5x) = 5x + 6ln(5x-1)
and F(4x) = 4x + 6ln(4x-1)

so F(5x) - F(4x) = x + 6ln(5x-1) - 6ln(4x-1)

which is the same as the previous method.