Question

The rate of water flow R(t) can be estimated by W(t) = ln(t2 + 7). Use W(t) to approximate the average rate of water flow during the 8-hour time period. Indicate units of measure.

Answer 1

@freckles @IrishBoy123 @zepdrix I would integrate correct? If so how would I integrate ln?

Answer 2

do you know integration by parts?

Answer 3

also what 8 hour time period are they talking about?

Answer 4

Leme give you a screenshot of that sorry. And I am familiar with the term but it would be nice if you could show me how it's done.

Answer 5

https://gyazo.com/96debaa25dbffb8f00cf7b37bdf6365b

Answer 6

Would u-substitution not work?

Answer 7

you will have to use integration by parts if you want to integrate W

though your less has been all about approximating integrals using midpoint, trapezoid, and other rules so you may can get away without doing integration by parts....

however if integration by parts is the way you want to choose to go...
then take this as an example
\[\int\limits 1 \cdot \ln(x) dx\]
notice the times 1 there
we need the 1
1 will be the one we choose to integrate
and ln(x) will be the one we differentiate
\[\int\limits \color{red}{1} \cdot \ln(x) dx\\ =\color{red}{x} \color{blue}{\ln(x)}- \int\limits \color{red}{x} \cdot \color{blue}{\frac{1}{x}} dx\]

Answer 8

\[\int\limits \ln(x) dx=x \ln(x)-\int\limits 1 dx \\ =x \ln(x)-x +C\]

Answer 9

you will do something similar here

Answer 10

you have you want to evaluate
\[\frac{ \int\limits _0^8 W(t) dt}{8-0}\]

Answer 11

\[\int\limits W(t) dt= \int\limits 1 \cdot \ln(t^2+7) dt\]
integrate 1 and differentiate t^2+7

you can plug into this formula:
\[\int\limits f'(x) g(x) dx=f(x) g(x)-\int\limits f(x) g'(x)dx\]
notice I chose f' to integrate
and g to differentiate

Answer 12

you will have a bit more work to do like a trig identity unless you know your integration table well

Answer 13

at the end

Answer 14

now you might decide to approximate that integral above using some method you have learned in this I guess its called lesson pack

Answer 15

integrate 1 and differentiate ln(t^2+7)
*
working in the dark

Answer 16

Can you show how it applies to ln? It'll help grasp it.

Answer 17

how what applies to ln?

Answer 18

Integration by parts.

Answer 19

oh you didn't see the example?

Answer 20

\[\int\limits f'(x)g(x)dx=f(x) g(x)-\int\limits f(x)g'(x) dx \\ \int\limits 1 \cdot \ln(x) dx= x \ln(x)-\int\limits x \frac{1}{x} dx \\ \int\limits 1 \cdot \ln(x) dx=x \ln(x)-\int\limits 1 dx=x \ln(x)-x+C \\ \text{ so } \int\limits \ln(x) dx=x \ln(x)-x+C\]
this example above

Answer 21

you have
\[\int\limits 1 \cdot \ln(x^2+7) dx\]

Answer 22

choose f'(x)=1 and the other function, the log function, to be g(x) just like I did in the example

Answer 23

have you plug into the integration by parts formula above?

Answer 24

f(x)=x and g'(x)=? if g(x)=ln(x^2+7)

Answer 25

Absolutely confused on how I should approach the part with ln(t^2+7)

Answer 26

Just wanted to mention you can use a substitution, but it doesn't prevent you from having to use integration by parts. Set \(t=\sqrt{e^s-7}\), so \(\mathrm{d}t=\dfrac{e^s}{2\sqrt{e^s-7}}\,\mathrm{d}s\), and \(s=\ln(t^2+7)\). This means you get
\[\int_0^8\ln(t^2+7)\,\mathrm{d}t=\frac{1}{2}\int_{\ln7}^{\ln71}\frac{se^s}{\sqrt{e^s-7}}\,\mathrm{d}s\]This way, you end up making more work for yourself.

Instead, using freckles' suggestion, you have
\[\int\ln(t^2+7)\,\mathrm{d}t=uv-\int v\,\mathrm{d}u\]where
\[\begin{matrix}u=\ln(t^2+7)&&&\mathrm{d}v=\mathrm{d}t\\[1ex]
\mathrm{d}u=\dfrac{2t}{t^2+7}\,\mathrm{d}t&&&v=t\end{matrix}\]so
\[\int_0^8\ln(t^2+7)\,\mathrm{d}t=\bigg[t\ln(t^2+7)\bigg]_0^8-2\int_0^8\frac{t^2}{t^2+7}\,\mathrm{d}t\]

Answer 27

I'm probably going to need to practice this a lot more barely got it. Thanks for your help guys.