Integral Help

[abilitiesz]

I need to find the underestimate, overestimate, and the average. I'm not sure how to start the problem so can anyone help me? Thank you.

 

[galactus]

The position function is the antiderivative of the velocity function. \(\displaystyle s(t)=\int v(t)dt\)

\(\displaystyle 9\int\frac{1}{4+t}dt=9ln(t+4)\)

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ f(t) \ = \ \frac{9}{4+t}, \ \Delta t \ = \ 0.2 \ = \ \frac{1}{5}, \ 0 \ \le \ t \ \le \ 1\)

\(\displaystyle Right \ Endpoint: \ \frac{1}{5}\sum_{i=1}^{5}\frac{45}{20+i} \ = \ 1.96396668549\)

\(\displaystyle Left \ Endpoint: \ \frac{1}{5}\sum_{i=1}^{5}\frac{45}{19+i} \ = \ 2.05396668549\)

\(\displaystyle Avg: \ \frac{[1.96396668549+2.05396668549]}{2} \ = \ 2.00896668549\)

\(\displaystyle Under \ estimate \ (Right \ Endpoint) \ = \ 1.96396668549\)

\(\displaystyle Over \ estimate \ (Left \ Endpoint) \ = \ 2.05396668549\)

\(\displaystyle Avg. \ of \ two \ = \ 2.00896668549\)

\(\displaystyle Actual \ (true) \ value: \ \int_{0}^{1}\frac{9}{4+t}dt \ = \ 9ln(5/4) \ = \ 2.00829196183\)

Here is the under estimation, right endpoint of f(t) = 9/(4+t).
[attachment=1:1vzk9lg0]bbb.jpg[/attachment:1vzk9lg0]

Here is the over estimation, left endpoint of f(t) = 9/(4+t).

[attachment=0:1vzk9lg0]ccc.jpg[/attachment:1vzk9lg0]

 

[abilitiesz]

Thank you for your help. I also re-did the problem following your steps and they were very helpful. One thing though, is that I tried to get the correct units for it, but it is somehow telling me it is wrong.

Do I have to calculate the answer you have given me to match the correct units? I know that velocity is measured in m/s but in this problem its given me m/hr so do i need to convert the hr to s and thus changing the answer to the problem?

Thank you once again.

 

[abilitiesz]

I figured out what my problem was. Thank you.