Question

CALCULUS HELP!!

Answer 1

my best guess is to integrate the difference from 0 to 10

Answer 2

can someone explain this step by step with me

Answer 3

\[\int_0^{10}\frac{t+6}{1+t}dt-\int_0^{10}\frac{\ln(t+1)}{t+1}dt\]

Answer 4

so i just solve that?

Answer 5

i believe so
the rate is the derivative, you want the amount

Answer 6

http://www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B0%7D%5E%7B10%7D%20%5Cfrac%7Bt%2B6%7D%7B1%2Bt%7Ddt%20-%20%5Cint_%7B0%7D%5E%7B10%7D%20%5Cfrac%7Bln%5Cleft(t%2B1%5Cright)%7D%7Bt%2B1%7Ddt/?origin=enterkey

Answer 7

@perl please help me!!!!

Answer 8

@perl

Answer 9

You need to solve what @satellite73 gave you, plus you will need to \(assume\) initial conditions, such as V=0 at t=0.
It is clear that the question wanted a numerical value at t=10 minutes, but forgot to give the initial condition.

Answer 10

Also you have to assume tank does not overflow before 10 minutes.
Are you sure you have posted the complete question? @wade123

Answer 11

@wade123 In you posted calculations, the numerator of the second integral should be ln(t+2) instead of ln(t+1)

Answer 12

no the problem is correct...

Answer 13

@satellite73

Answer 14

My bad, it's t+1.
You just have to evaluate the expression you got from your work, and it comes out to be about 19.