Question

Stewart 7th edition 6.5 number 10

Answer 1

sorry section 6.5

Answer 2

haha please write out the problem :)
stewart..hmm sounds like either a calculus textbook or probability

Answer 3

(a) Find the average value of f on the given interval
(b) Find c such that Fave=F(c)
(c) Sketch the graph of f and a rectangle whose area is the same as the area under the graph of f.

f(x)=1/x, [1,3]

Answer 4

and yeah it is a calculus textbook haha

Answer 5

for avg value within an interval [a,b]
\[=\frac{1}{b-a}\int\limits_{a}^{b} f(x) dx\]

Answer 6

It looks like par a is going to result in 1/2ln(3). Which is 0.54930

Answer 7

Ok, I got A to be the same, but then how do you start B?

Answer 8

well you know the avg f value is ln(3)/2 so:
\[f(c) = \frac{1}{c} = \frac{\ln 3}{2}\]
solve for c

Answer 9

Ok, I get .549 to be the answer for c

Answer 10

Where did f(c) come from?

Answer 11

@Enigmatron , its the notation used in the question
"c" refers to the x value corresponding to avg function value

Answer 12

@mikrol , .549 = ln3/2 which is avg "y" value
plus the interval for domain is [1,3]

Answer 13

@dumbcow I see now. Thank you for the clarification.

Answer 14

so it would be 2ln/3=c

Answer 15

yes.. essentially you take the reciprocal
2/ln3 = 1.82

Answer 16

Thank you very much that helped a lot!

Answer 17

for the rectangle part....length is always size of interval (b-a) and width is always avg function value

\[Area = 2*\frac{\ln 3}{2}\]

Answer 18

So the answer would be 1.82 because we took the reciprocal and not 0.549, correct?

Answer 19

yes...since f(x) = 1/x

f(1.82) = 1/1.82 = 0.549