Question

Integration help please. If you heat a piece of metal to a temperature of 520 degrees C and then place it in a roo where the temperature is 20 degrees C, the cooler room will cause the metal to cool down. The temperature of the metal after t minutes of being in the room is given by T(t)=20+500e^(-t/2) what is the avg temperature of the metal during the first have hour. I am not sure how to even attempt this question.

Answer 1

It looks like an average value question to me.

Answer 2

average value (if i am not mistaken) is the integral divided by the lenght of the path

Answer 3

which in this case is 1 for one hour, so you really just need the integral

Answer 4

I have a formula here y avg=1/(b-a)

Answer 5

\[\int_0^1 (20+500e^{-\frac{t}{2}}dt\]
\[\int_0^1 20dt +500\int _0^1e^{-\frac{t}{2}}dt \]

Answer 6

first integral is a nice easy one. it is 20

Answer 7

I am able to use the calculator to integraet too. You make things look so easy

Answer 8

don't really need one for this question i don't think.
lets see if we get an answer that makes sense

Answer 9

\[\int e^{-\frac{t}{2}}dt=-2e^{-\frac{t}{2}}\] by a mental u-sub

Answer 10

so second integral is
\[-1000e^{-\frac{t}{2}}\] evalutate at 1 and zero and subtract

Answer 11

mental u sub. I feel mental in general doing this stuff lol

Answer 12

\[-1000(e^{-\frac{1}{2}}-e^0)\]
\[-1000(.60653-1)=393.74\] then add 20 and i think that is it

Answer 13

using this equation

T(t)=20+500e^(-t/2)

if I put t=6o minutes

then answer will be 20'C...... and I think so this is the answer.....
why you guys are taking integral?

Answer 14

don't feel too mental
you have
\[e^{-\frac{t}{2}}\] and you want the antiderivative. it would just be
\[e^{-\frac{t}{2}}\] except if you take the derivative of that you would have to multiply by
\[-\frac{1}{2}\] so you adjust by multiplying by -2 to make it work

Answer 15

oh damn i can't read
we have to start again !!

Answer 16

the formula is for t in MINUTES!!

Answer 17

oh okay I will erase :) my paper and block that part out. yep t minutes of being in the room.

Answer 18

\[\frac{1}{60}\int_0^{60}(20+e^{-\frac{t}{2}}) dt\] damn damn damn

Answer 19

well anti derivative is still the same, and it is still an easy integral

Answer 20

no damn just great help to you both :)

Answer 21

\[\frac{1}{60}\int _0^{60} 20dt=\frac{1}{60}\times 60\times 20=20\] so that part hasn't changed

Answer 22

I have the same problem... why satellite, you are taking integral? just putting t in minutes will not give answer......?

Answer 23

where did my 500 go from the question?

Answer 24

\[ \frac{1}{60}\int_0^{60}(20+500e^{-\frac{t}{2}}) dt\] anti derivative is still

\[20+\frac{500}{60}\int_0^{60}e^{-\frac{t}{2}}dt\]

Answer 25

i forgot it, so i put it back

Answer 26

anti derivative is still
\[-2e^{-\frac{t}{2}}\]

Answer 27

so we get
\[20-\frac{2000}{60}(e^{\frac{60}{2}}-e^0)\]

Answer 28

Well you both said 20 degrees so just need to look to make sure I have it jotted down how to do it correctly so I can study it

Answer 29

whoa hold the phone
there is no way 20 degrees is the answer!!

Answer 30

20 degrees is the room temperature. it is a constant 20 degrees. we still have to compute the integral

Answer 31

another typo
we need to compute
\[20-\frac{1000}{60}(e^{\frac{60}{2}}-e^0)\]

Answer 32

What is the number in brackets I can see e^ I cannot see what the numerator is

Answer 33

god in heaven, does it say first HALF hour???

Answer 34

oh 60/2 sorry

Answer 35

yep during first half hour

Answer 36

fine
http://www.wolframalpha.com/input/?i=1%2F30%28integral+0+to+30+%2820%2B500e^%28-t%2F2%29%29

Answer 37

i used hour, not half hour so it was wrong ignore everything i wrote

Answer 38

ok ignoring :)

Answer 39

weird thing is this cools so fast, which is why the answer makes no sense to me

Answer 40

hahahaha..... after 30 minutes temperature will be 20.00015295'C
means 20'C

Answer 41

Hey I don't think so there is any need for integration....... just put time in formula
Don't make it complex

Answer 42

right, seems rather unlikely doensn't it?

Answer 43

in 15 minutes it is 20.28 so almost room temperature.
i think this was a rather feeble attempt at a word problem
should be something like
\[20+500e^{-.002t}\] would be more lifelike

Answer 44

in any case here is the answer, i didn't bother to ingtegrate by hand
about 53.3 degrees is the average
http://www.wolframalpha.com/input/?i=1%2F30%28integral+0+to+30+%2820%2B500e^%28-t%2F2%29%29

Answer 45

waqqassaddique so your format way up the page is correct? I just sub in time so staellite just what you put then right?
I am allowed to use a calculator

Answer 46

you have to write what i wrote in wolfram.
integral from 0 to 30 of your function, divided by the length of the path, which is 30

Answer 47

sorry i misread the question and confused you. it is just that it seems very unlikely something would cool that fast, but that is neither here nor there

Answer 48

Thats no problem. That is why they make word problems to confuse! Have a test next Friday and trying to get this stuff down pat. Some stuff I will be able to use a calc. other times it is totally long hand she said.

Answer 49

best of luck

Answer 50

@waqassaddique you need the integral because this is an average value question, average of something that is continuous

Answer 51

really? that is something new to me..... thank you very much @satellite73

Answer 52

so 53.3 degrees C is final answer?

Answer 53

how?

Answer 54

20'C

Answer 55

20'C is final answer dear

Answer 56

That is what Satellite had in his answer above in Wolfram

Answer 57

lemme check

Answer 58

ok ty

Answer 59

@satellite73 ......I can't understand bro how you solving this integral.... there is something fishy...... please check your answer again....

Answer 60

So if I wanted to put the integral in my calculator do I just sub in the 60 where it shows the e^-t/2 ?

Answer 61

Christina answer is 20'C..... that integration is wrong .....

Answer 62

Yep I believe you waqassaddique :)

Answer 63

e^(-60/2)

Answer 64

appricated @Christina1972

Answer 65

aww okay great thank you so much for your help.

Answer 66

my pleasure
dear