finding the temperature after a period of time












1












$begingroup$


I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.



A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.



I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
    $endgroup$
    – Curious
    Jun 15 '16 at 15:33










  • $begingroup$
    I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
    $endgroup$
    – google
    Jun 15 '16 at 15:38










  • $begingroup$
    The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
    $endgroup$
    – Curious
    Jun 15 '16 at 15:52










  • $begingroup$
    even trying that its saying its not right
    $endgroup$
    – google
    Jun 15 '16 at 15:53
















1












$begingroup$


I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.



A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.



I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
    $endgroup$
    – Curious
    Jun 15 '16 at 15:33










  • $begingroup$
    I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
    $endgroup$
    – google
    Jun 15 '16 at 15:38










  • $begingroup$
    The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
    $endgroup$
    – Curious
    Jun 15 '16 at 15:52










  • $begingroup$
    even trying that its saying its not right
    $endgroup$
    – google
    Jun 15 '16 at 15:53














1












1








1





$begingroup$


I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.



A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.



I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?






calculus







share|cite|improve this question











$endgroup$




I have a homework problem that I am really struggling with. I will put the problem below and then discuss what I have tried to do.



A science geek brews tea at $195$ °F, and observes that the temperature $T(t)$ of the tea after $t$ minutes is changing at the rate of $T′(t)=−4.5e^{−0.07t}$ °F/min. What is the average temperature of the tea during the first $5$ minutes after being brewed? Round your answer to the nearest hundredth of a degree.



I tried to integrate from $0$ to $5$ minutes but I couldn't get the right answer does anyone know how to do this?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 15 '16 at 16:06









Jennifer

8,44721837




8,44721837










asked Jun 15 '16 at 15:26









googlegoogle

805




805












  • $begingroup$
    You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
    $endgroup$
    – Curious
    Jun 15 '16 at 15:33










  • $begingroup$
    I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
    $endgroup$
    – google
    Jun 15 '16 at 15:38










  • $begingroup$
    The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
    $endgroup$
    – Curious
    Jun 15 '16 at 15:52










  • $begingroup$
    even trying that its saying its not right
    $endgroup$
    – google
    Jun 15 '16 at 15:53


















  • $begingroup$
    You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
    $endgroup$
    – Curious
    Jun 15 '16 at 15:33










  • $begingroup$
    I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
    $endgroup$
    – google
    Jun 15 '16 at 15:38










  • $begingroup$
    The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
    $endgroup$
    – Curious
    Jun 15 '16 at 15:52










  • $begingroup$
    even trying that its saying its not right
    $endgroup$
    – google
    Jun 15 '16 at 15:53
















$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33




$begingroup$
You have to integrate the equation from $0$ to $5$. Next you have to divide your result with $5$. Then you have the average temperature.
$endgroup$
– Curious
Jun 15 '16 at 15:33












$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
– google
Jun 15 '16 at 15:38




$begingroup$
I got 52.4472 and dividing it by 5 i got 10.48944 but i dont think thats possibly the answer can anyone help?
$endgroup$
– google
Jun 15 '16 at 15:38












$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52




$begingroup$
The equation starts at $195 °F$. So I think the equation is: $195 - 4.5e^{-0.07t}$. After integrating and dividing by 5, I get $173.08 °F$
$endgroup$
– Curious
Jun 15 '16 at 15:52












$begingroup$
even trying that its saying its not right
$endgroup$
– google
Jun 15 '16 at 15:53




$begingroup$
even trying that its saying its not right
$endgroup$
– google
Jun 15 '16 at 15:53










1 Answer
1






active

oldest

votes


















0












$begingroup$

Hint :

The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$



Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.





  1. Obtain an expression for $T$.

    $T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$

  2. Use the formula for the average value of the function :

    $$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
    e^{-0.35}-1)+145approx187.187$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what answer did you get? I got the same answer as the guy above and its wrong
    $endgroup$
    – google
    Jun 15 '16 at 16:02










  • $begingroup$
    added the answer
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:03










  • $begingroup$
    @google Did it answer your question ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:11










  • $begingroup$
    its still saying that its wrong. I dont know what it could be
    $endgroup$
    – google
    Jun 15 '16 at 16:17










  • $begingroup$
    @google I did a mistake in last line can you check the new result ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:22













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint :

The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$



Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.





  1. Obtain an expression for $T$.

    $T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$

  2. Use the formula for the average value of the function :

    $$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
    e^{-0.35}-1)+145approx187.187$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what answer did you get? I got the same answer as the guy above and its wrong
    $endgroup$
    – google
    Jun 15 '16 at 16:02










  • $begingroup$
    added the answer
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:03










  • $begingroup$
    @google Did it answer your question ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:11










  • $begingroup$
    its still saying that its wrong. I dont know what it could be
    $endgroup$
    – google
    Jun 15 '16 at 16:17










  • $begingroup$
    @google I did a mistake in last line can you check the new result ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:22


















0












$begingroup$

Hint :

The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$



Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.





  1. Obtain an expression for $T$.

    $T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$

  2. Use the formula for the average value of the function :

    $$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
    e^{-0.35}-1)+145approx187.187$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what answer did you get? I got the same answer as the guy above and its wrong
    $endgroup$
    – google
    Jun 15 '16 at 16:02










  • $begingroup$
    added the answer
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:03










  • $begingroup$
    @google Did it answer your question ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:11










  • $begingroup$
    its still saying that its wrong. I dont know what it could be
    $endgroup$
    – google
    Jun 15 '16 at 16:17










  • $begingroup$
    @google I did a mistake in last line can you check the new result ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:22
















0












0








0





$begingroup$

Hint :

The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$



Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.





  1. Obtain an expression for $T$.

    $T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$

  2. Use the formula for the average value of the function :

    $$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
    e^{-0.35}-1)+145approx187.187$$






share|cite|improve this answer











$endgroup$



Hint :

The average value of a function $f(x)$ on the intervall $[a,b]$ is given by :
$$frac{1}{b-a}int_a^bf(x)dx$$



Here $f=T$, so you need to integrate to get $T(t)$ then to obtain the average value.





  1. Obtain an expression for $T$.

    $T(0)=195$, so $T(t)=int_0^tT'(s)ds+T(0)=int_0^t−4.5e^{−0.07s}ds+195=−4.5int_0^te^{−0.07s}ds+195=frac{4,5}{0.07}[e^{-0.07s}]_0^t+195=frac{4,5}{0.07}(e^{-0.07t}-1)+195$

  2. Use the formula for the average value of the function :

    $$frac{1}{5-0}int_0^5T(x)dx=frac{1}{5}int_0^5frac{4,5}{0.07}(e^{-0.07t}-1)+195dt=10int_0^5e^{-0.07t}dt-50+195=-frac{10}{0.07}(
    e^{-0.35}-1)+145approx187.187$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 15 '16 at 16:21

























answered Jun 15 '16 at 15:32









JenniferJennifer

8,44721837




8,44721837












  • $begingroup$
    what answer did you get? I got the same answer as the guy above and its wrong
    $endgroup$
    – google
    Jun 15 '16 at 16:02










  • $begingroup$
    added the answer
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:03










  • $begingroup$
    @google Did it answer your question ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:11










  • $begingroup$
    its still saying that its wrong. I dont know what it could be
    $endgroup$
    – google
    Jun 15 '16 at 16:17










  • $begingroup$
    @google I did a mistake in last line can you check the new result ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:22




















  • $begingroup$
    what answer did you get? I got the same answer as the guy above and its wrong
    $endgroup$
    – google
    Jun 15 '16 at 16:02










  • $begingroup$
    added the answer
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:03










  • $begingroup$
    @google Did it answer your question ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:11










  • $begingroup$
    its still saying that its wrong. I dont know what it could be
    $endgroup$
    – google
    Jun 15 '16 at 16:17










  • $begingroup$
    @google I did a mistake in last line can you check the new result ?
    $endgroup$
    – Jennifer
    Jun 15 '16 at 16:22


















$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02




$begingroup$
what answer did you get? I got the same answer as the guy above and its wrong
$endgroup$
– google
Jun 15 '16 at 16:02












$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03




$begingroup$
added the answer
$endgroup$
– Jennifer
Jun 15 '16 at 16:03












$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11




$begingroup$
@google Did it answer your question ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:11












$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17




$begingroup$
its still saying that its wrong. I dont know what it could be
$endgroup$
– google
Jun 15 '16 at 16:17












$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22






$begingroup$
@google I did a mistake in last line can you check the new result ?
$endgroup$
– Jennifer
Jun 15 '16 at 16:22




















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