0
$\begingroup$

I'm trying to find the average value of the function:

$$p(t) = t7*sin0.2t^2+75 \quad dt \quad on[0,12]$$

So I wanted to start off by first finding the definite integral. I'm being thrown off by the +75 at the end and my answer is coming out differently than I'm finding elsewhere.

Work so far:

$$\int^{12}_0 7t*sin(0.2t^2)+75\quad dt$$

$$7* \int^{12}_0 t*sin(0.2t^2)+75 \quad dt $$

$$7* \int^{12}_0 t*sin(u)+75\quad dt$$ Next, multiplying inside and dividing outside by 0.4 to match the derivative of u:

$$\frac{7}{0.4} \int^{12}_0 0.4t*sin(u)+75 \quad dt $$ $\frac {du}{dx}$ = 0.4, and du = 0.4*dx, so substitute this in. The boundaries will also change from 0 to 12 and the u boundaries will be 0 to 28.8.

$$\frac{7}{0.4}* \int^{28.8}_0 sin(u)+75\quad du$$ $$\frac{7}{0.4}* \int^{28.8}_0 sin(u)+75\quad du$$ $$ = 7\frac{-cos(u)}{0.4}+75t$$

Here is where I get stuck. Since I converted the trig part to u-substitution, can I still treat the integral of 75 with the variable t from 0 to 12?

$$ 7\frac{-cos(28.8)}{0.4}+75(12) \quad - \quad 7\frac{-cos(0)}{0.4}+75(0) = definite \quad integral$$

The answer according to many different sources should be about 911, but I'm getting about 932. $$915.137 + 17.5 = 932.637.$$

Splitting the integral up into two parts at the beginning (thank you for the hint!):

$$\int^{12}_0 7t*sin(0.2t^2)\quad dt+\int 75\quad dt$$ $$\int^{12}_0 7t*sin(0.2t^2)\quad dt+ \int_0^{12} 75\quad dt$$

$$\frac{7}{0.4}*(-cos(0.2t^2))+75t \quad $$

$$\frac{7}{0.4} (-cos(0.2(12^2)) - (-cos(0.2(0^2)) + (75(12) - 75(0)) \quad $$ $$ 16.137 + (900) \quad $$

This answer of about 916 is still is off by a little bit I've been tearing my hair out looking for any mistakes but I haven't come across a problem like this so far.

Thank you for taking the time to help!!

$\endgroup$
3
  • 1
    $\begingroup$ Hint: At the very beginning, split the definite integral into two definite integrals. Also, if you are finding the average value of a function, don't you need to multiply by $\frac {1}{b-a}$? $\endgroup$ – Tdonut Apr 30 '15 at 16:30
  • $\begingroup$ Oh! That would do it :-) And to answer your question, yes, I need to multiply by 1/b-a, but I wanted to take it one step at a time. It turns out this first step took hours of scratch paper and lots of confusion! :-D Thank you for the hint!! $\endgroup$ – barney Apr 30 '15 at 16:58
  • $\begingroup$ I updated the problem showing your suggestion to split it up at the beginning, but I'm still off by a little bit! $\endgroup$ – barney Apr 30 '15 at 17:17
1
$\begingroup$

$\displaystyle \bar{p}=\frac{1}{12}\int_0^{12}\left(7t\sin(.2t^2)+75\right)dt=\frac{1}{12}\left[\int_0^{12}7t\sin(.2t^2)dt+\int_0^{12}75dt\right]$

$\displaystyle=\frac{1}{12}\cdot\frac{7}{.4}\int_0^{28.8}\sin u\; du+\frac{1}{12}\big(75\cdot12\big)=\frac{35}{24}\left(1-\cos 28.8\right)+75\approx77.720$

$\endgroup$
2
  • $\begingroup$ Wow! This is a great explanation. But I need to ask about the very last part involving the second integral (the 75t part). The calculation of $\frac{35}{24}(1-cos 28.8)$ evaluates to 2.71 for me. You need to distribute the 35/24 to both of the terms in parentheses, right? That got me $\frac{35}{24} + 2.522 = 2.719.$ $\endgroup$ – barney Apr 30 '15 at 19:04
  • $\begingroup$ Thanks for your comment, and I had an error in calculating the first term, as you pointed out. (My calculator was in degree mode, instead of radians.) $\endgroup$ – user84413 Apr 30 '15 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.