# Finding the definite integral using two variables - what am I doing wrong here?

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!!

• 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 $\frac {1}{b-a}$? – Tdonut Apr 30 '15 at 16:30
• 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!! – barney Apr 30 '15 at 16:58
• I updated the problem showing your suggestion to split it up at the beginning, but I'm still off by a little bit! – barney Apr 30 '15 at 17:17

$\displaystyle \bar{p}=\frac{1}{12}\int_0^{12}\left(7t\sin(.2t^2)+75\right)dt=\frac{1}{12}\left[\int_0^{12}7t\sin(.2t^2)dt+\int_0^{12}75dt\right]$
$\displaystyle=\frac{1}{12}\cdot\frac{7}{.4}\int_0^{28.8}\sin u\; du+\frac{1}{12}\big(75\cdot12\big)=\frac{35}{24}\left(1-\cos 28.8\right)+75\approx77.720$
• 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 $\frac{35}{24} + 2.522 = 2.719.$ – barney Apr 30 '15 at 19:04