6.5 Average Value of a Function/17: Difference between revisions

Revision as of 17:27, 29 November 2022

1. Use the Average Value from a to b:

$f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx$

${\frac {1}{12}}\int _{0}^{12}50+14\sin({\frac {\pi }{12}}t)\,dt={\frac {1}{12}}[50t-$

${\begin{aligned}u&={\frac {\pi }{12}}t\\\quad dt\cdot {\frac {du}{dt}}&=dt\\{\frac {12}{\pi }}du&=dt\end{aligned}}$

@@ Line 3: / Line 3: @@
 <math>
 f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx
-\frac{1}{12}\int_{0}^{12} 50 + 14\sin(\frac{\pi}{12}t)\,dt =\frac{1}{12}[50t-
+</math>
 <math>
-\begin{align}
+\frac{1}{12}\int_{0}^{12} 50 + 14\sin(\frac{\pi}{12}t)\,dt =\frac{1}{12}[50t-
-u&=\frac{\pi}{12}
-\end{align}
 </math>
@@ Line 15: / Line 12: @@
 <math>
 \begin{align}
-u= ln(p) \quad \quad  dv=p^5dp \\[2ex]
-du = \frac{1}{p} \quad \quad \quad v= \frac{1}{6}p^6  \\[2ex]
+u &= \frac{\pi}{12}t  \\
+\quad dt\cdot\frac{du}{dt} &= dt \\
+\frac{12}{\pi}du &= dt
 \end{align}
 </math>