Question

Find the average value fave of the function f on the given interval. f() = 5 sec2(/6), 0, 3 2

Answer 1

Answer:

$f_{ave}$ = ${\frac{10 }{3\pi }$

Step-by-step explanation:

To find - Find the average value fave of the function f on the given interval. f(x) = 5 sec²(x/6), [0, 3$\pi$/2]

Proof -

We know that,

Average value of f in the interval [a, b] is -

$f_{ave}$ = $\frac{1}{b - a}\int\limits^b_a {f(x)} \, dx$

Now,

Here a = 0, b = $\frac{3\pi }{2}$, f(x) = $5sec^{2}(\frac{x}{6} )$

Now,

$f_{ave}$ = $\frac{1}{\frac{3\pi }{2} - 0}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

      = ${\frac{2 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {5sec^2 ({\frac{x}{6} }) } \, dx$

      = ${\frac{10 }{3\pi }}\int\limits^{\frac{3\pi }{2} }_0 {sec^2 ({\frac{x}{6} }) } \, dx$

      = ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) }|^{\frac{3\pi }{2} }_0 \, ]$

      = ${\frac{10 }{3\pi }}[\ {tan({\frac{x}{6} }) - tan({0) } \, ]$

      = ${\frac{10 }{3\pi }}[ {1 - 0 } ]$

      = ${\frac{10 }{3\pi }$

⇒$f_{ave}$ = ${\frac{10 }{3\pi }$