What is the average value of a function # y=sec^2 x# on the interval #[0,pi/4]#?

Question

Isabella Ackerman · Accepted Answer

#4/pi# The average value of a function # y=f(x) # over an interval # [a,b] # is given by # 1/(b-a)int_a^b f(x) dx # So for # y=sec^2 x # we have: # Avg = 1/(pi/4-0) int_0^(pi/4) sec^2x dx# # = 1/(pi/4) int_0^(pi/4) sec^2x dx # # = 4/pi [tanx]_0^(pi/4) # # = 4/pi {tan (pi/4) - tan0 } # # = 4/pi {tan (pi/4) - tan0 } # # = 4/pi {1-0 } # # = 4/pi #

Emma Aldridge · Answer

undefined To find the average value of the function $ y = \sec^2(x) $ on the interval $[0, \frac{\pi}{4}]$, you can use the formula for average value of a function on a closed interval $ [a, b] $:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

Applying this formula to the given function $ y = \sec^2(x) $ over the interval $[0, \frac{\pi}{4}]$:

$$ \text{Average value} = \frac{1}{\frac{\pi}{4} - 0} \int_{0}^{\frac{\pi}{4}} \sec^2(x) \, dx $$

$$ = \frac{4}{\pi} \int_{0}^{\frac{\pi}{4}} \sec^2(x) \, dx $$

$$ = \frac{4}{\pi} [\tan(x)]_{0}^{\frac{\pi}{4}} $$

$$ = \frac{4}{\pi} \left(\tan\left(\frac{\pi}{4}\right) - \tan(0)\right) $$

$$ = \frac{4}{\pi} \left(1 - 0\right) $$

$$ = \frac{4}{\pi} $$

So, the average value of $ y = \sec^2(x) $ on the interval $[0, \frac{\pi}{4}]$ is $ \frac{4}{\pi} $.