Since the upper curve is the sine curve and the bottom one is cosine, we are going to do upper - lower, thus the integral of sine-cosine.
\[\int_{\pi/4}^{5 \pi/4} sinx-cosx \,dx \ \] \[=-cosx-sinx \]
\[ -cos(5\pi/4)-(-cos(\pi/4))-(sin(5pi/4)-sin(\pi/4))=(-(\frac{-1}{\sqrt{2}})+(\frac{1}{\sqrt{2}})) - (\frac{-1}{\sqrt{2}}+(\frac{1}{\sqrt{2}}) = \]
\[ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}= \frac{4}{\sqrt{2}} \] Have to remove the square root from the denominator, so multiply numerator & denominator by the square root:
\[ \frac{4}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}\]
top <- function(x){sin(x)}
bottom <- function(x){cos(x)}
# Calculate area over the range [pi/4, 5pi/4]
areaTop <- integrate(top, lower = pi/4, upper = 5*pi/4)
areaBottom <- integrate(bottom, lower = pi/4, upper = 5*pi/4)
# Take the difference of the areas
(areaTop$value - areaBottom$value )
## [1] 2.828427
2*sqrt(2)==areaTop$value - areaBottom$value
## [1] TRUE