If #D=int_a^b(sin(kx)+2)dx# where #b=a+4# and #k# is unknown, then #D# must be greater than _____ and less than ______?

1 Answer
Steve M
Feb 12, 2018

Approximately, we have:

#4 lt D lt 12 #

Explanation:

We have:

# D = int_a^b \ sin(kx)+2 \ dx #

But also, we have #b=a+4# so we have:

# D = int_a^(a+4) \ sin(kx)+2 \ dx #

# \ \ \ = [-1/kcos(kx)+2x]_a^(a+4) #

# \ \ \ = (-1/kcos(ak+4k)+2a+8) - (-1/kcos(2k)+2a) #

# \ \ \ = -1/kcos(ak+4k)+2a+8 +1/kcos(ak)-2a #

# \ \ \ = (cos(ak)-cos(ak+4k))/k+8 #

So we have a function of #2# parameters:

# D(a,k) = (cos(ak)-cos(ak+4k))/k+8 #

Now, we can use the trigonometric identity:

# cosA-cosB -= -2sin((A+B)/2)sin((A-B)/2) #

Or, Equivalently:

# cosB-cosA -= 2sin((A+B)/2)sin((A-B)/2) #

With #B=ak# and #A=ak+4k#, we can write

# D(a,k) = (2sin((ak+4k+ak)/2)sin((ak+4k-ak)/2))/k + 8 #

# \ \ \ \ \ \ \ \ \ \ \ \ = (2sin((2ak+4k)/2)sin((4k)/2))/k + 8 #

# \ \ \ \ \ \ \ \ \ \ \ \ = (2sin(ak+2k)sin(2k))/k + 8 #

As this is a function of #2# variables the analysis is more complex, and I will add further details if time permits, but we can establish that we have a minima of slightly more than #4 #, and maxima slightly less than #12#.

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