Sup do you have the photo of your hw?
Answer:
Step-by-step explanation:
a)

this can be done by substitute lnx = u
dx/x = du
When x =1, u =0 and when x =2, u = ln 2
So integral = 
We find that this integral value is not definid for p =1
Hence for values of p other than 1, this converges.
When we substitute limits

and converges for p ≠1
b) 
So not converging for p =1
But ln x is defined only for x >0
So integral 0 to 1 makes this integral not valid and hence not convergent.
Recall the double angle identity,
sin(2x) = 2 sin(x) cos(x)
Then we can write
sin(9x) cos(9x) = 1/2 sin(2 • 9x) = 1/2 sin(18x)
Then
∫ sin(9x) cos(9x) dx
= 1/2 ∫ sin(18x) dx
= -1/2 • 1/18 cos(18x) + C
= -1/36 cos(18x) + C
though you could continue with another double angle identity,
cos(2x) = cos²(x) - sin²(x)
to rewrite the antiderivative as
= -1/36 (cos²(9x) - sin²(9x)) + C
= 1/36 (sin²(9x) - cos²(9x)) + C
Look at my attachment to see the four functions on the graph.
f --> g . . . . . shift the graph down one unit
f --> h . . . . . shift down 2 units, and double the slope
f --> d . . . . . shift up 2 units, and reduce the slope by half
Answer:
(-4,2)
Step-by-step explanation: