Question

This question is about f(x= -(x+1)sin(5x). ?

Answer 1

See below.

Explanation:

a). If we test the end points of the interval `[(7pi)/5 , (8pi)/5]`

`-((7pi)/5+1)sin(7pi)=0` , `-((8pi)/5+1)sin(8pi)=0`

We can see that these are zeros, if they are consecutive zeros and above the x axis then a point in the interval will give a positive output:

`((7pi)/5 +(8pi)/5)/2=(3pi)/2`

`-((3pi)/2+1)sin((15pi)/2)=24.562` ( above the x axis )

b).

We have already ascertained the interval is above the x axis, so our integral will be:

`color(blue)(int_((7pi)/5)^((8pi)/5)[sin(5x)-xsin(5x)] dx)`

c). If our area is solely above the x axis, we do not need to integrate in parts.

`A=int_((7pi)/5)^((8pi)/5)[sin(5x)-xsin(5x)] =[-1/25sin(5x)+1/5cos(5x)x+1/5cos(5x)]_((7pi)/5)^((8pi/5)`

#[-1/25sin(5x)+1/5cos(5x)x+1/5cos(5x)]^((8pi)/5)-[-1/25sin(5x)+1/5cos(5x)x+1/5cos(5x)]_((7pi)/5) #

Plugging in upper and lower bounds:

#[-1/25sin(8pi)+1/5cos(8pi)((8pi)/5)+1/5cos(8pi)]^((8pi)/5)-[-1/25sin(7pi)+1/5cos(7pi)((7pi)/5)+1/5cos(7pi)]_((7pi)/5) #

#[0+1/5((8pi)/5)+1/5]^((8pi)/5)-[0-1/5((7pi)/5)-1/5]_((7pi)/5) #

`color(blue)((3pi+2)/5~~2.285)` units squared.

Graph:

I haven't included the calculations for finding the integral, in order to keep the answer from being too long.