Find the area bounded by the given curves y= 7cos(x) and y=7cos^2(x) between x=0 and x= π/2.

Answer 1

first we need their intersection:
7cos^2(x) = 7cos(x)
cos^2 x - cosx = 0
cosx(cosx- 1)
cosx = 0 or cosx = 1
x = π/2, or x = 0
ahhh, our boundaries happen to be their intersection.
let's look at the graph:
www.wolframalpha.com/input/?i=plot+y%3D+7cos(x)+,+y%3D7cos%5E2(x)+from+-.2+to+(%CF%80%2F2+%2B+1)

looks like 7cos^2 x is above 7cosx
so area =∫ (7cos^2x -7cosx) dx from 0 to π/2
= (1/2)(x + sinx coxx) + 7sinx ] from 0 to π/2
= ..... (you so the substitution)

(the integrals of sin^2 x and cos^2 x should be in your repertoire of standard integrals)

Answer 2

I think you make a mistake with which one above and below

Answer 3

You are right, so just turn the terms around to
area =∫ (7cosx - 7cos^2x) dx from 0 to π/2

I should have looked at my own graph a bit closer.

Answer 4

This are how I solve it. Do you think my answer correct?
[0,π/2]∫7cos(x)-7cos^2(x)dx
[0,π/2]∫7cos(x)-7(1-sin^2(x))dx
[0,π/2]∫7cos(x)-7(1-(1/2(1-cos(2x)))dx
[0,π/2]7sin(x)-7x/2 -7sin(2x)/2]
[0,π/2]7sin(x)-7x/2 -7*2sin(x)cos(x)/2]
7*π/2(-7*1-7*1*0)]
=7π/2

Answer 5

In your step:
[0,π/2]∫7cos(x)-7cos^2(x)dx
[0,π/2]∫7cos(x)-7(1-sin^2(x))dx

why did you change cos^2 x to (1 - sin^2 x)
to integrate sin^2 x is just as difficult as cos^2 x

did you mean to use: cos(2x) = 2cos^2 x - 1 ??
then 2cos^2 x = cos(2x) + 1
cos^2 x = (1/2)cos(2x) + 1/2
∫7cos^2 x = ∫7(1/2)cos(2x) + 7/2
= (7/4)sin(2x) + 7x/2

so [0,π/2]∫7cos(x)-7cos^2(x)dx
= [7sinx - (7/4)sin(2x) - 7x/2 ] from 0 to π/2
= (7(1) - (0 - 7π/4)) - (0 - ((7/4)(0) - 0))
= 7 + 7π/4

Wolfram shows 7 - 7π/4, I can't seem to find my arithmetic error

www.wolframalpha.com/input/?i=%E2%88%AB7cos(x)-7cos%5E2(x)dx+from+0+to+%CF%80%2F2