Find the area bounded by the given curves y= 7cos(x) and y=7cos^2(x) between x=0 and x= π/2.
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)
I think you make a mistake with which one above and below
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.
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
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
To find the area bounded by the given curves, we can set up an integral and evaluate it.
First, let's find the points of intersection between the two curves:
Set y = 7cos(x) equal to y = 7cos^2(x):
7cos(x) = 7cos^2(x)
Divide both sides by 7:
cos(x) = cos^2(x)
Now, we can solve for x:
cos^2(x) - cos(x) = 0
Factor out cos(x):
cos(x) (cos(x) - 1) = 0
Now, we have two possibilities:
1. cos(x) = 0
To find the intersections in the interval [0, π/2], we know that cos(x) = 0 when x = π/2.
2. cos(x) - 1 = 0
To find the intersections in the interval [0, π/2], we know that cos(x) - 1 = 0 when x = 0.
Now that we have the points of intersection, we can set up the integral to find the area:
The area bounded by the curves y = 7cos(x) and y = 7cos^2(x) between x = 0 and x = π/2 can be calculated using the following integral:
A = ∫[0,π/2] (7cos^2(x) - 7cos(x)) dx
Simplify the integral:
A = 7∫[0,π/2] (cos^2(x) - cos(x)) dx
Use the trigonometric identity cos^2(x) = (1 + cos(2x))/2:
A = 7∫[0,π/2] ((1 + cos(2x))/2 - cos(x)) dx
Expand the integral:
A = (7/2)∫[0,π/2] (1/2 + cos(2x)/2 - cos(x)) dx
Now, evaluate the integral:
A = (7/2) [(1/2)x + (1/4)sin(2x) - sin(x)] from x = 0 to x = π/2
Substitute the upper limit:
A = (7/2) [(1/2)(π/2) + (1/4)sin(2(π/2)) - sin(π/2)]
Simplifying further:
A = (7/2) [(1/4)π + (1/4)sin(π) - 1]
Since sin(π) = 0, we can simplify the expression:
A = (7/2) [(1/4)π - 1]
Finally, compute the value of the expression:
A = (7/8)π - 7/2
The area bounded by the given curves between x = 0 and x = π/2 is (7/8)π - 7/2 square units.