Find the area of the region between the curves?

Question

y = sin(pix/2) and y=x?

so i know one of the steps is to solve x

sin(pix/2) = x
im not sure how you would solve for that.


i know after you find the intersection points you would find a number in between [a,b] that would tel you which of teh two y's are bigger.

then you fine the area by taking the integral of the bigger-smaller and then taking the derivative of that and solving F(b)-F(a) but i cant do anything without knowing the values of x and i just dont know how to solve with sin in the problem

it is not 2/pi - 0.5 and its not -2/pi -0.5

yea im still not getting it.


i tried

takin the derivative and got
-cos (pi/4)x^2 - x^2/2 and am still not getting the answer.

ahh!!

Answer 1

sin(pi*x/2) = x is not easily solvable, so we find
an "intuitive" way.
sin(pi/2) = 1, so if x = 1, this is the solution.
There is another solution too, x = -1.
And both curves go through the origin.

You'll only need to find the area between the curves
that are above the X-axis, because the area below
the X-axis is the same, so just multiply your result by 2.

EDIT: you can solve the equation using Newton's iterative
method. Let f(x) = sin(pi*x/2) - x. Alternatively, plug f(x) into
a graphing calculator and the zeros will become evident.

Answer 2

First discover the place the curves intersect by ability of putting y1 = y2: x^2+2x+a million = 2x+5 x^2+a million = 5 x^2 - 4 = 0 (x+2)(x-2) = 0 So the curves intersect at x=-2, and x=2. Now ensure which curve is greater interior the bounded area. picking a variety between -2 and a couple of, i exploit 0. y1 = 0^2 + 2*0 + a million = a million. y2 = 2*0 + 5 =5. So y2 is greater than y1 between -2 and a couple of. Now, take the indispensable of y2-y1 from -2 to 2. indispensable[ (2x+5) - (x^2 + 2x + a million) ] = indispensable[ 4 - x^2 ] = 4x - x^3/3 Evaluated from -2 to 2, we get (4*2 - 2^3/3) - (4*(-2) - (-2)^3/3) = (8 - 8/3) - (-8 + 8/3) = sixteen - sixteen/3 = 32/3.

Answer 3

first plot the graphs of these 2 functions

You know that sin func does not exceed 1 so these 2 curves will meet at (1,1) and in the interval of x [0,1] sin(pi/2 x) is greater than x verify this by differentiating [sin(pi/2 x) - x ] and the derivative is >0.Then find the integral of sin from 0 to 1 and subtract it from integral of X from 0 to 1

Answer 4

Well you have to do some intepretation

If x is say, 0

sin(0*pi/2) = 0

This is true

-So we know that 0 is one of our boundaries

For the other one, lets go with 1 maybe?

sin(1*pi/2) = 1

This checks out so there we go! We have our region

so we want the integral of sin(pi*x/2) - x

-2/pi * cos(pi*x/2) - 0.5*x^2

Evaluated at 1 we get

-0.5*1^2 = -0.5

Evaluated at 0 we get

-2/pi

So our area is 2/pi - 0.5

0.1366197724

Answer 5

To find the limits of integration solve for x.

sin(πx/2) = x
x = 0, 1

Integrate from 0 to 1.

Area = ∫[sin(πx/2) - x] dx

= (-2/π)cos(πx/2) - x²/2 | [Evaluated from 0 to 1]

= (0 - 1/2) - (-2/π - 0) = -1/2 + 2/π ≈ 0.1366197