given the graph of f(x) = x sinx, 0<=x<=2pi
assuming that a quantity y changes at a rate of y' = xsinx, find by how much it will increase or decrease over 3pi/2 <= x <= 2pi
the area under the graph from 0 to pi/2 is 1
the area under the graph from pi/2 to pi is pi-1
the area under the graph from pi to 3pi/2 is pi+1
and the area under the graph from 3pi/2 to 2pi is 2pi-1
To find how much the quantity y changes over the interval 3pi/2 <= x <= 2pi, we need to find the area under the graph of f(x) = x sin(x) over this interval.
To do this, we can integrate the function f(x) over the interval [3pi/2, 2pi].
The definite integral of f(x) from a to b is denoted by ∫[a,b] f(x) dx. In this case, we want to find ∫[3pi/2, 2pi] x sin(x) dx.
To evaluate this integral, we can use integration techniques like integration by parts or integration by substitution. In this case, integration by parts would be suitable.
The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Let's assign the following values for u and dv:
u = x
dv = sin(x) dx
Differentiating u with respect to x gives du/dx = 1, and integrating dv with respect to x gives v = -cos(x).
Now let's apply the integration by parts formula:
∫[3pi/2, 2pi] x sin(x) dx = -x cos(x) - ∫[3pi/2, 2pi] (-cos(x)) dx
Evaluating the integral of -cos(x) dx gives -sin(x), so the expression becomes:
-x cos(x) + sin(x) evaluated from 3pi/2 to 2pi
To evaluate this, we substitute the upper and lower limits into the expression:
(-(2pi) cos(2pi) + sin(2pi)) - (-(3pi/2) cos(3pi/2) + sin(3pi/2))
The cosine of 2pi is 1 and the sine of 2pi is 0, so the first term becomes:
(-(2pi) * 1 + 0)
And the cosine of 3pi/2 is 0 and the sine of 3pi/2 is -1, so the second term becomes:
-(-(3pi/2) * 0 + (-1))
Simplifying further, we have:
-2pi
Therefore, the quantity y will decrease by 2pi over the interval 3pi/2 <= x <= 2pi.
To find how much the quantity y will increase or decrease over the interval 3π/2 ≤ x ≤ 2π, we need to find the definite integral of y' = x sin(x) over this interval.
Integrating y' with respect to x gives us the antiderivative or primitive function of y:
y = ∫ (x sin(x)) dx
Now, let's find the integral of x sin(x) over the interval 3π/2 ≤ x ≤ 2π:
∫ (x sin(x)) dx = [ -x cos(x) + sin(x) ] evaluated from 3π/2 to 2π
Plugging in the upper limit (2π) into the antiderivative:
[ -2π cos(2π) + sin(2π) ]
And plugging in the lower limit (3π/2) into the antiderivative:
[ -3π/2 cos(3π/2) + sin(3π/2) ]
Simplifying, we have:
[ -2π cos(2π) + sin(2π) ] - [ -3π/2 cos(3π/2) + sin(3π/2) ]
Since cos(2π) = cos(0) = 1 and sin(2π) = sin(0) = 0, the first term becomes:
[ -2π(1) + 0 ]
And since cos(3π/2) = 0 and sin(3π/2) = -1, the second term becomes:
[ -3π/2(0) + (-1) ]
Simplifying further, we get:
[ -2π(1) + 0 ] - [ -3π/2(0) + (-1) ]
= [ -2π ] - [ -1 ]
= -2π + 1
= -2π + 6π/6
= 4π/6
= 2π/3
Therefore, over the interval 3π/2 ≤ x ≤ 2π, the quantity y will decrease by 2π/3.