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
This is just a problem in integrals. We have y', and we want to add up all the small changes given by that function.
Int[x sinx dx] can be solved using integration by parts.
Recall the product rule of derivatives:
(uv)' = u'v + uv'
so
uv' = (uv)' - u'v
Going the other direction,
Int[u dv] = uv - Int[v du]
So, we want to split up the integrand into two factors, where one part (u) gets simpler after differentiation, and the other part (dv) can be easily integrated.
Here, we have x sinx
If u = x, then du = dx
If dv = sinx, then v = -cosx
Int[x sinx dx] = uv - Int[v du]
= -x cosx - Int[-cos x * dx]
Now we have a simple integrand, -cosx
Int cos x dx = sin x
So, the final integration is
Int[x sinx dx] = uv - Int[v du]
= -x cosx - Int[-cos x * dx]
= -x cosx + sin x
Evaluating at 3pi/2 and 2pi, we have
(-2pi * 1 + 0) - (-3pi/2 * 0 + -1) = -2pi + 3pi/2 + 1 = 1 + 7pi/2
To find how much the quantity y will increase or decrease over the interval 3π/2 ≤ x ≤ 2π, we need to calculate the definite integral of y' = xsinx over this interval.
Step 1: Calculate the antiderivative of y' = xsinx.
The antiderivative (also known as the indefinite integral) of xsinx can be found using integration by parts. Integrating by parts twice will give us the final answer. The integration by parts formula is:
∫ u dv = u*v - ∫ v du
Start by assigning u = x and dv = sinx. Then, find du and v:
du = dx
v = -cosx
Using the integration by parts formula, integrate xsinx:
∫ x*sinx dx = x*(-cosx) - ∫ (-cosx) dx
= -x*cosx + ∫ cosx dx
= -x*cosx + sinx + C
Step 2: Apply the Fundamental Theorem of Calculus.
According to the Fundamental Theorem of Calculus, to find the definite integral of y' = xsinx over the interval 3π/2 ≤ x ≤ 2π, we can subtract the antiderivative evaluated at the lower bound from the antiderivative evaluated at the upper bound:
∫[3π/2, 2π] xsinx dx = (-x*cosx + sinx)|[3π/2, 2π]
Step 3: Evaluate the antiderivative at the upper and lower bounds.
Substitute the upper bound (2π) into the antiderivative formula:
(-2π*cos(2π) + sin(2π))
= (-2π*cos(0) + sin(0))
= -2π(1) + 0
= -2π
Now substitute the lower bound (3π/2) into the antiderivative formula:
(-3π/2*cos(3π/2) + sin(3π/2))
Since cos(3π/2) = 0 and sin(3π/2) = -1:
= (-3π/2*0 + (-1))
= -(-1)
= 1
Step 4: Calculate the difference.
Now subtract the value at the lower bound from the value at the upper bound:
(-2π) - (1) = -2π - 1
Therefore, the quantity y will decrease by -2π - 1 over the interval 3π/2 ≤ x ≤ 2π.