PLZ show work,
find the rieman sum of 6 intervals of sinx from (0,pie)
also find the absolute min and max of ln(2+sinx) for pie<(or equal to) x >(or equal to) 2pie
Sure, let's break down the problem step by step.
1. Riemann Sum of sin(x):
To find the Riemann sum of the function sin(x) over 6 intervals in the interval [0, π], we need to determine the width of each interval and evaluate the function at specific points within each interval.
First, let's calculate the width of each interval. The interval [0, π] has a total length of π - 0 = π. Since we need to divide this interval into 6 equal parts, the width of each interval would be (π - 0)/6 = π/6.
Next, we need to choose a representative point within each interval to evaluate the function. A common approach is to take the left endpoint of each interval.
So the left endpoints of the intervals are:
Interval 1: x = 0
Interval 2: x = π/6
Interval 3: x = 2π/6 = π/3
Interval 4: x = 3π/6 = π/2
Interval 5: x = 4π/6 = 2π/3
Interval 6: x = 5π/6
Now, we evaluate sin(x) at each of these points:
sin(0) = 0
sin(π/6) = 1/2
sin(π/3) = √3/2
sin(π/2) = 1
sin(2π/3) = √3/2
sin(5π/6) = 1/2
Finally, we calculate the Riemann sum by summing the products of the function values and the interval widths:
Riemann sum = (π/6)(0) + (π/6)(1/2) + (π/6)(√3/2) + (π/6)(1) + (π/6)(√3/2) + (π/6)(1/2)
= π/12 + π√3/12 + π/6 + π√3/12 + π/12
= (2π + π√3)/6
Therefore, the Riemann sum of sin(x) over 6 intervals in the interval [0, π] is (2π + π√3)/6.
2. Absolute min and max of ln(2+sin(x)) on [π, 2π]:
To find the absolute minimum and maximum values of ln(2+sin(x)) in the interval [π, 2π], we need to find the critical points and evaluate the function at those points, as well as the endpoints of the interval.
Let's find the derivative first. The derivative of ln(2+sin(x)) with respect to x can be calculated using the chain rule:
d/dx [ln(2+sin(x))] = (1/(2+sin(x))) * cos(x)
To find the critical points, we need to solve the equation (1/(2+sin(x))) * cos(x) = 0. Notice that this equation can be satisfied if either (1) cos(x) = 0 or (2) 2+sin(x) = 0.
(1) cos(x) = 0:
We know that cos(x) = 0 when x = π/2.
(2) 2+sin(x) = 0:
Subtracting 2 from both sides gives sin(x) = -2. However, sin(x) always lies between -1 and 1, so there are no real solutions for this equation.
Next, we evaluate the function at the critical point obtained:
ln(2+sin(π/2)) = ln(2+1) = ln(3)
Finally, evaluate the function at the endpoints of the interval:
ln(2+sin(π)) = ln(2+0) = ln(2) = 0
ln(2+sin(2π)) = ln(2+0) = ln(2) = 0
Therefore, the absolute minimum value of ln(2+sin(x)) on the interval [π, 2π] is 0, and the absolute maximum value is ln(3).