Use the midpoint rule with n=4 to approximate the region bounded by y=-x^3 and y=-x
Show that the function f(x)= (integral from 2x to 5x) (1)/(t) dt is constant on the interval (0,+infinity)
(-1,-1), (0,0) and (1,1). The four midpoints are
-3/4, -1/4, +1/4, and +3/4. Each vertical strip has a width of 1/2. The lengths of the 4 strips are:
-27/64-(-3/4); -1/64-(-1/4); 1/4 - 1/64; and 3/4-27/64;
simplifying these, you get
21/64, 15/64, 15/64, and 21/64.
So the area is something like
(1/2)*(21+15+15+21)/64 = 36/64 = 9/16. MY ANSWER TO question 1
For the first one i think the integral is from -1 to 1 for the two functions, because they intersect at (-1,1) (0,0) and (1,-1)
since -x^3 < -x on (-1,0), you should be subtracting (-x)-(-x^3). the actual math should be
-(-3/4) - (-(-3/4)^3) = 3/4 - 27/64 = 21/64
and so on. It comes out the same.
Assuming your intermediate results are good, the answer looks good.
Q1:
There are three intersection points: (-1,1), (0,0), (1,-1)
Thus the endpoints are: (-1,1), (-0.5,0.5), (0,0), (0.5,0.5), (1,1)
Thus, the midpoints are: (-0.75,0.75), (-0.125,0.125), (0.125,-0.125), (0.75,-0.75)
The area is as follows:
0.75(-0.5) + (0.25(-0.5)) + -0.25(0.5) + -0.75(0.5)
= -1
I took the area of the whole thing instead of taking the area of the 2 regions separately because I assumed that is what the question is asking
Q2:
If we evaluate the derivative of the function, we get:
int of [ln(t)] from 2x to 5x which results in
ln|5x| - ln|2x|
Taking the derivative of this, we get:
d/dx (ln(5|x|) - ln(2|x|))
d/dx (ln(5|x|/2|x|))
d/dx (ln(5/2))
This equals 0
Since the derivative of a function gives the slope of the tangent line at any given point, we know the above function is constant on the given interval because the derivative is 0 which indicates the function neither decreases nor increases which means it remains constant
To approximate the region bounded by y=-x^3 and y=-x using the midpoint rule, we divide the interval [a, b] into n subintervals of equal width. In this case, a = -1 and b = 0. So, the interval is [-1, 0].
Step 1: Calculate the width of each subinterval using the formula ∆x = (b - a) / n.
∆x = (0 - (-1)) / 4
∆x = 1/4
Step 2: Calculate the midpoints of each subinterval by adding half of the width to each starting point.
x1 = -1 + (1/4) * (1/2)
x2 = -1 + (1/4) * (3/2)
x3 = -1 + (1/4) * (5/2)
x4 = -1 + (1/4) * (7/2)
Simplifying, we get:
x1 = -3/8
x2 = -1/8
x3 = 1/8
x4 = 3/8
Step 3: Evaluate the function at each midpoint.
f(x1) = (-3/8)^3 = -27/512
f(x2) = (-1/8)^3 = -1/512
f(x3) = (1/8)^3 = 1/512
f(x4) = (3/8)^3 = 27/512
Step 4: Calculate the sum of all the function values multiplied by the width of each subinterval.
Approximation = ∆x * (f(x1) + f(x2) + f(x3) + f(x4))
= (1/4) * (-27/512 - 1/512 + 1/512 + 27/512)
= (1/4) * (0)
= 0
Therefore, the approximation of the region bounded by y=-x^3 and y=-x using the midpoint rule with n=4 is 0.
Now, let's move on to the second question:
To show that the function f(x) = ∫(2x to 5x) (1/t) dt is constant on the interval (0,+infinity), we need to evaluate the function for different values of x within the given interval and check if the values remain the same.
Step 1: Calculate the definite integral.
∫(2x to 5x) (1/t) dt = ln|t| from 2x to 5x
Simplifying, we get:
f(x) = ln|5x| - ln|2x|
Step 2: Evaluate the function for different values of x within the interval (0,+infinity).
f(1) = ln|5| - ln|2| = ln(5/2)
f(2) = ln|10| - ln|4| = ln(10/4) = ln(5/2)
f(3) = ln|15| - ln|6| = ln(15/6) = ln(5/2)
...
We can see that for any value of x within the interval (0,+infinity), the value of f(x) remains the same, ln(5/2). Therefore, the function f(x) is constant on the interval (0,+infinity).