1) Find the area of the region bounded by the curves y=arcsin (x/4), y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.
2)Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -1.
3)The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x^2. Cross sections of the solid perpendicular to the x-axis are squares. What is the volume, in cubic units, of the solid?
4)Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 - 9x on the interval [-1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
I've been trying to use the notes and past problems as a stepping stool to help me figure these out but I'm so lost. Especially number 4
Mean Value Theorem for Integrals. A variation of the mean value theorem which guarantees that a continuous function has at least one point where the function equals the average value of the function.
Mean value= (f(1)+f(-1))/2
= (-8+8)/2=0
if f(x)=0=x^3-9x, the the one point is x=0
#1
∫[0,4] arcsin(x/4) dx = 2π-4
To do this one, let
x = 4sin(u)
arcsin(x/4) = arcsin(sin(u)) = u
dx = 4cos(u) du
Now you have
∫[0,4] 4u cos(u) du
That you can easily do using integration by parts. Note that in changing variables,
∫[0,4] f(x) dx = ∫[0,π/2] g(u) du
#2
Using discs of thickness dx,
v = ∫[1,e] π(R^2-r^2) dx
where R=2 and r=lnx+1
v = ∫[1,e] π(4-(lnx + 1)^2) dx
Using shells of thickness dy,
v = ∫[0,1] 2πrh dy
where r = y+1 and h=x-1=e^y-1
v = ∫[0,1] 2π(y+1)(e^y-1) dy
#3 Each square has base equal to the distance between the curves, so adding up all the areas,
v = ∫[0,1] (x-x^2)^2 dx = 1/30
1) To find the area of the region bounded by the curves y = arcsin(x/4), y = 0, and x = 4 by integrating with respect to y, we need to set up the definite integral.
Step 1: Draw a graph of the given curves to visualize the region.
Step 2: Determine the limits of integration in terms of y. The region is bounded by y = 0 and y = arcsin(x/4). We need to find the x-values that correspond to these y-values.
For y = 0, we can see that x takes values from -4 to 4. So the limits of integration for y = 0 are x = -4 and x = 4.
For y = arcsin(x/4), it is convenient to rewrite it in terms of x as x = 4*sin(y). This implies that y takes values from 0 to the maximum value of arcsin(x/4). To find this maximum value, we set x = 4 and solve for y:
4*sin(y) = 4
sin(y) = 1
y = π/2
So the limits of integration for y = arcsin(x/4) are x = 4*sin(y) and x = 4.
Step 3: Set up the definite integral. The area can be found by integrating the difference between the upper curve (arcsin(x/4)) and the lower curve (y = 0) with respect to y over the specified limits.
∫[from y = 0 to y = π/2] [(4*sin(y)) - 0] dy
Step 4: Integrate the function with respect to y. The antiderivative of 4*sin(y) is -4*cos(y).
∫[from y = 0 to y = π/2] -4*cos(y) dy
2) To set up the integral that gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = -1, we can follow these steps:
Step 1: Draw a graph of the given curves to visualize the region.
Step 2: Determine the limits of integration in terms of x. The region is bounded by y = 1, y = Ln(x), and x = 1. We need to find the y-values that correspond to these x-values.
For x = 1, we can see that y takes values from -∞ to 1. So the limits of integration for y = 1 are y = -∞ and y = 1.
For y = Ln(x), we can rewrite it in terms of x as x = e^y. This implies that y takes values from 0 to the maximum value of Ln(x). To find this maximum value, we set x = 1 and solve for y:
e^y = 1
y = 0
So the limits of integration for y = Ln(x) are y = 0 and y = Ln(x).
Step 3: Set up the definite integral. The volume can be found by revolving the region between the curves around the line y = -1, so we need to use the method of cylindrical shells.
The integral can be set up as:
∫[from y = -∞ to y = 1] 2π(r)(h) dy
where r represents the distance from the axis of revolution (y = -1) to the curve Ln(x), and h represents the height of the cylindrical shell.
Step 4: Evaluate the integral using the limits of integration and the appropriate functions for r and h.
3) To find the volume of the solid with square cross sections, we can follow these steps:
Step 1: Draw a graph of the given curves to visualize the solid.
Step 2: Determine the limits of integration in terms of x. The solid is bounded by y = x and y = x^2 in the first quadrant. We need to find the x-values that correspond to these y-values.
For y = x, we can see that x takes values from 0 to the intersection point with y = x^2. Setting the equations equal to each other gives:
x = x^2
x^2 - x = 0
x(x - 1) = 0
The intersection points are x = 0 and x = 1. So the limits of integration for y = x are x = 0 and x = 1.
Step 3: Set up the definite integral. The volume can be found by integrating the area of each cross section perpendicular to the x-axis.
The area of a square is given by side length squared, so the area of each cross section is (x - x^2)^2.
The integral can be set up as:
∫[from x = 0 to x = 1] (x - x^2)^2 dx
Step 4: Evaluate the integral using the limits of integration.
4) To determine if the Mean Value Theorem for Integrals applies to the function f(x) = x^3 - 9x on the interval [-1, 1], we need to check if the function satisfies the conditions of the theorem.
The Mean Value Theorem for Integrals states that if a function f(x) is continuous over a closed interval [a, b], then there exists a number c in the interval (a, b) such that the definite integral of f(x) over the interval [a, b] is equal to f(c) multiplied by the length of the interval (b - a).
Step 1: Check if the function f(x) = x^3 - 9x is continuous over the interval [-1, 1].
The function is a polynomial, which is continuous everywhere. Therefore, it is continuous over the interval [-1, 1].
Step 2: Find the average value of f(x) over the interval [-1, 1].
The average value of a function f(x) over the interval [a, b] is given by:
average value = (integral of f(x) from a to b) / (b - a)
In this case, the average value of f(x) over the interval [-1, 1] is:
average value = (integral of f(x) from -1 to 1) / (1 - (-1))
Step 3: If the average value matches the function value at some point c in the interval, then the Mean Value Theorem for Integrals applies.
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we solve the equation:
f(c) = average value
In this case, we need to solve the equation:
c^3 - 9c = average value
If there exists a solution to this equation within the interval [-1, 1], then the Mean Value Theorem for Integrals applies.