Let M be the region under the graph of f(x) = 3/e^x from x=0 to x=5.
A. Find the area of M.
B. Find the value of c so that the line x=c divides the region M into two pieces with equal area.
C. M is the base of a solid whose cross sections are semicircles whose diameter lies in the xy plane. The cross sections are perpendicular to the x-axis. Find the volume of this solid.
A. I could ask my mathematician friend for help, but why bother when I can just make a joke. The area of M is so small that it's practically microscopic. You might need a microscope to see it!
B. Ah, the elusive value of c. It's like trying to find Waldo in a sea of striped shirts. But fear not, my friend! With some clever math, you'll find c and divide that region like a boss.
C. A solid with semicircular cross sections? Now we're in the realm of circular logic. I mean, literally, we're talking about circles here. So imagine the volume of this solid as a circle of never-ending jokes – it just keeps getting bigger and bigger. The volume is like laughter, it never truly ends!
A. To find the area of region M, we need to evaluate the definite integral of the function f(x) over the interval [0, 5].
∫(0 to 5) 3/e^x dx
To solve this integral, we can apply the power rule of integration:
∫e^k dx = e^k + C,
where C is the constant of integration.
Applying this rule, we get:
∫(0 to 5) 3/e^x dx = 3 * ∫(0 to 5) e^(-x) dx
Now let u = -x. Then, du = -dx.
This changes the limits of integration and the integral becomes:
-3 * ∫(-5 to 0) e^u du
= -3 * [e^u] from -5 to 0
= -3 * (e^0 - e^-5)
= -3 * (1 - 1/e^5)
= -3 + 3/e^5.
Therefore, the area of region M is given by -3 + 3/e^5 (since the area cannot be negative, we disregard the negative sign).
B. To find the value of c that divides region M into two equal areas, we need to find the value of c that satisfies the equation:
∫(0 to c) 3/e^x dx = ∫(c to 5) 3/e^x dx.
Let's first evaluate both integrals:
∫(0 to c) 3/e^x dx = -3 + 3/e^c,
∫(c to 5) 3/e^x dx = -3 + 3/e^5.
Now, equating the two integrals:
-3 + 3/e^c = -3 + 3/e^5.
To simplify, we can cancel out the terms -3:
3/e^c = 3/e^5.
Now, we can cross multiply:
e^c = e^5,
which implies c = 5.
Therefore, the value of c that divides region M into two pieces with equal area is c = 5.
C. The solid described is a solid of revolution. To find its volume, we need to integrate the areas of the cross sections.
Since the cross sections are semicircles with diameters lying in the xy plane and perpendicular to the x-axis, the radius of each semicircle is given by f(x) = 3/e^x.
The volume of the solid can be found by integrating the area of each cross section over the interval [0, 5]:
V = ∫(0 to 5) A(x) dx,
where A(x) represents the area of each cross section.
The area of each cross section can be calculated by finding the area of a full circle and dividing it by 2:
A(x) = 1/2 * π * (f(x))^2.
Substituting the value of f(x) into the equation:
A(x) = 1/2 * π * (3/e^x)^2,
A(x) = 1/2 * π * 9/e^(2x).
Now, we can integrate A(x) over the interval [0, 5]:
V = ∫(0 to 5) 1/2 * π * 9/e^(2x) dx.
To solve this integral, let u = 2x. Then, du = 2 dx.
This changes the limits of integration and converts the integral to:
V = 1/4 * π * 9 ∫(0 to 10) e^(-u) du.
Using the power rule of integration, we get:
V = 1/4 * π * 9 * [- e^(-u)] from 0 to 10,
V = 1/4 * π * 9 * (- e^(-10) + 1).
Thus, the volume of the solid is given by 9/4 * π * (1 - e^(-10)).
To find the area of region M, we need to integrate the function f(x) = 3/e^x from x=0 to x=5.
A. Area of M:
First, we calculate the integral of f(x) over the given interval:
∫[0,5] (3/e^x) dx
Integrating this function, we get:
-3e^(-x) from 0 to 5
Now, we substitute the upper and lower limits into the function:
-3e^(-5) - (-3e^(0))
This simplifies to:
-3e^(-5) + 3
So, the area of M is -3e^(-5) + 3 square units.
B. To find the value of c such that the line x=c divides the region M into two equal areas, we need to set up an equation and solve it.
Let's denote the area of the two parts as A1 and A2. We want to find c such that A1 = A2.
We have:
A1 = ∫[0,c] (3/e^x) dx
A2 = ∫[c,5] (3/e^x) dx
Using the same formula for integration as before, we have:
A1 = -3e^(-x) from 0 to c
A2 = -3e^(-x) from c to 5
Now, set up the equation:
A1 = A2
Either by substitution or simplification, we get:
-3e^(-c) + 3 = -3e^(-5) + 3
Simplifying this equation, we get:
e^(-c) = e^(-5)
Taking the natural logarithm (ln) of both sides:
-ln(e^c) = -ln(e^5)
Simplifying further, we have:
-c = -5
Finally, we solve for c:
c = 5
Therefore, the value of c that divides region M into two equal areas is c = 5.
C. To find the volume of the solid, we use the given information that the cross sections are semicircles perpendicular to the x-axis, with their diameters lying in the xy-plane.
The volume of the solid can be expressed as the integral of the areas of the semicircles with respect to x, over the interval [0, 5]. Let's denote the radius of each semicircle as r.
Since the cross sections are semicircles, the area of each cross section is given by the formula 1/2 * π * r^2.
We need to find an expression for r in terms of x. The diameter of each semicircle lies in the xy-plane and is given by the function f(x) = 3/e^x.
Since the diameter is twice the radius, we have:
2r = f(x)
Simplifying the equation above, we get:
r = (3/e^x) / 2
Now, we can find the volume of the solid by integrating the areas of the semicircles over the interval [0, 5]:
V = ∫[0,5] (1/2 * π * [(3/e^x) / 2]^2) dx
Simplifying and evaluating this integral gives us the volume.
A
a = ∫[0,5] 3/e^x dx = 3 - 3/e^5
B now you want c such that
∫[0,c] 3/e^x dx = ∫[c,5] 3/e^x dx
3 - 3/e^c = 3/e^c - 3/e^5
6/e^c = 3 + 3/e^5
e^c = 2/(1+e^-5)
c = ln 2/(1+e^-5)
C
each semicircle has diameter equal to y. Adding up all those thin slices of thickness dx, we have
v = ∫[0,5] π/4 9/e^(-2x) dx = 9π/8 (1-e^-10)