Can someone check my answers:
1) Use geometry to evaluate 6 int 2 (x) dx where f(x) = { |x|, -2 <= x <= 2}
{2, 2 < x <= 4}
{-x+4 4 < x <= 6}
for my answer I got 0.512
2) R is the first quadrant region enclosed by the x axis, the curve y = 2x+a and the line x = a where a > 0. Find the value of a so that the are of the region R is 18 square units.
a) 3
b) 3.772
c) 4.242 <-- my answer
d) 9
3) The base of a solid is bounded by y = √x+2, the x-axis and the line x = 1. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid.
a) 0.5
b) 3.464 <-- my answer
c) 4.5
d) None of the above
4) Find the domain of the particular solution to the differential equation dx/dy = 1/2x, with the initial condition y(1) = 1
a) x > 0
b) x < 0
c) x =/= 0 <-- my answer
d) All real numbers
5) The function f is continuous on the interval [4,15], with some of its values given in the table. Estimate the average value of the function using trapezoidal approximation.
x: 4 9 11 14 15
f(x): -6 -11 -18 -21 -25
a) -12.727
b) -11.546 <-- my answer
c) -16.273
d) -13.909
6) If the graph of f"(x) is continuous and has a relative maximum of x = c, which of the following can be true"
a) The graph of f has an x-intercept of x = c
b) The graph of f has an inflection point at x = c
c) The graph of x has a relative maximum at x = c <-- my answer
d) None of the above is necessarily true
7) Which of the following shows that g(x) grows faster than f(x)?
a) lim x--> inf f(x)/g(x) = 1000 <-- my answer
b) lim x --> inf f(x)/g(x) = 0
c) lim x--> inf f(x)/g(x) = inf
d) None of these
8) Find F'(x) for F(x) = 2intx^3 sin(t^4) dt
a) sin(2^4) - sin(x^12)
b) sin(x^7)
c) -sin(x^12) <-- my answer
d) -3x^2sin(x^12)
Thank you!
1) To evaluate the integral of f(x) = |x|, -2 <= x <= 2, you need to break it down into different intervals based on the given piecewise function.
- In the interval -2 <= x <= 2, the function is |x|. Since |x| is an absolute value function, it can be split into two cases:
- For x >= 0, |x| = x.
- For x < 0, |x| = -x.
By graphing the function, you can see that |x| has a V-shaped graph centered at x = 0. The interval -2 <= x <= 2 covers the entire V shape, so you can split the integral into two parts:
Integrating x from -2 to 2:
∫[from -2 to 2] x dx = x^2/2 [from -2 to 2] = (2^2/2 - (-2^2/2)) = 4/2 - 4/2 = 0.
- In the interval 2 < x <= 4, the function is f(x) = 2.
Since the function is constant, the integral is simply the value of the function multiplied by the width of the interval:
∫[from 2 to 4] 2 dx = 2x [from 2 to 4] = (2*4) - (2*2) = 8 - 4 = 4.
- In the interval 4 < x <= 6, the function is f(x) = -x + 4.
To integrate this linear function, you can use the power rule for integration:
∫[from 4 to 6] (-x + 4) dx = -x^2/2 + 4x [from 4 to 6] = (-6^2/2 + 4*6) - (-4^2/2 + 4*4) = (-36/2 + 24) - (-16/2 + 16) = -18 + 24 + 8 - 8 = 6.
Now, add up the results from each interval:
0 + 4 + 6 = 10.
So, the correct answer to the first question is 10, not 0.512.
2) To find the value of a that gives the region R an area of 18 square units, you need to set up and solve an integral.
The region R is bounded above by the curve y = 2x + a and below by the x-axis. It is also bounded on the right by the line x = a. To find the area of R, you need to set up an integral and solve it for a.
The integral representing the area of R is:
∫[from 0 to a] (2x + a) dx.
To find the area, integrate the function:
∫[from 0 to a] (2x + a) dx = x^2 + ax [from 0 to a] = (a^2 + a*a) - (0^2 + 0*a) = a^2 + a^2 = 2a^2.
Set this expression equal to 18 and solve for a:
2a^2 = 18
a^2 = 9
a = ±√9.
Since a needs to be greater than 0 (as stated in the question), the value of a is 3.
So, the correct answer to the second question is a) 3.
3) To find the volume of the solid, you need to use the method of cross-sections.
The base of the solid is bounded by the curve y = √x + 2, the x-axis, and the line x = 1. The cross-sections, taken perpendicular to the x-axis, are squares.
To find the volume, you integrate the areas of these squares over the interval where x varies. Each cross-section is a square, so the area of each cross-section is just the square of a side length.
At each value of x, the side length of the square is given by the value of the function √x + 2. The integral is set up as follows:
∫[from 0 to 1] (√x + 2)^2 dx.
Simplifying, you get:
∫[from 0 to 1] (x + 4√x + 4) dx.
= x^2/2 + 4/3 * x^(3/2) + 4x [from 0 to 1]
= (1/2 + 4/3 + 4) - (0 + 0 + 0)
= 15/6 + 8/6 + 24/6
= 47/6.
So, the correct answer to the third question is: d) None of the above (47/6 is not equal to 3.464).
The explanation for the fourth, fifth, sixth, seventh, and eighth questions will be provided in the following responses.