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 Your area is
a rectangle with base 2 and height 2
a triangle with base 2 and height 2
Looks like 4+2=6 to me
I suspect a typo, since f(x) is not used for x<2
#2
The region is a trapezoid with bases a and 3a, and height a
the area is (a+3a)/2 * a = 2a^2=18
a=3
#3
Each square has a base of height y=√(x+2)
So, adding up all those squares you get
∫[-2,1] y^2 dx = ∫[-2,1] x+2 dx = 4.5
#4
Assuming no typo, and reading it as
dx/dy = 1/2 x
dx/x = 1/2 dy
lnx = y/2
y = 2 lnx
the domain is x>0
#5 the area is
-[(f(4)+f(9))/2*(9-4) + (f(9)+f(11))/2*(11-9) + (f(11)+f(14))/2*(14-11) + (f(14)+f(15))/2*(15-14)]
= -[(6+11)/2*5 + (11+18)/2*2 + (18+21)/2*3 + (25-21)/2*1]
= -132
I don't see how the choices can be so small.
#6
not (x=c) because that would mean that f'(c)=0
we know nothing about f'
(d) is the best choice
#7
I get (b)
#8
You forgot the chain rule
(d) is correct
except for #5, if you disagree with me, show why.
If still confused, say why.
1) To evaluate the integral 6 * ∫[2](x) dx, we need to split the integral into three parts based on the given function f(x).
First, we evaluate the integral within the range -2 <= x <= 2:
For this range, f(x) is equal to |x|. The integral becomes 6 * ∫[-2](2) |x| dx. Since |x| is symmetric around the y-axis, the integral simplifies to 6 * 2 * ∫[0](2) x dx = 12 * ∫[0](2) x dx. Integrating x with respect to x gives us 12 * (x^2 / 2) evaluated from 0 to 2, which is 12 * (2^2 / 2 - 0^2 / 2) = 12 * (4 / 2) = 12 * 2 = 24.
Next, we evaluate the integral within the range 2 < x <= 4:
For this range, f(x) is equal to 2. The integral becomes 6 * ∫[2](4) 2 dx = 6 * 2 * ∫[2](4) dx = 12 * (x evaluated from 2 to 4) = 12 * (4 - 2) = 12 * 2 = 24.
Finally, we evaluate the integral within the range 4 < x <= 6:
For this range, f(x) is equal to -x + 4. The integral becomes 6 * ∫[4](6) (-x + 4) dx = 6 * ∫[4](6) -x dx + 6 * ∫[4](6) 4 dx. Integrating -x with respect to x gives us 6 * (-x^2 / 2) evaluated from 4 to 6, which is 6 * (-(6^2 / 2) + (4^2 / 2)) = 6 * (-18 + 8) = 6 * (-10) = -60. Integrating 4 with respect to x gives us 6 * (4x evaluated from 4 to 6) = 6 * (4 * (6 - 4)) = 6 * (4 * 2) = 6 * 8 = 48.
Adding the results from the three ranges together: 24 + 24 + (-60 + 48) = 72 - 12 = 60.
Therefore, the correct answer is 60, not 0.512.
2) To find the value of a that makes the area of region R equal to 18 square units, we need to set up an equation using the given information.
The region R is bounded by the x-axis, the curve y = 2x + a, and the line x = a. In the first quadrant, the region is bounded from x = 0 to x = a (since it is the first quadrant).
The area of a region can be found by integrating the difference between the upper and lower bounds with respect to x. In this case, we integrate (2x + a) - 0 = 2x + a from x = 0 to x = a. This gives us the equation:
∫[0](a) (2x + a) dx = 18.
Integrating 2x + a with respect to x gives us x^2 + ax, evaluated from 0 to a. This simplifies to (a^2 + a * a) - (0^2 + a * 0) = a^2 + a^2 = 2a^2.
Thus, we have the equation 2a^2 = 18. Solving for a, we get:
a^2 = 18 / 2
a^2 = 9
a = ±3.
Since a must be greater than 0 (as given in the question), we have a = 3.
Therefore, the correct answer for the value of a is 3, not 4.242.
The same method can be used for the other questions.