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 answer
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.