Math 125 Quiz 1
1. Let f(t) =
ln(cos(t)) sin(t)
0
. Compute f (t).
2
t +1
2. Suppose f(x) = 8x3 −5xπ +7 . Find a particular antiderivative of f(x).
1
ln(cos(t)) sin(t)
0
. Compute f (t).
2
t +1
Let g(t) = ln(cos(t)), h(t) = sin(t) and l(t) = t2 + 1.
1. Let f(t) =
Then
0
• g (t) = −
sin(t)
= − tan(t).
cos(t)
0
• h (t) = cos(t).
0
• l (t) = 2t
0
0
0
(g (t) h(t) + g(t) h (t))l(t) − g(t) h(t) l (t)
.
• f (t) =
l2(t)
0
Therefore, putting it all together
0
f (t) =
(− tan(t) sin(t) + ln(cos(t)) cos(t)) (t2 + 1) − (ln(cos(t)) sin(t)) 2t
(t2 + 1)2
2. Suppose f(x) = 8x3 −5xπ +7 . Find a particular antiderivative of f(x).
F (x) = 2x4 −
5xπ+1
π+1
1
A rocket is taking off, going straight up. At time t, its height s(t) is given
by the formula
Z
q
πt
4 sin2 x + 1 dx
s(t) =
0
1. Estimate the height s( 21 ) of the rocket at time t =
hand sum with n = 3 equal subintervals.
1
2
by using a right-
2. Find the velocity v(t) of the rocket as a function of time t.
1
A rocket is taking off, going straight up. At time t, its height s(t) is given
by the formula
Z
q
πt
4 sin2 x + 1 dx
s(t) =
0
1. Estimate the height s( 21 ) of the rocket at time t =
hand sum with n = 3 equal subintervals.
Z π q
1
2
s( ) =
4 sin2 x + 1 dx.
2
0
π
π
∆x = 2 = .
3
6
x0 = 0, x1 = π6 , x2 = π3 , x3 = π2 .
If f(x) =
=
q
q
1
2
by using a right-
4 sin2 x + 1 then R3 = f(x1) π6 + f(x2 ) π6 + f(x3 ) π6 =
4 sin2 ( π6 ) + 1 ·
π
6
+
q
4 sin2 ( π3 ) + 1 ·
π
6
+
q
4 sin2( π2 ) + 1 ·
2. Find the velocity v(t) of the rocket as a function of time t.
√
v(t) = s0 (t) = 4 sin2 t + 1 · π
1
π
6
≈ 2.96
Calculate the following integrals.
1.
Z
2
1
2.
Z
0
3.
Z
√
x x − 1 dx
4
|x2 − 9| dx
cot x dx
1
Calculate the following integrals.
1.
Z
2
1
√
x x − 1 dx
2. Use substitution:
u=x−1
du = dx
x=u+1
the integral becomes:
Z 1
Z 1
Z 1
√
√
√
3
1
u u + u du =
u 2 + u 2 du =
(u + 1) u du =
0
0
=
3.
Z
2 5 2 3
u2 + u2
5
3
4
1
=
0
0
16
15
|x2 − 9| dx The function y = x2 − 9 is a parabola with graph :
0
Therefore we need to break the integral and calculate
Z
0
4.
Z
3
2
9 − x dx +
Z
3
4
3
4
x3 x3
64
x − 9 dx = (9x − ) + ( − 9x) =
3 0
3
3
3
2
cot(x) dx
Rewrite the integral as
Z
cos x
dx and use the subsitution:
sin x
u = sin x
du = cos x dx
The integral becomes
1
1
du = ln |u| + C
u
Therefore our answer is ln | sin x| + C
Z
2
A bag of sand originally weights 160 lb. It is lifted at a constant rate of
4 ft/min. The sand leaks out of the bag at a constant rate so that when it
has been lifted 20ft only half the sand is left. How much work is done lifting
the bag 20 ft ?
1
A bag of sand originally weights 160 lb. It is lifted at a constant rate of
4 ft/min. The sand leaks out of the bag at a constant rate so that when it
has been lifted 20ft only half the sand is left. How much work is done lifting
the bag 20 ft ?
Solution: W = F ∗ d d = 20; F is the weight of the bag , but it it not
constant, so we need to ”slice” and set up a Rieman sum.
Let’s denote by F (xi) the weight of the bag xi feet above the ground . We
want to find a formula for F (xi); we know that F (0) = 160 and F (20) = 80.
So the bag has leaked 80 lb in 20 feet.We know that the bag leaks sand at a
constant rate, so this rate must be 80/20 lb per foot or 4 lb /ft. Therefore
F (xi) = 160 − 4xi (160 is the original weight, 4xi is the weight of the sand
leaked out while the bag has been lifted xi feet).
Let’s calculate the work done to lift the bag a small height ∆x from xi
feet above the ground to xi + ∆x feet above the ground. We can make the
approximation that the weight F (xi) of the sand bag stays constant during
this small movement, therefore the small amount of work done is
∆W = F (xi) · ∆xi = (160 − 4xi ) · ∆x. So the work done is approximately
Pn
i=1 (160 − 4xi ) · ∆x.
ToZ find the total work we need to integrate:
20
W =
0
160 − 4x dx = 2400 lb · ft.
2