MATH 220: WORKSHEET FOR 10/30 SOLUTIONS
(1) Evaluate
Z 3 the following integrals:
(x2 + 3/x2 ) dx
(a)
2
All parts of this problem use the FTC, part 2.Z
3
2
2
3
x + 3/x has antiderivative (1/3)x − 3/x. So
(x2 + 3/x2 ) dx = ((1/3)33 − 3/3) −
2
((1/3)23 − 3/2) = (9 − 1) − (8/3 − 3/2) = 41/6.
Z π/4
sec2 (θ) dθ
(b)
0
π/4
Z
2
sec2 (θ) dθ = tan(π/4)−tan(0) = 1−0 = 1
sec (θ) has antiderivative tan(θ). So
0
Z
1
(c)
0
1
t2 +1
t2
dt
+1
1
Z
has antiderivative arctan(t). So
0
t2
dt
= arctan(1) − arctan(0) = π/4 − 0 =
+1
π/4.
Z 1
(d)
5y dy
−1
y
y
Z
1
5 has antiderivative 5 / ln(5). So
−1
5y dy =
1
1
(5 − ).
ln(5)
5
(2) EvaluateZ the following derivatives:
x
d
3
(a)
et +sin(t) dt
dx 0
All Zparts of this problem use the FTC, part 1.
x
d
3
3
et +sin(t) dt = ex +sin(x) (using the FTC).
dx 0
Z
d tan(3t)
(b)
cos2 (x) dx
dt 1
t
Z
cos2 (x) dx. Then the function we are trying to take the derivative
Let g(t) =
1
of is g(tan(3t)). By the chain rule, this has derivative 3g 0 (tan(3t)) sec2 (3t). By the
FTC, this is just 3 cos2 (tan(3t)) sec2 (3t).
Z 10
d
(c)
ln(8s) ds
dy (ln(y))2
Z
10
Z
(ln(y))2
ln(8s) ds.
ln(8s) ds = −
10
R (ln(y))2
d
Now, we can proceed as we did in the previous step: dy
− 10
ln(8s) ds =
2
− ln(8 ln(y) )(2 ln(y))/y.
Z cos2 (θ)
d
xsec(x) dx
(d)
dθ sin2 (θ)
We first need the property of integrals
(ln(y))2
This is the trickiest thus far. We again rely on our property of integrals:
Z sin2 (θ)
Z cos2 (θ)
Z cos2 (θ)
Z 1/2
Z cos2 (θ)
sec(x)
sec(x)
sec(x)
sec(x)
xsec(x) dx
x
dx−
x
dx =
x
dx+
x
dx =
sin2 (θ)
sin2 (θ)
1/2
1/2
1/2
(one thing we should techincally be careful about is that our integrand is continuous
over our bounds of integration; it is).
Z cos2 (θ)
d
2
So, using the strategy of the last few problems,
xsec(x) dx = (cos2 (θ))sec(cos (θ)) (2 cos(θ)(−
dθ 1/2
Z sin2 (θ)
d
2
while
xsec(x) dx = (sin2 (θ))sec(sin (θ)) (2 sin(θ))(cos(θ)). Putting these two
dθ 1/2
together gives us the total derivative.
(3) Water is being drained out of a pond at a rate of 1600 − t2 gallons per hour at t hours.
Find the number of gallons drained from the pond between 1 and 3 hours.
If V (t) represents the volume of water in the pool at time t, then V 0 (t) = 1600 − t2 .
So we want
Z 3
V (3) − V (1) =
V 0 (t) dt = (1600 · 3 − (1/3)33 ) − (1600 · 1 − (1/3)13 ) ≈ 3191.3
1
(4) Suppose the volume of an irregular solid measured from it’s bottom at height 0 to height
1
h is changing at a rate of V 0 (h) = h3 − h+2
+ e2h . If the total height of the solid is 3, find
the total volume of the solid (this is the root of an idea we will explore more in depth
later later).
Z 3
1
Since V (0) = 0, this is given by V (3) = V (3) − V (0) =
(h3 −
+ e2h ) dh =
h
+
2
0
(1/4)34 − ln(5) + (1/2)e6 + ln(2) − (1/2).
(5) By measuring the heat produced by the reaction, you surmise that the rate of reaction
d
t = 1/(5t + 1) for t measured in seconds. Find how much
producing a chemical C is d[C]
the concentration [C] changes between 2 seconds and 10 seconds.
Z 1
This is given via the net change theorem by
01/(5t + 1) dt = (1/5) ln(5 · 10 + 1) −
2
r !
5 51
(1/5) ln(5 · 2 + 1) = ln
.
11
(6) Suppose a particle travels with a velocity v(t) = sin(2πt) in meters per second.
(a) Find the net change in position of the particle between t = 0 and t = 2.
Z 2
The net change is given by
v(t) dt = −(1/2) cos(4π) − (−(1/2) cos(0)) = 0.
0
(b) Find the total distance the particle travels between t = 0 and t = 2.
To do this, we need to know where the velocity is negative and where it is positive.
sin(θ) is positive for 0 < θ < π and 2π < θ < 3π and negative for π < θ < 2π and
3π < θ < 4π. So, when 0 < θ < π and 2π < θ < 3π, the position is increasing
and when 0 < θ < π and 2π < θ < 3π, the position is decreasing; we need to
Z 1/2
integrate each piece separately;
sin(2πt) dt = −(1/2) cos(π) + (1/2) cos(0) = 1.
0
Z 1
sin(2πt) dt = −(1/2) cos(2π) + (1/2) cos(π) = −1, so the particle travels a
1/2
distance of 1 meter in this time period.
Z
3/2
sin(2πt) dt = 1
Using the fact that sin(2π(t + 1)) = sin(2πt), we know that
1
Z 2
sin(2πt) dt − 1, so the particle travels a total of 4 meters (though, by part
and
3/2
a, we know it ends up exactly where it started.)