_________________
Test Name: #4
Integrals
In-class portion
No Calculators
1. The easiest thing to do here would be to split it up. The first part needs u-substitution, but not the
second part.
4
4
4
t
t
0 2 t 21 2 1 dt =0 2 t 21 2 dt 0 1 dt
2
u = 2 t 1 ; du =4 t dt
1
9 u1
4
4
4
4
t =4
4
4
14
4t
1 t =4 1
1
dt 0 1 dt = t =0
du 0 1 dt = t=0 t0 = 2 t 21 1 t0
0 2 t 21 2
2
4
4
4
u
0
1
1
1
1
1
8
8
=
4 0 = 1 4 = 4 =4
4 2 4 2 1
4 33
33
33
2 0 2 1
Find the average value of this function over the interval [0, 4]
Divide the definite intergral we just found by the length of the interval (4-0 = 4), and we get 1
2. Find the general solution to the equation
Separate the variables:
1
3
dy
=x2 y
dx
dy
= x 2 dx and integrate:
y
1
3
1
2
33
dy
1
= x 2 dx ln y = x 3C
y
3
3
Solve for y: y =e 3 x C =e 3 x e C =Ce 3 x This is the general form.
Find the particular solution to the equation that passes through the point (2, 5e2).
2
5e
1
8
2
C = 8 =5 e 28 /3= 5 e 2 /3
Plug in values: 5 e 2=Ce 3 =Ce 3 and
e3
3
What is the slope of the graph at that point? The slope is given by the derivative (given),
2
2
=2 5 e = 20 e
2
dy
= x2 y
dx
3. Consult the following table of integrals for two continuous functions f and g over [a, b]:
Interval
[a, b]
b
[b, c]
Definite integral
of f(x)
a
Definite integral
of g(x)
a g x dx =8
d
f x dx =10
b
b
c
A.
a
B.
a g x dx =
f x dx =12
c
b g x dx =15
d
c
[d, e]
f x dx =10
d
c g x dx =18
e
d
f x dx = 4
e
d g x dx =9
d
c g x dx =b g x dx =15
C.
c
b
[c, d]
f x dx =a f x dx b f x dx c f x dx =1012 10=32
b
c
a
0
D. Let h(x) be defined as f(x) g(x).
d
d
a h x dx =a
f x g x dx = 10 1210 81518 =9
E. Can we determine at what points f(x) = g(x)? If so, when? If not, why not?
Since f x dx g x dx on [a,b] and g x dx f x dx on [b,c], we know that, on
average, f > g on on [a,b] and, on average, g > f on on [b,c]. This means that somewhere on on [a,c],
the two cross, but we don't know exactly where. (There may be more than one intersection, but we
know there's at least one.) Likewise, there's at least one intersection on x [c,e].
4.
x
9 dx =
x
We will need u substitution for this one. u = x + 9; du = dx
x
x 9 dx = u du =
However, when we substitute, we still have an x left, so we need to get
that in terms of u. We use our original substitution and get u 9 = x.
x
u 9
x 9 dx = u du = u 9 u1/ 2 du = u 1/ 29 u1/ 2 du after re-writing as a negative
exponent and distributing.
u 1 /2 9 u1 /2 du =2 / 3 u3 /2 2 9 u 1/ 2C = 2 / 3 x 9 3/ 218 x 9 1 /2 C
5. Consider the following graph
5
4
3
2
1
-6
.
-4
.-2
.
2
.
4
6
-1
What is the maximum error in estimating the area under the curve over the interval [-6, 4] for L10 or
R10?
Answer these questions for a hint, and partial credit:
b a
error
What is the formula for finding the maximum error? f b f a
n
What do you get if you apply that formula to this curve? Since f(b) = f(a) = 0, this is 0
What is the problem with that? This would imply that any approximation would have no error,
which is not the case.
How might you fix the problem? The formula only applies for increasing or decreasing functions.
This both increases and decreases, so we can split it into two pieces, from [-6,0] where it
increases, and from [0,4] where it decreases.
So, what is the maximum error? 5*6 = 30 for the left, and 5*4 = 20 for the right, for a total max
error of 50.
EC: The function f(t) is defined as the area under the
5
curve to the right between 0 and t:
t
f t =0 x 21 dx
4
What is f(1)?
1
3
2
f 1 =0 x 1 dx =2 / 3
2
f(5)?
5
1
f 5 =0 x 21 dx =125 / 35=110 / 3
f(-2)?
2
.
2
.
4
6
-1
f 2 =0 x 2 1 dx =8 / 3 2 =2 / 3
At what point(s) is f(x) = 0?
t3
x 21 dx =0 t =0
0
3
t
Solving for t (factoring out a t), we get that this is 0 at t = 0, and t = 3
What is the instantaneous rate of change at t = 2?
(How much is f changing by at that exact point?)
At t = 2, the amount the area is increasing by is the value of the function at that point. That small,
infitessimal slice with a height of f(2) is being added to the area, so the rate of change at that point is
2 21=3
What is f '(t)? The amount of change at any point t is equal to the height of the curve at that point,
2
t 21 . So, f ' t =t 1