Sample Solved Problems for the Final Exam
Basic Properties of Integrals Through this section we assume that all functions are continuous on a
closed interval I = [a,b]. Below r is a real number, f and g are functions.
Basic Properties of Integrals
c
1
∫ f ( x ) dx = 0
3
a
b
c
b
∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
a
5
a
∫ f ( x ) dx = −∫ f ( x ) dx
2
c
b
b
a
4
c
b
b
b
a
a
a
b
b
a
a
∫ r f ( x ) dx = r ∫ f ( x ) dx
∫ ( f ( x ) + g ( x ) ) dx = ∫ f ( x ) dx + ∫ g ( x ) dx
These properties of integrals follow from the definition of integrals as limits
of Riemann sums.
Solved Problems for the Final Exam by M. Seppälä
Table of Indefinite Integrals
1
3
⎧⎪ ln ( x ) + C 1
x > 0
dx
=
∫ x ⎪⎩⎨ ln ( − x ) + C 2 x < 0
dx
Abbreviation:
∫ x = ln x + C
x r +1
∫ x dx = r + 1 + C,
r ≠ −1
r
∫ k dx
= kx + C , k ∈ \
5
4
∫ s in ( x ) d x
6
∫e
x
2
= − cos (x ) + C
dx = e x + C
dx
1+ x
= a rc ta n ( x
9
)+
8
∫
10
∫ (a f ( x ) + b g ( x )) d x
2
7
C 10
∫ cos (x )dx
= s in ( x ) + C
ax
∫ a d x = ln ( a ) + C , a > 0, a ≠ 1
dx
∫ 1 − x 2 = a rc s in ( x ) + C
x
dx
∫ cos2 x = tan x + C
= a ∫ f ( x ) d x + b ∫ g ( x ) d x , a, b ∈ R
Solved Problems for the Final Exam by M. Seppälä
Fundamental Theorem of Calculus
Theorem
Let f be a continuous function. The function
x
F ( x ) = ∫ f ( t ) dt
a
is an antiderivative of the function f, i.e., F’(x) = f(x).
Conversely, if F is any antiderivative of f,
b
∫ f ( x ) dx = F ( b ) − F ( a) .
a
Solved Problems for the Final Exam by M. Seppälä
Summary of Formulae
b
Area bounded by the Graph of a Function
A=
∫ f ( x ) dx
a
b
Area between the Graphs of Functions
A=
∫ f ( x ) − g ( x ) dx
a
b
Volume of a Solid of Revolution
V = π ∫ f ( x ) dx
2
a
Volume of the solid of revolution by letting the domain
between the graphs of the non-negative functions f and
g rotate about the x-axis.
b
V = π ∫ f ( x ) − g ( x ) dx
Solved Problems for the Final Exam by M. Seppälä
a
2
2
Problems
x
1
For which values of x the function f ( x ) = ∫ sin2 t dt is
0
increasing?
x
2
For which values of x the graph of the function f ( x ) = ∫ e−t dt is
0
concave down?
n
3
4
Express the limit lim ∑
n →∞
k =1
n2 − k 2
as an integral.
n2
⎧ sin t
⎪
Define the function g setting g ( t ) = ⎨ t
⎪⎩ 1
Find the values of x for which the function
has a local extreme value?
Solved Problems for the Final Exam by M. Seppälä
for t ≠ 0
for t = 0
.
2
Problems
x2
5
Let F ( x ) = ∫
x
sin ( t )
t
dt. Compute F′ ( x ) .
x2
6
The function f is continuous, and
Determine f(1).
x
7
Determine
∫
lim 0
x →0
8
9
∫
f ( t ) dt = e− x .
2
0
1 − t 3 dt
x
.
Find the volume of the solid of revolution obtained by letting
the domain bounded by the graph of the function
f(x) = x - x2 and the x-axis rotate around the x-axis.
Find the area of the domain bounded by the curve
y2 – x2 + x4 = 0.
Solved Problems for the Final Exam by M. Seppälä
Problem 1
x
1
For which values of x the function f ( x ) = ∫ sin2 t dt is
0
increasing?
Solution
There are two ways to solve this problem.
Using the double angle formula for
Cosine we may compute f:
Complicated Solution
x
f ( x ) = ∫ sin2 t dt =
0
x
∫
1 − cos (2t )
0
Differentiation yields
f′ ( x ) =
2
x
⎛ t sin (2t ) ⎞
x sin (2 x )
dt = ⎜ −
.
⎟ = −
2
4
2
4
⎝
⎠0
1 cos (2 x )
−
= sin2 (2 x ) .
2
2
Here we used the
double angle
formula again.
Conclude: f’(x) > 0 if x ≠ nπ, n integer. This means that f is
increasing for all x.
Solved Problems for the Final Exam by M. Seppälä
Problem 1
x
1
For which values of x the function f ( x ) = ∫ sin2 t dt is
0
increasing?
Solution (cont’d)
Simple Solution
Using Fundamental Theorem
of Calculus we get
f ′ ( x ) = sin2 ( x ) .
Graph of f
Conclude: f’(x) > 0 if x ≠ nπ, n integer. This means that f is
increasing for all x.
Solved Problems for the Final Exam by M. Seppälä
Problem 2
x
2
For which values of x the graph of the function f ( x ) = ∫ e−t dt is
0
concave down?
Solution
2
The graph of a function f is concave down if f’’(x) < 0.
By the Fundamental Theorem of
2
Calculus
f ′ ( x ) = e− x .
Differentiating again we get
( ) = −2x e
f ′′ ( x ) = D e− x
2
− x2
Since the Exponential Function
takes always positive values,
f’’(x) < 0 for x > 0.
Answer
.
The graph of f.
The graph of the function f is concave down for x > 0.
Solved Problems for the Final Exam by M. Seppälä
Problem 3
n
Express the limit lim ∑
3
n →∞
k =1
n2 − k 2
as an integral.
2
n
Solution
The sum in question does not, at first sight, look like a Riemann sum for an
integral because the summand is not of the form
(function value) (length of a subinterval). A rewriting is therefore necessary.
n
∑
k =1
n2 − k 2
n2
Now we have the
length of a
subinterval here.
=
n
∑
k =1
n
=∑
k =1
n2 − k 2 ⎛ 1 ⎞
⎜n⎟
n
⎝ ⎠
2
⎛k ⎞ ⎛1⎞
1−⎜ ⎟ ⎜ ⎟
⎝ n⎠ ⎝n⎠
=
n
∑
k =1
n2 − k 2
n2
1
⎛1⎞
⎜n⎟
⎝ ⎠
⎯⎯⎯⎯
→ ∫ 1 − x 2 dx
n →∞
0
Solved Problems for the Final Exam by M. Seppälä
Problem 3
n
Express the limit lim ∑
3
n →∞
Solution
n
lim ∑
n →∞
k =1
k =1
n2 − k 2
as an integral.
2
n
The conclusion of the previous slide was that
n2 − k 2
=
2
n
1
∫
1 − x 2 dx.
0
To compute the value of this integral using the
Fundamental Theorem of Calculus is a rather
technical computation.
An easy way to compute the value of the integral is to observe that the
graph of the function 1 − x 2 is the upper half of a circle of radius 1.
n
Hence the value of the integral is
the area of the blue domain.
lim ∑
n →∞
k =1
n2 − k 2
=
2
n
Solved Problems for the Final Exam by M. Seppälä
1
∫
0
1 − x 2 dx =
π
4
Problem 4
4
⎧ sin t
⎪
Define the function g setting g ( t ) = ⎨ t
⎪⎩ 1
for t ≠ 0
.
for t = 0
x
Find the values of x for which the function f ( x ) = ∫ g ( t ) dt
0
has a local extreme value?
Solution
The function f has an extreme value at points
where the derivative f’ changes its sign.
f′ ( x ) =
By the FTC we get
f′ ( x ) = 0 ⇔
sin x
.
x
sin x
= 0 ⇔ x = nπ , n ∈ ], n ≠ 0.
x
Since f’ changes its sign at all non-zero
integer multiples of π, all of these points
are local extreme values of the function f.
Solved Problems for the Final Exam by M. Seppälä
Graph of f
Problem 5 x2
Let F ( x ) = ∫
5
sin ( t )
dt. Compute F′ ( x ) .
t
The function F is defined in terms of an integral whose both
Solution
lower and upper bound depends on x. This means that a
rewriting is needed before we can apply the Fundamental
Theorem of Calculus (FTC).
x2
x2
x
sin ( t )
sin ( t )
sin ( t )
The
function
G
x
=
(
)
F (x) = ∫
dt − ∫
dt
∫1 t dt is
t
t
x
1
By the
Chain
Rule.
1
a composed function U ( V ( x ) ) where
By the
FTC.
U (v ) =
v
∫
1
F′ ( x ) = U′ ( V ( x ) ) V′ ( x ) −
sin ( x )
x
=
sin ( t )
t
dt and V ( x ) = x 2.
( ) (2x ) − sin ( x ) = 2 sin ( x ) − sin ( x ) .
x
x
x
sin x 2
2
Solved Problems for the Final Exam by M. Seppälä
2
Problem 6 x2
6
The function f is continuous, and
Determine f(1).
Solution
Let G ( x ) =
∫
f ( t ) dt = e− x .
2
0
We need to differentiate the given equation to get a
formula for the function f.
x2
∫ f (t ) dt. The function
0
G ( x ) is a composed function
v
U ( V ( x ) ) where U (v ) = ∫ f ( t ) dt and V ( x ) = x 2.
0
So we get:
On the other hand:
Conclude
( )
G′ ( x ) = U′ ( V ( x ) ) V′ ( x ) = f x 2 2 x
( )
Using FTC and
the Chain Rule.
G ( x ) = e− x . Hence G′ ( x ) = −2 x e− x .
2
2
2
( )
2 x f x 2 = −2 x e− x ⇒ f x 2 = − e− x
2
and f (1) = − e−1 .
Solved Problems for the Final Exam by M. Seppälä
Problem 7 x
7
∫
Determine
lim 0
x →0
Solution
1 − t 3 dt
x
.
Use l’Hospital’s Rule:
x
∫
lim 0
x →0
Next use the FTC:
⎛x
⎞
3
D ⎜ ∫ 1 − t dt ⎟
1 − t dt
⎠.
= lim ⎝ 0
x →0
D(x)
x
3
⎛x
⎞
D ⎜ ∫ 1 − t 3 dt ⎟ = 1 − x 3 .
⎝0
⎠
Conclude
x
∫
lim 0
x →0
1 − t 3 dt
x
1 − x3
= lim
= 1.
x →0
1
Solved Problems for the Final Exam by M. Seppälä
Problem 8
8
Find the volume of the solid of revolution obtained by letting
the domain bounded by the graph of the function
f(x) = x - x2 and the x-axis rotate around the x-axis.
Solution
The pictures below illustrates the curve and the solid
in question.
The volume can be
computed by straightforward
application of the formula.
The limits for integration can
be obtained by solving the
equation x – x2 = 0.
To integrate expand the product
Conclude
1
(
V = π ∫ x − x2
0
)
2
1
1
⎛x
π
x
x ⎞
dx = π ∫ x 2 − 2 x 3 + x 4 dx = π ⎜
.
−
+
=
⎟
3
2
5
30
⎝
⎠0
0
(
)
3
Solved Problems for the Final Exam by M. Seppälä
4
5
Problem 9
9
Find the area of the domain bounded by the curve
y2 – x2 + x4 = 0.
Solution
It is useful to draw a picture first. To that end solve
y in terms of x of the equation for the curve to get
y = ± x 1 − x2 .
y = −x 1 − x2
y = x 1 − x2
-1
1
The graph of the curve y2 – x2 + x4 = 0
Solved Problems for the Final Exam by M. Seppälä
Problem 9
9
Find the area of the domain bounded by the curve
y2 – x2 + x4 = 0.
Solution
(cont’d)
Observe that, by symmetry, the area of the domain in
question is four times the area of the yellow domain.
y = x 1 − x2
Substitute
t = 1 – x2.
1
Conclude
1
0
A = 4∫ x 1 − x dx = 4∫
0
2
1
0
1
3
2
t
⎛ dt ⎞
t ⎜−
=
−
tdt
=
tdt
=
2
2
2
⎟
∫1
∫0
3
⎝ 2 ⎠
2
Solved Problems for the Final Exam by M. Seppälä
1
=
0
4
.
3