Definition A function F is called an antiderivative of f on an interval I if F0(x) =f(x) for all x inI.
TheoremIfF is an antiderivative off on an intervalI, then the most general antiderivative of f on I isF(x) +C where C is an arbitrary constant.
Examples Find the most general antiderivative of each of the following functions.
(1) f(x) = sinx (2) f(x) = 1
x
(3) f(x) =xn, n 6=−1 (4) f(x) = 4 sinx+ 2x5 −√
x x Examples
1. The following are graphs of a functionf and a rough sketch of its antiderivativeF satisfying F(0) = 2.
2. A particle moves in a straight line and has acceleration given by a(t) = 6t+ 4. Its initial velocity is v(0) = −6 cm/s and its initial displacement is s(0) = 9 cm. Find its position functions(t).
Definitions
(a) Let I = [a, b]⊂R. A set P ={x0, x1, . . . , xn} ⊂I is called a partition of I if a=x0 < x1 < . . . < xn=b.
(b) Let P be a partition of I. Define
∆xi = xi −xi−1 = the width of the ith interval [xi−1, xi] kPk = max
1≤i≤n ∆xi = max {∆xi |1≤i≤n}= the norm of partition P (c) Let f be a function defined on I and let P be a partition of I. Then
R(f, P) =
n
X
i=1
f(x∗i) ∆xi, where x∗i ∈[xi−1, xi] for each 1≤i≤n, is called a Riemann sum of f with respect to the partition P of I.
(d) Let f be a function defined on I = [a, b]. The definite integral of f from a to b, denoted Z b
a
f(x)dx is defined to be Z b
a
f(x)dx= lim
kPk→0 n
X
i=1
f(x∗i) ∆xi provided that the limit exists,
where x∗i ∈[xi−1, xi] and ∆xi =xi−xi−1 for each 1≤i≤n.
(e) Let f be a bounded function defined on I and letP be a partition of I. Then U(f, P) =
n
X
i=1
sup
x∈[xi−1,xi]
f(x) ∆xi,
where sup
x∈[xi−1,xi]
f(x) = the least upper bound of {f(x) | x ∈ [xi−1, xi]}, is called an upper sum of f with respect to the partition P of I, and
L(f, P) =
n
X
i=1
x∈[xinfi−1,xi]f(x) ∆xi, where inf
x∈[xi−1,xi]f(x) = the greatest lower bound of {f(x) |x ∈ [xi−1, xi]}, is called a lower sum of f with respect to the partition P of I.
(f) Let P ={a =x0 < x1 < . . . < xn =b}, Q ={a=t0 < t1 < . . . < tm =b} be partitions of I. Q issaid to be finer than P,denoted P ⊆Q, if
P ={a=x0 < x1 < . . . < xn=b} ⊆ {a =t0 < t1 < . . . < tm =b}=Q
=⇒ kQk= max
1≤j≤m ∆tj ≤ kPk= max
1≤i≤n ∆xi.
(g) Let f be a function defined on I = [a, b]. Then f is said to be (Riemann) integrable on I if f is bounded on [a, b] and the definite integral
Z b a
f(x)dx exists, i.e. if f is bounded on [a, b] and the definite integral
Z b
a
f(x)dx= lim
kPk→0 R(f, P)= lim
1≤i≤nmax ∆xi→0 n
X
i=1
f(x∗i) ∆xi exists.
Proposition Let
• f be bounded on [a, b],
• P = {a = x0 < x1 < . . . < xn =b}, Q ={a =t0 < t1 < . . . < tm =b} be partitions of I such thatQ is finer than P.
Then inf
x∈[a,b]f(x) (b−a)≤L(f, P)≤L(f, Q)≤U(f, Q)≤U(f, P)≤ sup
x∈[a,b]
f(x) (b−a).
Proof For each 1≤j ≤m, since P ⊆Q, there exists 1≤i≤n such that [tj−1, tj]⊆[xi−1, xi] =⇒ inf
x∈[xi−1,xi]f(x)≤ inf
x∈[tj−1,tj]f(x)≤ sup
x∈[tj−1,tj]
f(x)≤ sup
x∈[xi−1,xi]
f(x) Definitions Let f be a function defined on an interval I. Then
(a) f is said to becontinuous onI if for eachε >0 and for eachx∈I,there existsδ =δ(ε, x)>
0, such that
if y∈I and if |y−x|< δ then |f(y)−f(x)|< ε.
(b) f is said to be uniformly continuous on I if for each ε > 0 and for each x∈I, there exists δ=δ(ε)>0, such that
if x, y ∈I and if |y−x|< δ then |f(y)−f(x)|< ε.
(c) f is not uniformly continuous on I if there exists ε0 > 0, such that for each n ∈ N = {1,2, . . .}, there existxn, yn∈I such that
n→∞lim |yn−xn|= 0 while lim
n→∞|f(yn)−f(xn)| ≥ε0 >0.
Examples
(1) Let f be defined byf(x) = 1
x forx∈I = (0,1), a bounded, not closed interval.
For each a∈I, since a6= 0,
x→alim f(x) =f(a), f is continuous on I.
However, ifxn = 1
n+ 1, yn = 1
2(n+ 1) for each n ∈N={1,2, . . .}, then xn, yn∈I,
n→∞lim |yn−xn|= lim
n→∞
1
2(n+ 1) − 1 n+ 1
= lim
n→∞
1
2(n+ 1) = 0 while
n→∞lim |f(yn)−f(xn)|= lim
n→∞
1 yn − 1
xn
= lim
n→∞
1 1 2(n+ 1)
− 1 1 n+ 1
= lim
n→∞(n+1)≥1 = ε0 >0.
This implies that f(x) = 1
x is not uniformly continuous on (0,1). Note that f(x) = 1 x is uniformly continuous on [a, b] for any a >0.
(2) Let f be defined byf(x) =x2 for x∈R= (−∞,∞),an unbounded interval.
For each a∈R,since
x→alim f(x) =f(a), f is continuous on R.
However, ifxn =n+ 1
n, yn=n for each n∈N={1,2, . . .},then xn, yn ∈(−∞,∞),
n→∞lim |yn−xn|= lim
n→∞
n−n− 1 n
= lim
n→∞
1
n(n+ 1) = 0 while
n→∞lim |f(yn)−f(xn)|= lim
n→∞
y2n−x2n
= lim
n→∞
n2−
n+ 1 n
2
= lim
n→∞
−2 + 1 n2
= 2 =ε0 >0.
This implies thatf(x) = x2 is not uniformly continuous on (−∞,∞).Note that ifa, b∈R, then f(x) =x2 is uniformly continuous on [a, b].
Theorem Letf be continuous on a closed interval I = [a, b].Then
• the range of f is a bounded, closed interval, i.e. there exist m≤M ∈R such that min
x∈[a,b]f(x) = m, max
x∈[a,b]f(x) = M and f(I) = {f(x)|x∈[a, b]}= [m, M].
ProofUse the Extreme Value Theorem and the Intermediate Value Theorem.
• f is uniformly continuous on I.
Proof Use the Bolzano Weierstrass Theorem (i.e. every bounded sequence in I = [a, b]
contains a convergent subsequence) and a contradictory argument.
• f is (Riemann) integrable on I.
Proof Since f is continuous on a closed interval I = [a, b], f is uniformly continuous on I = [a, b].Given ε >0, there exists δ >0 such that
if x, y ∈I and if |y−x|< δ then |f(y)−f(x)|< ε b−a. LetP ={a =x0 < x1 < . . . < xn=b} be a partition of I = [a, b] such that
kPk= max
1≤i≤n∆xi < δ.
Since
0≤U(f, P)−L(f, P) =
n
X
i=1
x∈[xmaxi−1,xi] f(x)− min
x∈[xi−1,xi]f(x)
∆xi < ε
b−a · (b−a) = ε, f is (Riemann) integrable on I.
Properties of Integral
(a) If cis a constant, then cis integrable on [a, b] and Z b
a
c dx=c(b−a).
(b) If f, g are integrable on [a, b], then cf +g is integrable on [a, b] for any constantc and Z b
a
cf(x) +g(x)
dx=c Z b
a
f(x)dx+ Z b
a
g(x)dx.
(c) If f is integrable on [a, b] and if [a, b] = [a, c]∪[c, b], then Z c
c
f(x)dx = 0, Z b
a
f(x)dx= Z c
a
f(x)dx+
Z b c
f(x)dx and Z a
b
f(x)dx=− Z b
a
f(x)dx.
(d) If f, g are integrable on [a, b] andf(x)≥g(x) for x∈[a, b], then Z b
a
f(x)dx≥ Z b
a
g(x)dx.
In particular,
• if m≤f(x)≤M for x∈[a, b], then m(b−a)≤
Z b a
f(x)dx ≤M(b−a),
• if f(x)≥0 for x∈[a, b],then Z b
a
f(x)dx = the area of the region bounded by y=f(x), y = 0, x=a and x=b.
By the Property (c) of Integral, we have the following
The Fundamental Theorem of Calculus Iff is continuous on [a, b],then (a) the function g defined by
g(x) = Z x
a
f(t)dt is continuous on [a, b] and differentiate on (a, b) with
g0(x) = f(x) ⇐⇒ lim
h→0
g(x+h)−g(x)−f(x)h
h = lim
h→0
Rx+h
x [f(t)−f(x)]dt
h = 0
⇐⇒ d dx
Z x a
f(t)dt=f(x) for x∈(a, b), i.e. g(x) =
Z x a
f(t)dt is an antiderivative of f on (a, b).
(b) Z b
a
f(x)dx=F(b)−F(a),whereF is any antiderivative off on (a, b),that is,F0(x) = f(x) for x∈(a, b).
Examples
(1) Let g(x) = Z x
0
√
1 +t2dt. Find g0(x).
(2) Let S(x) = Z x
0
sin πt2 2
dt. FindS0(x).
(3) Evaluate the integral Z 3
1
exdx.
Combining the Property (c) of Integral, the Fundamental Theorem of Calculus and the Chain Rule, we have the following
Leibniz Rule If f is continuous on [a, b] and if u = u(x), ` = `(x) are differentiable functions with values in [a, b], then
d dx
Z u(x)
`(x)
f(t)dt = d dx
Z a
`(x)
f(t)dt+ Z u(x)
a
f(t)dt
= d dx
Z u(x)
a
f(t)dt− Z `(x)
a
f(t)dt
= d
du Z u
a
f(t)dt · du dx− d
d`
Z ` a
f(t)dt · d`
dx by the Chain Rule
= f(u(x))du
dx −f(`(x)) d`
dx by the Fundamental Theorem of Calculus
Average Value of a Function Letf be a function defined on [a, b]. The average value off on the interval [a, b] is defined by
favg = 1 b−a
Z b a
f(x)dx.
The Mean Value Theorem for IntegralIffis continuous on [a, b],then there exists a number cin [a, b] such that
F(b)−F(a) =F0(c)(b−a) ⇐⇒ f(c) = 1 b−a
Z b a
f(x)dx=favg, where F(x) =
Z x a
f(t)dt.
Definition An indefinite integral Z
f(x)dx is a family of functions F(x) +C such that d
dx F(x) +C
=F0(x) =f(x).
Table of Indefinite Integrals Z
cf(x)dx=c Z
f(x)dx
Z
[f(x) +g(x)]dx= Z
f(x)dx+ Z
g(x)dx Z
k dx=kx+C Z
xndx= xn+1
n+ 1 +C if n6= 1
Z 1
xdx= ln|x|+C Z
exdx=ex+C
Z
bxdx= bx
lnb +C for 0< b6= 1 Z
sinx dx=−cosx+C
Z
cosx dx= sinx+C Z
sec2x dx= tanx+C
Z
csc2x dx=−cotx+C Z
secxtanx dx= secx+C
Z
cscxcotx dx=−cscx+C Z 1
x2+ 1dx= tan−1x+C
Z 1
√1−x2 dx= sin−1x+C Z
sinhx dx= coshx+C
Z
coshx dx= sinhx+C
The Substitution Rule If u = g(x) is a differentiable function whose range is an interval I and f is continuous onI, then
Z
f(g(x))g0(x)dx= Z
f(u)du.
The Substitution Rule for Definite IntegralsIfg0 is continuous on [a, b] andf is continuous on the range ofu=g(x),then
Z b a
f(g(x))g0(x)dx= Z g(b)
g(a)
f(u)du.
Examples
(1) Find
Z cosθ sin2θdθ.
(2) The integral of a rate of change F0(x) is the net change of F fromx=a tox=b:
Z b a
F0(x)dx=F(b)−F(a)
• If V(t) is the volume of water in a reservoir at time t, then its derivative V0(t) is the rate at which water flows into the reservoir at time t. So
Z t2
t1
V0(t)dt=V(t2)−V(t1)
is the change in the amount of water in the reservoir between timet1 and time t2.
• If [C](t) is the concentration of the product of a chemical reaction at time t, then the rate of reaction is the derivative d[C]
dt .So Z t2
t1
d[C]
dt dt = [C](t2)−[C](t1) is the change in the concentration ofC from timet1 to time t2.
• If the mass of a rod measured from the left end to a pointx ism(x),then the linear density isρ(x) = m0(x). So
Z b a
ρ(x)dx=m(b)−m(a)
is the mass of the segment of the rod that lies between x=a and x=b.
• If the rate of growth of a population is dn dt, then Z t2
t1
dn
dt dt=n(t2)−n(t1)
is the net change in population during the time period fromt1 tot2.
• If C(x) is the cost of producing x units of a commodity, then the marginal cost is the derivativeC0(x). So
Z x2
x1
C0(x)dx=C(x2)−C(x1)
is the increase in cost when production is increased fromx1 units tox2 units.
• If an object moves along a straight line with position function s(t), then its velocity is v(t) =s0(t), so
Z t2
t1
v(t)dt=s(t2)−s(t1)
is the net change of position, or displacement, of the object during the time period from t1 tot2,and
Z t2
t1
|v(t)|dt= total distance travelled
• The acceleration of the object isa(t) =v0(t),so Z t2
t1
a(t)dt=v(t2)−v(t1) is the change in velocity from timet1 to timet2.
(3) A particle moves along a line so that its velocity at time tis v(t) =t2−t−6.(measured in meters per second).
• Show that the displacement of the particle during the time period 1 ≤t≤4 is −9 2 m.
• Show that the distance traveled during this time period is 61 6 m.
Examples
(1) Find Z
x3cos(x4+ 2)dx.
(2) Show that Z 4
0
√2x+ 1dx= 26 3 .
(3) Suppose that f is continuous and even on [−a, a], i.e. f(−x) = f(x) for x∈ [−a, a]. Show that
Z a
−a
f(x)dx= 2 Z a
0
f(x)dx.
(4) Suppose that f is continuous and odd on [−a, a],i.e. f(−x) =−f(x) for x∈[−a, a]. Show that
Z a
−a
f(x)dx= 0.
(5) Show that Z 2
−2
x6+ 1
dx= 2 Z 2
0
x6+ 1
dx= 284 7 . (6) Show that
Z 1
−1
tanx
1 +x2+x4 dx= 0.
Integration by PartsThe Product Rule states that iff andg are differentiable functions, then d
dx[f(x)g(x)] = f(x)g0(x) +g(x)f0(x)
Letu=f(x) andv =g(x). Then the differentials aredu=f0(x)dx and dv=g0(x)dx, so, by the Substitution Rule, the formula for integration by parts becomes
Z
u dv=uv− Z
v du ⇐⇒
Z
f(x)g0(x)dx=f(x)g(x)− Z
g(x)f0(x)dx
If we combine the formula for integration by parts with the Fundamental Theorem of Calculus, we can evaluate definite integrals by parts.
Z b a
f(x)g0(x)dx=f(x)g(x)|ba− Z b
a
g(x)f0(x)dx
Examples
(1) Find Z
xsinx dx, Z
(x2+ 3x+ 1) sinx dxand Z
(x2+ 3x+ 1)(cosx+e2x)dx.
In summary, if we want to find Z
Pn(x)
cosax or sinax or eax
dx, where Pn(x) =
n
X
k=0
akxk is a polynomial of degree n, we set
u=Pn(x) dv=
cosax sinax eax
dx =⇒
du=Pn0(x)dx v = 1a
sinax
−cosax eax
and apply the integration by parts formulan times.
(2) Use the integration by parts to calculate Z 1
0
tan−1x dx and Z 1
0
xtan−1x dx.
(3) Use the integration by parts to prove the reduction formula Z
sinnx dx=−1
ncosxsinn−1x+n−1 n
Z
sinn−2x dx where n≥2 is an integer.
(4) Find Z
sin2x dx, Z
sin3x dx and Z
sin4x dx.
(5) Let a6= 0, b6= 0. Then Z
eaxcosbx dx Set u=eax, dv = cosbx dx =⇒ du=aeaxdx, v = sinbx b
= eaxsinbx
b −a
b Z
eaxsinbx dx Set u=eax, dv = sinbx dx =⇒ du=aeaxdx, v =−cosbx b
= eaxsinbx
b +aeaxcosbx b2 −a2
b2 Z
eaxcosbx dx Combine coefficients of Z
eaxcosbx dx
= b2 a2+b2
eaxsinbx
b +aeaxcosbx b2
+C
Integration of Rational Functions by Partial Fractions
LetQ(x) be an irreducible polynomial of degree≤2. Then we can find
Z 1
Q(x)dx by a method of substitution as follows.
• Ifa6= 0, then
Z 1
ax+bdx= 1
aln|ax+b|+C.
• Ifa∈R and a6= 0, then
Z 1
x2+a2 dx= 1
atan−1x a
+C.
• Ifa, b∈Rand a2−4b <0, then
Z 1
x2+ax+bdx=
Z 1
(x+a2)2+4b−a4 2 dx= 2
√4b−a2 tan−1
2x+a
√4b−a2
+C.
For a general rational function f(x) = P(x)
Q(x), we need the following algebraic facts to establish the method of partial fractions.
The Factorization of Polynomials with Real Coefficients
• Proposition Let p(x) = x2 +ax+b be a quadratic polynomial with real coefficients, i.e.
a, b∈R. Then
p(x) = x2+ax+b = (x+a
2)2−a2−4b
4 = (x+a 2−
√a2−4b
2 )(x+a 2+
√a2−4b
2 ) = (x−r1)(x−r2) and
r1, r2 =−a 2±
√a2−4b
2 are
(two real roots (counting multiplicity) ⇐⇒ a2−4b ≥0 two conjugate complex roots ⇐⇒ a2−4b <0 where we let i2 =−1 ⇐⇒ i=√
−1.
• Theorem Let p(x) be a non-constant polynomial with real coefficients. Then p(x) has a unique factorization
p(x) =c(x−r1)(x−r2)· · ·(x−rm)(x2+a1x+b1)(x2+a2x+b2)· · ·(x2+amx+bm) where r1, r2, . . . , rm ∈ R are the real roots of p(x) and c, a1, a2, . . . , an, b1, b2, . . . , bn ∈ R, i.e. a non-constant polynomial with real coefficients can be uniquely factorized into product of irreducible polynomials.
• Let p(x) = an−1xn−1+an−2xn−2+· · ·+a1x+a0 be a polynomial with real coefficients of degreen−1.
For anya∈R, there existck, k= 0, 1,2, . . . , n−1,such that p(x) =
n−1
X
k=0
ck(x−a)k =cn−1(x−a)n−1+cn−2(x−a)n−2+· · ·+c1(x−a) +c0
=⇒ p(x)
(x−a)n = cn−1
(x−a) + cn−2
(x−a)2 +· · ·+ c1
(x−a)n−1 + c0 (x−a)n
=⇒ p(x)
(x−a)n = A1
(x−a) + A2
(x−a)2 +· · ·+ An−1
(x−a)n−1 + An
(x−a)n (∗) where An−k=ck = p(k)(a)
k! = dkp
dxk(a) fork = 0,1, . . . , n−1.
• Letp(x) =a2n−1x2n−1+a2n−2x2n−2 +· · ·+a1x+a0 be a polynomial with real coefficients of degree≤2n−1.
For anya, b∈B satisfying a2−4b <0,there exist ck, dk, k= 0, 1,2, . . . , n−1, such that p(x) =
n−1
X
k=0
(ckx+dk)(x2+ax+b)k = (cn−1x+dn−1)(x2 +ax+b)n−1+· · ·+ (c0x+d0)
=⇒ p(x)
(x2+ax+b)n = cn−1x+dn−1
x2 +ax+b +· · ·+ c1x+d1
(x2+ax+b)n−1 + c0x+d0 (x2+ax+b)n
=⇒ p(x)
(x2+ax+b)n = A1x+B1
x2+ax+b +· · ·+ An−1x+Bn−1
(x2+ax+b)n−1 + Anx+Bn
(x2+ax+b)n (∗∗) where ck = An−k, dk =Bn−k ∈ R for k = 0, 1, . . . , n−1, are uniquely determined by the division or by the method of comparing the coefficients.
The Method of Partial Fractions Consider a rational function
f(x) = P(x) Q(x),
whereP and Qare polynomials. Its possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree ofQ. Such a rational function is called proper.
If f is improper, that is, deg(P)≥ deg(Q), then we must take the preliminary step of dividing Q intoP (by long division) until a remainder R(x) is obtained such that deg(R)<deg(Q).
The result is
f(x) = P(x)
Q(x) =S(x) + R(x) Q(x),
where the quotient S(x) and the remainder R(x) are also polynomials.
• Case IThe denominator Q(x) is a product of distinct linear factors, i.e.
Q(x) = (a1x+b1)(a2x+b2)· · ·(akx+bk),
where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1, A2, . . . , Ak such that
R(x)
Q(x) = A1
a1x+b1 + A2
a2x+b2 +· · ·+ Ak akx+bk
• Case IIQ(x) is a product of linear factors, some of which are repeated.
Suppose that the first linear factor (a1x+b1) is repeatedr times; that is (a1x+b1)r occurs in the factorization ofQ(x). Then instead of
Pr−1 k=0dkxk (a1x+b1)r in the partial fraction equation of R(x)
Q(x),by (∗), we would use A1
a1x+b1 + A2
(a1x+b1)2 +· · ·+ Ar (a1x+b1)r
• Case IIIQ(x) contains irreducible quadratic factors, none of which is repeated.
If Q(x) has the factor ax2 +bx+c, where b2 −4ac < 0, then, in addition to the partial fractions in Equations inCase I and Case II, the expression for R(x)
Q(x) will have a term of the form
Ax+B ax2+bx+c, where A and B are constants to be determined.
• Case IVQ(x) contains a repeated irreducible quadratic factor.
IfQ(x) has the factor(ax2+bx+c)r, where b2−4ac < 0, then instead of P2r−1
k=0 dkxk (a1x+b1)r
in the partial fraction of R(x)
Q(x), by (∗∗),we would use A1x+B1
ax2+bx+c + A2x+B2
(ax2+bx+c)2 +· · ·+ Arx+Br
(ax2+bx+c)r
Note that each of the terms can be integrated by using a substitution or by first completing the square if necessary.
Examples
(1)
Z x3+x x−1 dx=
Z
x2+x+ 2 + 2 x−1
dx= x3 3 + x2
2 + 2x+ 2ln|x−1|+C.
(2)
Z x2+ 2x−1
2x3+ 3x2−2xdx =
Z x2+ 2x−1
x(2x−1)(x+ 2)dx = Z
A1
x + A2
2x−1 + A3 x+ 2
dx, where x2+x−1 =A1(2x−1)(x+2)+A2x(x+2)+A3x(2x−1).So,A1 = 1/2, A2 = 1/5, A3 =−1/10 by setting x= 0, 1/2, −2, respectively.
(3)
Z x4−2x2+ 4x+ 1 x3−x2−x+ 1 dx=
Z
x+1+ 4x
(x−1)2(x+ 1)dx= Z
x+ 1 + A1
x−1 + A2
(x−1)2 + A3 x+ 1
dx, where 4x = A1(x−1)(x+ 1) +A2(x+ 1) +A3(x−1)2. So, A3 = −1, A2 = 2, A1 = 1 by
settingx=−1,1, 0, respectively.
(4)
Z 4x2−3x+ 2 4x2−4x+ 3dx=
Z
1 + x−1 (2x−1)2 + 2
dx=
Z "
1 +
1
2 2x−1
− 12 (2x−1)2+ 2
# dx
= Z
1+ 2· 2x−1√2 4√
2h (2x−1√
2 )2 + 1i− 1
4h (2x−1√
2 )2+ 1idx=x+
lnh (2x−1√
2 )2+ 1i 4√
2 −
√2 tan−1(2x−1√2 )
8 +C.
(5)
Z 1−x+ 2x2−x3 x(x2+ 1)2 dx=
Z A1 x +
A2x+B2
x2+ 1 +A3x+B3 (x2+ 1)2
dx
=A1ln|x|+A2
2 ln(x2+ 1) +B2tan−1x− A3
2(x2+ 1) +B3
tan−1x
2 + x
2(x2+ 1)
+C,where
−x3+ 2x2−x+ 1 =A1(x2+ 1)2+ (A2x+B2)x(x2+ 1) + (A3x+B3)x= (A1+A2)x4+B2x3+ (2A1+A2+A3)x2+ (B2+B3)x+A1. So, A1 = 1, A2 =−1, B2 =−1, A3 = 1, B3 = 0.
(6) Z √
x+ 4
x dxu2=x+4=
Z 2u2
u2−4du = Z
2 + 8 u2−4
du =
Z
2 + A1
u−2+ A2
u+ 2
du = 2√
x+ 4 +A1ln|√
x+ 4−2|+A2ln|√
x+ 4 + 2|+C, where 8 =A1(u+ 2) +A2(u−2). So, A1 = 2, A2 =−2 by setting u= 2, −2,respectively.
(7) Find
Z dx
sinx+ cosx+ 2dx by setting u= tanx2. Then sinx = 2 sinx2cosx2 = 2u 1 +u2, cosx= (cosx2)2−(sinx2)2 = 1−u2
1 +u2, dx= 2du 1 +u2 and
Z dx
sinx+ cosx+ 2dx=
Z 2du (u+ 1)2+ 2. (8) Find
Z 1
x1/2+x1/3dx by setting x=u6, where 6 is the least common multiple of 2 and 3.
Then
Z 1
x1/2 +x1/3dx=
Z 6u5du u3+u2 =
Z 6u3du u+ 1 = 6
Z
u2−u+ 1− 1
u+ 1du= 2u3−3u2+ 6u−6ln|u+ 1|+C = 2x1/2−3x1/3+ 6x1/6−ln|x1/6+ 1|+C.
(9) Find
Z 1
1 +exdx =
Z e−x/2dx
e−x/2+ex/2 by setting u=e−x/2. Then du = −12e−x/2dx =⇒ e−x/2dx=−2du and
Z 1
1 +exdx =
Z e−x/2dx e−x/2+ex/2 =
Z −2du u+u1 =−
Z 2udu
u2+ 1 = −ln(u2+ 1) +C =−ln(e−x+ 1) +C.
Integrals of nonnegative Powers of Trigonometric Functions Examples
(1) Find Z
sin5xcos2x dx.
(2) Evaluate Z π
0
sin2x dx.
(3) Find Z
sinmxcosnx dx if eitherm orn is odd.
(4) Find Z
sinmxcosnx dx if bothm and n are even.
(5) Find Z
tanx dx, Z
tan2x dx and the reduction formula for Z
tannx dx for n∈N. (6) Find
Z
tan3x dx.
(7) Find Z
secx dx, Z
sec2x dx and the reduction formula for Z
secnx dxfor n ∈N. (8) Find
Z
tanmxsecnx dx= Z
tanmx(1 + tan2x)k−1dtanx if n = 2k, k≥1.
(9) Find Z
tanmxsecnx dx = Z
(sec2x−1)ksecn−1x dsecx if m = 2k+ 1 and n = 2` + 1, k, `≥0.
Integrals of the form Z
sinmxcosnx dx, Z
sinmxsinnx dx, or Z
cosmxcosnx dx Recall the identities:
sinAcosB = 1 2
sin(A−B)+sin(A+B) , 1
2
cos(A−B)±cos(A+B)
=
(cosAcosB when + sinAsinB when − ExampleFind
Z
sin 4xcos 5x dx.
Trigonometric Substitution for integrals of functions of√
a2−x2,√
x2−a2,or√
x2+a2. Examples
(1) Find Z √
9−x2 x2 dx.
(2) Find
Z 1 x2√
x2+ 4dx.
(3) Find
Z dx
√x2−a2, where a >0.
(4) Evaluate Z 3√
3/2 0
x3
(4x2+ 9)3/2dx.
Area Between Curves Consider the region S shown in Figure below that lies between two curvesy=f(x) andy =g(x) and between the vertical lines x=aand x=b, where f andg are continuous functions and f(x)≥g(x) for all xin [a, b].
We divide S into n strips of equal width and then we approximate the ith strip by a rectangle with base ∆x and height f(x∗i)−g(x∗i). (See Figure above. If we like, we could take all of the sample points to be right endpoints, in which case x∗i =xi.)
The Riemann sum
n
X
i=1
f(x∗i)−g(x∗i)
∆x
is therefore an approximation to what we intuitively think of as the area of S.
This approximation appears to become better and better asn→ ∞.Therefore we define the area Aof the regionS as the limiting value of the sum of the areas of these approximating rectangles.
A= Z b
a
f(x)−g(x)
dx= lim
n→∞
n
X
i=1
f(x∗i)−g(x∗i)
∆x
Furthermore, ifS is a region bounded between the continuous curvesy =f(x) andy=g(x) and between x=a and x=b, then the area A of S is
A= Z b
a
|f(x)−g(x)|dx
Examples
(1) Find the area of the region bounded above byy=ex,bounded below byy=x,and bounded on the sides by x= 0 andx= 1.
(2) Find the area of the region bounded by the curves y= sinx, y = cosx, x= 0, and x= π 2. (3) Find the area enclosed by the ellipse x2
a2 +y2 b2 = 1.
(4) If a region is bounded by curves with equations x=f(y), x=g(y), y =c,andy =d,where f and g are continuous and f(y)≥g(y) forc≤y≤d, then its area is
A= Z d
c
f(y)−g(y) dy.
Find the area enclosed by the line y=x−1 and the parabola y2 = 2x+ 6.
Volumes
Recall that if the base is a disk with radiusr,then the cylinder is a circular cylinder with volume V = πr2h, and if the base is a rectangle with length l and width w, then the cylinder is a rectangular box (also called a rectangular parallelepiped) with volumeV =lwh.
For a solid S that isnt a cylinder we first cut S into pieces and approximate each piece by a cylinder. We estimate the volume ofS by adding the volumes of the cylinders. We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large.
We start by intersectingSwith a plane and obtaining a plane region that is called a cross-section of S.
Let A(x) be the area of the cross-section of S in a plane Px perpendicular to the x-axis and passing through the point x, where a ≤ x ≤ b. (Think of slicing S with a knife through x and computing the area of this slice.) The cross-sectional area A(x) will vary as x increases from a tob.
Definition of VolumeLetS be a solid that lies betweenx=aandx=b.If the cross-sectional area of S in the plane Px, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is
V = Z b
a
A(x)dx= lim
n→∞
n
X
i=1
A(x∗i) ∆x
ExampleShow that the volume of a sphere of radius r is V = 4 3πr3.
Remark If we revolve a region about a line, we obtain a solid of revolution. In general, we calculate the volume of a solid of revolution by using the basic defining formula
V = Z b
a
A(x)dx or Z d
c
A(y)dy
and we find the cross-sectional area A(x) or A(y) in one of the following ways:
• If the cross-section is adisk, we find the radius of the disk (in terms of x ory) and use A =π(radius)2.
• If the cross-section is a washer, we find the inner radius rin and outer radius rout from a sketch and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk:
A=π(outer radius)2−π(inner radius)2.
Examples
(1) Figure below shows a solid with a circular base of radius 1. Parallel cross-sections perpen- dicular to the base are equilateral triangles. Find the volume of the solid.
[Hint: Let’s take the circle to bex2+y2 = 1.The solid, its base, and a typical cross-section at a distance x from the origin are shown below. Then |AB| = 2y = 2√
1−x2 and the cross-sectional area is A(x) = 1
2·2√
1−x2·√ 3√
1−x2 for −1≤x≤1.]
(2) Find the volume of the solid obtained by rotating the region about the line x=−1.
[Hint: It is a washer with inner radius 1 +y, outer radius 1 +√
y and the cross-sectional area isA(y) =π[(1 +√
y)2−(1 +y)2] for 0≤y≤1.]
The Method of Cylindrical Shells
Consider the problem of finding the volume of the solid obtained by rotating about the y-axis the region bounded by y= 2x2−x3 and y= 0.
If we slice perpendicular to the y-axis, we get a washer. But to compute the inner radius and the outer radius of the washer, we would have y= 2x2−x3 for x in terms ofy; that’s not easy.
Fortunately, there is a method, called the method of cylindrical shells, that is easier to use in such a case. Figure above shows a cylindrical shell with inner radius r1, outer radius r2, and height h.Its volume V is calculated by subtracting the volumeV1 of the inner cylinder from the volumeV2 of the outer cylinder, i.e.
V =πr22h−πr12h=π(r2+r1)(r2−r1)h= 2πr2+r1
2 h(r2−r1).
LetS be the solid obtained by rotating about they-axis the region bounded byy=f(x) [where f(x)≥0],y = 0, x=a, and x=b, where b > a≥0.
We divide the interval [a, b] into n subintervals [xi−1, xi] of equal width ∆x and let ¯xi be the midpoint of the ith subinterval. If the rectangle with base [xi−1, xi] and height f( ¯xi) is rotated about the y-axis, then the result is a cylindrical shell with average radius ¯xi, height f( ¯xi) and thickness ∆x, so its volume is
Vi = 2πx¯i f( ¯xi)
∆x
The volume of the solid in figure above, obtained by rotating about they-axis the region under the curve y=f(x) from a to b, is
V = Z b
a
2π x
| {z }
circumference
f(x)
| {z }
height
dx
thickness|{z}
= lim
n→∞
n
X
i=1
2πx¯if( ¯xi)∆x.
Examples
(1) Find the volume of the solid obtained by rotating about the y-axis the region bounded by y= 2x2−x3 and y= 0.
(2) LetR be the region in the first quadrant bounded by the curvesy=x2 and y= 2x. A solid is formed by rotating the region about the linex=−1.Find the volume of the solid using (a)x as the variable of integration and (b)y as the variable of integration.
Improper Integrals Definition The integral
Z b a
f(x)dx is called animproper integral if
• either the interval is infinite, i.e. |b−a|=∞,
• or f has an infinite discontinuity in [a, b], i.e. there exists a c ∈ [a, b] such that either
| lim
x→c−f(x)|=∞ or| lim
x→c+f(x)|=∞.
For the case if the interval has infinite length and
• if a∈R and Rt
af(x)dx exists for every number t≥a, then Z ∞
a
f(x)dx= lim
t→∞
Z t
a
f(x)dx provided this limit exists (as a finite number).
• if b∈R and Rb
t f(x)dx exists for every number t≤b, then Z b
−∞
f(x)dx= lim
t→−∞
Z b
t
f(x)dx provided this limit exists (as a finite number).
The improper integrals Z ∞
a
f(x)dx and Z b
−∞
f(x)dx are called convergent if the corre- sponding limit exists and divergentif the limit does not exist.
• if both Z ∞
a
f(x)dx and Z a
−∞
f(x)dx are convergent for any a ∈R, then Z ∞
−∞
f(x)dx= Z a
−∞
f(x)dx+ Z ∞
a
f(x)dx.
For the case if f has an infinite discontinuity in [a, b] and
• if f is continuous on [a, b) and lim
x→b−|f(x)|=∞,then Z b
a
f(x)dx= lim
t→b−
Z t
a
f(x)dx provided this limit exists (as a finite number).
• if f is continuous on (a, b] and lim
x→a+|f(x)|=∞, then Z b
a
f(x)dx= lim
t→a+
Z b
t
f(x)dx provided this limit exists (as a finite number).
The improper integral Z b
a
f(x)dx is called convergentif the corresponding limit exists and divergentif the limit does not exist.
• if f has a discontinuity at c, where a < c < b, and both Z c
a
f(x)dx and Z b
c
f(x)dx are convergent then
Z b a
f(x)dx= Z c
a
f(x)dx+ Z b
c
f(x)dx
Examples
(1) Z ∞
1
1
xp dx= lim
t→∞
Z t 1
1
xp dx= lim
t→∞
x1−p 1−p
t
1
if p6= 1 lnx|t1 if p= 1
=
1
p−1 if p > 1
∞ if p≤1
(2) Z 1
0
1
xp dx= lim
t→0+
Z 1 t
1
xp dx= lim
t→0+
x1−p 1−p
1
t
if p6= 1 lnx|1t if p= 1
=
1
1−p if p < 1
∞ if p≥1 (3)
Z 5 2
√ 1
x−2dx= lim
t→2+
Z 5 t
√ 1
x−2dxu=x−2=
s=t−2 lim
s→0+
Z 3 s
√1
udu= lim
s→0+2√
u|3s = 2√ 3.
Comparison Theorem Suppose that f and g are continuous functions with f(x)≥ g(x) ≥0 for x≥a.
(a) If Z ∞
a
f(x)dx is convergent, then Z ∞
a
g(x)dx is convergent.
(b) If Z ∞
a
g(x)dx is divergent, then Z ∞
a
f(x)dx is divergent.
Remarks
(a) Limit Comparison Test (Type I)Suppose thatf andgare continuous, positive functions for all values of x,and
x→∞lim f(x)
g(x) =`≥0 .
• If 0< ` <∞, then Z ∞
a
f(x)dx converges if and only if and Z ∞
a
g(x)dx converges.
• If` = 0 and Z ∞
a
g(x)dx converges, then Z ∞
a
f(x)dx converges.
• If` =∞ and Z ∞
a
f(x)dx converges, then Z ∞
a
g(x)dx converges.
(b) Limit Comparison Test (Type II) Suppose that f and g are continuous, positive func- tions for all values of x,and
lim
x→a+
f(x)
g(x) =` ≥0 .
• If 0< ` <∞, then Z b
a
f(x)dx converges if and only if and Z b
a
g(x)dx converges.
• If` = 0 and Z b
a
g(x)dx converges, then Z b
a
f(x)dx converges.
• If` =∞ and Z b
a
f(x)dx converges, then Z b
a
g(x)dx converges.
Examples
(1) By the Comparison Theorem, Z ∞
0
e−x2dx converges since e−x ≥ e−x2 > 0 for x ≥ 1 and Z ∞
1
e−xdx=e−1 <∞ converges.
(2) By the Comparison Theorem, Z ∞
1
1−e−x
x dx diverges since 1−e−x
x ≥ 3/4
x for x ≥2 and Z ∞
2
3/4
x dx=∞diverges.
(3) By the Limit Comparison Test, Z ∞
1
x x2+√
x+ 1dx diverges since lim
x→∞
x x2+√
x+1 1 x
= 1 and Z ∞
1
1
xdx=∞ diverges.
(4) Since Z 2
1
x2+x+ 1 (x2−1)1/3 dx=
Z 2 1
(x2−1) + 12(2x) + 2 (x2−1)1/3 dx =
Z 2 1
(x2−1)2/3dx +
u=x2−1
Z 3 0
1/2 u1/3 du+
Z 2 1
2
(x−1)1/3(x+ 1)1/3 dx = Z 2
1
(x2 − 1)2/3dx + Z 3
0
1/2
u1/3 du +
w=x−1
Z 1 0
2
w1/3(w+ 2)1/3 dw, lim
w→0+
2/w1/3(w+ 2)1/3
1/w1/3 = 2 and R1 0
1
w1/3 dw = 3
2 converges, Z 2
1
x2+x+ 1
(x2−1)1/3 dx converges by the Limit Comparison Test.
(5) Since Z ∞
1
5−2 sin(x)
x3/2 dx= 5 Z ∞
1
1
x3/2 dx−2 Z ∞
1
sin(x)
x5/4x1/4 dx, lim
x→∞
sinx/x3/2
1/x5/4 = lim
x→∞
sinx x1/4 = 0 and both
Z ∞ 1
1
x3/2 dx and Z ∞
1
1
x5/4 dx converge, Z ∞
1
5−2 sin(x)
x3/2 dx converges by the Limit Comparison Test.
(6) Since Z π/2
0
sin(x) x3/2 dx=
Z π/2 0
sin(x)
xx1/2 dx, lim
x→0+
sinx/x3/2
1/x1/2 = lim
x→0+
sinx
x = 1 and Z π/2
0
1 x1/2 dx converges,
Z π/2 0
sin(x)
x3/2 dx converges by the Limit Comparison Test.
