Riemann Integral: Motivation
Riemann Integral: Motivation
Partition of [a, b]: A finite set {x0 , x1 , ..., xn } ⊂ [a, b] such that
a = x0 < x1 < · · · < xn = b.
Riemann Integral: Motivation
Partition of [a, b]: A finite set {x0 , x1 , ..., xn } ⊂ [a, b] such that
a = x0 < x1 < · · · < xn = b.
Upper sum & Lower sum: Let f : [a, b] → R be bounded.
Riemann Integral: Motivation
Partition of [a, b]: A finite set {x0 , x1 , ..., xn } ⊂ [a, b] such that
a = x0 < x1 < · · · < xn = b.
Upper sum & Lower sum: Let f : [a, b] → R be bounded.
For a partition P = {x0 , x1 , ..., xn } of [a, b], let
Mi = sup{f (x) : x ∈ [xi−1 , xi ]},
mi = inf{f (x) : x ∈ [xi−1 , xi ]} for i = 1, 2, ..., n
Riemann Integral: Motivation
Partition of [a, b]: A finite set {x0 , x1 , ..., xn } ⊂ [a, b] such that
a = x0 < x1 < · · · < xn = b.
Upper sum & Lower sum: Let f : [a, b] → R be bounded.
For a partition P = {x0 , x1 , ..., xn } of [a, b], let
Mi = sup{f (x) : x ∈ [xi−1 , xi ]},
mi = inf{f (x) : x ∈ [xi−1 , xi ]} for i = 1, 2, ..., n
U(f , P) =
n
P
i=1
Mi (xi − xi−1 ) – Upper sum
Riemann Integral: Motivation
Partition of [a, b]: A finite set {x0 , x1 , ..., xn } ⊂ [a, b] such that
a = x0 < x1 < · · · < xn = b.
Upper sum & Lower sum: Let f : [a, b] → R be bounded.
For a partition P = {x0 , x1 , ..., xn } of [a, b], let
Mi = sup{f (x) : x ∈ [xi−1 , xi ]},
mi = inf{f (x) : x ∈ [xi−1 , xi ]} for i = 1, 2, ..., n
U(f , P) =
L(f , P) =
n
P
i=1
n
P
i=1
Mi (xi − xi−1 ) – Upper sum
mi (xi − xi−1 ) – Lower sum
Ex. Let f (x) = x 4 − 4x 3 + 10 for all x ∈ [1, 4]. Calculate U(f , P)
and L(f , P) for the partition P = {1, 2, 3, 4} of [1, 4].
Ex. Let f (x) = x 4 − 4x 3 + 10 for all x ∈ [1, 4]. Calculate U(f , P)
and L(f , P) for the partition P = {1, 2, 3, 4} of [1, 4].
m(b − a) ≤ L(f , P) ≤ U(f , P) ≤ M(b − a), where
M = {f (x) : x ∈ [a, b]} and m = inf{f (x) : x ∈ [a, b]}.
Ex. Let f (x) = x 4 − 4x 3 + 10 for all x ∈ [1, 4]. Calculate U(f , P)
and L(f , P) for the partition P = {1, 2, 3, 4} of [1, 4].
m(b − a) ≤ L(f , P) ≤ U(f , P) ≤ M(b − a), where
M = {f (x) : x ∈ [a, b]} and m = inf{f (x) : x ∈ [a, b]}.
Upper integral:
Rb̄
a
f = inf U(f , P)
P
Ex. Let f (x) = x 4 − 4x 3 + 10 for all x ∈ [1, 4]. Calculate U(f , P)
and L(f , P) for the partition P = {1, 2, 3, 4} of [1, 4].
m(b − a) ≤ L(f , P) ≤ U(f , P) ≤ M(b − a), where
M = {f (x) : x ∈ [a, b]} and m = inf{f (x) : x ∈ [a, b]}.
Upper integral:
Rb̄
f = inf U(f , P)
Rb
f = sup L(f , P)
a
Lower integral:
a
P
P
Ex. Let f (x) = x 4 − 4x 3 + 10 for all x ∈ [1, 4]. Calculate U(f , P)
and L(f , P) for the partition P = {1, 2, 3, 4} of [1, 4].
m(b − a) ≤ L(f , P) ≤ U(f , P) ≤ M(b − a), where
M = {f (x) : x ∈ [a, b]} and m = inf{f (x) : x ∈ [a, b]}.
Upper integral:
Rb̄
f = inf U(f , P)
Rb
f = sup L(f , P)
a
Lower integral:
a
P
P
Riemann integral: If Upper integral = Lower integral, then f is
Riemann integrable on [a, b] and the common value is the
Rb
Riemann integral of f on [a, b], denoted by f .
a
Examples:
(i) f (x) = k for all x ∈ [0, 1].
Examples:
(i) f (x) = k for all x ∈ [0, 1].
0 if x ∈ (0, 1],
(ii) Let f (x) =
1 if x = 0.
Examples:
(i) f (x) = k for all x ∈ [0, 1].
0 if x ∈ (0, 1],
(ii) Let f (x) =
1 if x = 0.
1 if x ∈ [0, 1] ∩ Q,
(iii) Let f (x) =
0 if x ∈ [0, 1] ∩ (R \ Q).
Examples:
(i) f (x) = k for all x ∈ [0, 1].
0 if x ∈ (0, 1],
(ii) Let f (x) =
1 if x = 0.
1 if x ∈ [0, 1] ∩ Q,
(iii) Let f (x) =
0 if x ∈ [0, 1] ∩ (R \ Q).
(iv) f (x) = x for all x ∈ [0, 1].
Examples:
(i) f (x) = k for all x ∈ [0, 1].
0 if x ∈ (0, 1],
(ii) Let f (x) =
1 if x = 0.
1 if x ∈ [0, 1] ∩ Q,
(iii) Let f (x) =
0 if x ∈ [0, 1] ∩ (R \ Q).
(iv) f (x) = x for all x ∈ [0, 1].
(v) f (x) = x 2 for all x ∈ [0, 1].
Examples:
(i) f (x) = k for all x ∈ [0, 1].
0 if x ∈ (0, 1],
(ii) Let f (x) =
1 if x = 0.
1 if x ∈ [0, 1] ∩ Q,
(iii) Let f (x) =
0 if x ∈ [0, 1] ∩ (R \ Q).
(iv) f (x) = x for all x ∈ [0, 1].
(v) f (x) = x 2 for all x ∈ [0, 1].
Ex. Let f : [a, b] → R be bounded. Let there exist a sequence
(Pn ) of partitions of [a, b] such that L(f , Pn ) → α and
Rb
U(f , Pn ) → α. Show that f ∈ R[a, b] and that f = α.
a
Ex. Let f (x) =
x if x ∈ [0, 1] ∩ Q,
0 if x ∈ [0, 1] ∩ (R \ Q).
Examine whether f is Riemann integrable on [0, 1].
R1
Also, find f , if it exists.
0
Ex. Let f (x) =
x if x ∈ [0, 1] ∩ Q,
0 if x ∈ [0, 1] ∩ (R \ Q).
Examine whether f is Riemann integrable on [0, 1].
R1
Also, find f , if it exists.
0
Ex. Let f : [0, 1] → R be defined by
1 if x = n1 for some n ∈ N,
f (x) =
0 otherwise.
Examine whether f is Riemann integrable on [0, 1].
R1
Also, find f , if it exists.
0
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Some Riemann integrable functions:
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Some Riemann integrable functions:
(i) A continuous function on [a, b]
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Some Riemann integrable functions:
(i) A continuous function on [a, b]
(ii) A bounded function on [a, b] which is continuous except at
finitely many points in [a, b]
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Some Riemann integrable functions:
(i) A continuous function on [a, b]
(ii) A bounded function on [a, b] which is continuous except at
finitely many points in [a, b]
(iii) A monotonic function on [a, b]
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Some Riemann integrable functions:
(i) A continuous function on [a, b]
(ii) A bounded function on [a, b] which is continuous except at
finitely many points in [a, b]
(iii) A monotonic function on [a, b]
Properties of Riemann integrable functions:
Riemann’s criterion for integrability: A bounded function
f : [a, b] → R is Riemann integrable on [a, b] iff for each ε > 0,
there exists a partition P of [a, b] such that U(f , P) − L(f , P) < ε.
Some Riemann integrable functions:
(i) A continuous function on [a, b]
(ii) A bounded function on [a, b] which is continuous except at
finitely many points in [a, b]
(iii) A monotonic function on [a, b]
Properties of Riemann integrable functions:
Ex. Show that
1
√
3 2
≤
R1
0
2
√x
1+x
dx ≤ 13 .
Ex. If f : [a, b] → R is continuous, then show that there exists
Rb
c ∈ [a, b] such that f (x) dx = (b − a)f (c).
a
(This result is called the mean value theorem of Riemann
integrals.)
Ex. If f : [a, b] → R is continuous, then show that there exists
Rb
c ∈ [a, b] such that f (x) dx = (b − a)f (c).
a
(This result is called the mean value theorem of Riemann
integrals.)
Ex. Let f : [a, b] → R and g : [a, b] → R be continuous and let
g(x) ≥ 0 for all x ∈ [a, b]. Show that there exists c ∈ [a, b] such
Rb
Rb
that f (x)g(x) dx = f (c) g(x) dx.
a
a
(This result is called the generalized mean value theorem of
Riemann integrals.)
First fundamental theorem of calculus: Let f : [a, b] → R be
Rx
Riemann integrable on [a, b] and let F (x) = f (t) dt for all
a
x ∈ [a, b]. Then F : [a, b] → R is continuous.
First fundamental theorem of calculus: Let f : [a, b] → R be
Rx
Riemann integrable on [a, b] and let F (x) = f (t) dt for all
a
x ∈ [a, b]. Then F : [a, b] → R is continuous.
Also, if f is continuous at x0 ∈ [a, b], then F is differentiable at
x0 and F 0 (x0 ) = f (x0 ).
First fundamental theorem of calculus: Let f : [a, b] → R be
Rx
Riemann integrable on [a, b] and let F (x) = f (t) dt for all
a
x ∈ [a, b]. Then F : [a, b] → R is continuous.
Also, if f is continuous at x0 ∈ [a, b], then F is differentiable at
x0 and F 0 (x0 ) = f (x0 ).
Second fundamental theorem of calculus: Let f : [a, b] → R be
Riemann integrable on [a, b]. If there exists a differentiable
function F : [a, b] → R such that F 0 (x) = f (x) for all x ∈ [a, b],
Rb
then f (x) dx = F (b) − F (a).
a
Riemann sum: S(f , P) =
n
P
i=1
f (ci )(xi − xi−1 ),
where f : [a, b] → R is bounded,
P = {x0 , x1 , ..., xn } is a partition of [a, b],
and ci ∈ [xi−1 , xi ] for i = 1, 2, ..., n.
Riemann sum: S(f , P) =
n
P
i=1
f (ci )(xi − xi−1 ),
where f : [a, b] → R is bounded,
P = {x0 , x1 , ..., xn } is a partition of [a, b],
and ci ∈ [xi−1 , xi ] for i = 1, 2, ..., n.
Result: A bounded function f : [a, b] → R is Riemann integrable
on [a, b] iff lim S(f , P) exists in R.
kPk→0
Riemann sum: S(f , P) =
n
P
i=1
f (ci )(xi − xi−1 ),
where f : [a, b] → R is bounded,
P = {x0 , x1 , ..., xn } is a partition of [a, b],
and ci ∈ [xi−1 , xi ] for i = 1, 2, ..., n.
Result: A bounded function f : [a, b] → R is Riemann integrable
on [a, b] iff lim S(f , P) exists in R.
kPk→0
Also, in this case,
Rb
a
f = lim S(f , P).
kPk→0
Riemann sum: S(f , P) =
n
P
i=1
f (ci )(xi − xi−1 ),
where f : [a, b] → R is bounded,
P = {x0 , x1 , ..., xn } is a partition of [a, b],
and ci ∈ [xi−1 , xi ] for i = 1, 2, ..., n.
Result: A bounded function f : [a, b] → R is Riemann integrable
on [a, b] iff lim S(f , P) exists in R.
kPk→0
Also, in this case,
Rb
a
f = lim S(f , P).
kPk→0
1
Ex. Evaluate lim [ n+1
+
n→∞
1
n+2
+ ··· +
1
n+n ].
Improper integrals:
Improper integrals:
(i) Type I : The interval of integration is infinite
Improper integrals:
(i) Type I : The interval of integration is infinite
(ii) Type II : The integrand is unbounded in the (finite) interval
of integration
Improper integrals:
(i) Type I : The interval of integration is infinite
(ii) Type II : The integrand is unbounded in the (finite) interval
of integration
Also, combination of Type I and Type II is possible.
Improper integrals:
(i) Type I : The interval of integration is infinite
(ii) Type II : The integrand is unbounded in the (finite) interval
of integration
Also, combination of Type I and Type II is possible.
Convergence of Type I improper integrals:
Rx
Let f ∈ R[a, x] for all x > a. If lim f (t) dt exists in R, then
R∞
a
f (t) dt converges and
R∞
a
x→∞ a
f (t) dt = lim
Rx
x→∞ a
f (t) dt.
Improper integrals:
(i) Type I : The interval of integration is infinite
(ii) Type II : The integrand is unbounded in the (finite) interval
of integration
Also, combination of Type I and Type II is possible.
Convergence of Type I improper integrals:
Rx
Let f ∈ R[a, x] for all x > a. If lim f (t) dt exists in R, then
R∞
f (t) dt converges and
a
Otherwise,
R∞
x→∞ a
f (t) dt = lim
a
R∞
a
f (t) dt is divergent.
Rx
x→∞ a
f (t) dt.
Improper integrals:
(i) Type I : The interval of integration is infinite
(ii) Type II : The integrand is unbounded in the (finite) interval
of integration
Also, combination of Type I and Type II is possible.
Convergence of Type I improper integrals:
Rx
Let f ∈ R[a, x] for all x > a. If lim f (t) dt exists in R, then
R∞
f (t) dt converges and
a
Otherwise,
R∞
x→∞ a
f (t) dt = lim
a
R∞
Rx
x→∞ a
f (t) dt.
f (t) dt is divergent.
a
Similarly, we define convergence of
Rb
−∞
f (t) dt and
R∞
−∞
f (t) dt.
Example:
R∞ 1
1
tp
dt converges iff p > 1.
Example:
R∞ 1
1
tp
dt converges iff p > 1.
Ex. Examine whether the following improper integrals converge.
R∞ t
R∞ 1
R∞
R∞ −t 2
2
(i)
e dt
(ii) 1+t
(iii) e−t dt
(iv)
te dt
2 dt
−∞
0
0
−∞
Example:
R∞ 1
1
tp
dt converges iff p > 1.
Ex. Examine whether the following improper integrals converge.
R∞ t
R∞ 1
R∞
R∞ −t 2
2
(i)
e dt
(ii) 1+t
(iii) e−t dt
(iv)
te dt
2 dt
−∞
0
0
−∞
Comparison test: Let 0 ≤ f (t) ≤ g(t) for all x ≥ a. If
converges, then
R∞
a
f (t) dt converges.
R∞
a
g(t) dt
Example:
R∞ 1
1
tp
dt converges iff p > 1.
Ex. Examine whether the following improper integrals converge.
R∞ t
R∞ 1
R∞
R∞ −t 2
2
(i)
e dt
(ii) 1+t
(iii) e−t dt
(iv)
te dt
2 dt
−∞
0
0
−∞
Comparison test: Let 0 ≤ f (t) ≤ g(t) for all x ≥ a. If
converges, then
R∞
R∞
g(t) dt
a
f (t) dt converges.
a
Limit comparison test: Let f (t) ≥ 0 let g(t) > 0 for all t ≥ a and
f (t)
let lim g(t)
= ` ∈ R.
t→∞
Example:
R∞ 1
1
tp
dt converges iff p > 1.
Ex. Examine whether the following improper integrals converge.
R∞ t
R∞ 1
R∞
R∞ −t 2
2
(i)
e dt
(ii) 1+t
(iii) e−t dt
(iv)
te dt
2 dt
−∞
0
−∞
0
Comparison test: Let 0 ≤ f (t) ≤ g(t) for all x ≥ a. If
converges, then
R∞
R∞
g(t) dt
a
f (t) dt converges.
a
Limit comparison test: Let f (t) ≥ 0 let g(t) > 0 for all t ≥ a and
f (t)
let lim g(t)
= ` ∈ R.
t→∞
(i) If ` 6= 0, then
(ii) If ` = 0, then
R∞
a
R∞
a
f (t) dt converges iff
f (t) dt converges if
R∞
a
R∞
a
g(t) dt converges.
g(t) dt converges.
Ex. Examine whether the following improper integrals converge.
R∞ 2
R∞ dt
R∞ −t p
√
(i) sint 2 t dt
(ii)
(iii)
e t dt
2
1
1 t
1+t
1
Ex. Examine whether the following improper integrals converge.
R∞ 2
R∞ dt
R∞ −t p
√
(i) sint 2 t dt
(ii)
(iii)
e t dt
2
1
1 t
1+t
Absolute convergence: If
converges.
R∞
a
1
|f (t)| dt converges, then
R∞
a
f (t) dt
Ex. Examine whether the following improper integrals converge.
R∞ 2
R∞ dt
R∞ −t p
√
(i) sint 2 t dt
(ii)
(iii)
e t dt
2
1 t
1
1+t
Absolute convergence: If
a
converges.
Example:
R∞ cos t
0
R∞
1+t 2
1
|f (t)| dt converges, then
dt converges.
R∞
a
f (t) dt
Ex. Examine whether the following improper integrals converge.
R∞ 2
R∞ dt
R∞ −t p
√
(i) sint 2 t dt
(ii)
(iii)
e t dt
2
1 t
1
1+t
Absolute convergence: If
Example:
R∞ cos t
0
R∞
a
converges.
1+t 2
1
|f (t)| dt converges, then
R∞
f (t) dt
a
dt converges.
Integral test for series: Let f : [1, ∞) → R be a positive
∞
R∞
P
decreasing function. Then
f (n) converges iff f (t) dt
converges.
n=1
1
Dirichlet’s test: Let f : [a, ∞) → R and g : [a, ∞) → R such that
(i) f is decreasing and lim f (t) = 0, and
t→∞
(ii) g
and there exists M > 0 such that
xis continuous
R
g(t) dt ≤ M for all x ≥ a.
Then
a
R∞
a
f (t)g(t) dt converges.
Dirichlet’s test: Let f : [a, ∞) → R and g : [a, ∞) → R such that
(i) f is decreasing and lim f (t) = 0, and
t→∞
(ii) g
and there exists M > 0 such that
xis continuous
R
g(t) dt ≤ M for all x ≥ a.
Then
a
R∞
f (t)g(t) dt converges.
a
Examples: (i)
R∞ sin t
1
t
dt
(ii)
R∞
0
sin(x 2 ) dx
Dirichlet’s test: Let f : [a, ∞) → R and g : [a, ∞) → R such that
(i) f is decreasing and lim f (t) = 0, and
t→∞
(ii) g
and there exists M > 0 such that
xis continuous
R
g(t) dt ≤ M for all x ≥ a.
Then
a
R∞
f (t)g(t) dt converges.
a
Examples: (i)
R∞ sin t
1
t
dt
(ii)
R∞
sin(x 2 ) dx
0
Convergence of Type II and mixed type improper integrals:
Dirichlet’s test: Let f : [a, ∞) → R and g : [a, ∞) → R such that
(i) f is decreasing and lim f (t) = 0, and
t→∞
(ii) g
and there exists M > 0 such that
xis continuous
R
g(t) dt ≤ M for all x ≥ a.
Then
a
R∞
f (t)g(t) dt converges.
a
Examples: (i)
R∞ sin t
1
t
dt
(ii)
R∞
sin(x 2 ) dx
0
Convergence of Type II and mixed type improper integrals:
Example:
R1
0
1
tp
dt converges iff p < 1.
Ex. Examine the convergence of the improper integral
R1
0
√ dx
x−x 2
.
Ex. Examine the convergence of the improper integral
R1
0
Ex. Determine all real values of p for which the integral
R∞ e−x −1
dx is convergent.
xp
0
√ dx
x−x 2
.
Lengths of smooth curves:
(i) Let y = f (x), where f : [a, b] → R is such that f 0 is
continuous.
Rb p
Then L =
1 + (f 0 (x))2 dx
a
Lengths of smooth curves:
(i) Let y = f (x), where f : [a, b] → R is such that f 0 is
continuous.
Rb p
Then L =
1 + (f 0 (x))2 dx
a
(ii) Let x = ϕ(t), y = ψ(t), where ϕ : [a, b] → R and
ψ : [a, b] → R are such that ϕ0 and ψ 0 are continuous.
Rb p
Then L =
(ϕ0 (t))2 + (ψ 0 (t))2 dt
a
Lengths of smooth curves:
(i) Let y = f (x), where f : [a, b] → R is such that f 0 is
continuous.
Rb p
Then L =
1 + (f 0 (x))2 dx
a
(ii) Let x = ϕ(t), y = ψ(t), where ϕ : [a, b] → R and
ψ : [a, b] → R are such that ϕ0 and ψ 0 are continuous.
Rb p
Then L =
(ϕ0 (t))2 + (ψ 0 (t))2 dt
a
(iii) Let r = f (θ), where f : [α, β] → R is such that f 0 is
continuous.
Rβ p
Then L =
r 2 + (f 0 (θ))2 d θ
α
Ex. Find the length of the curve y =
Rx √
cos 2t dt, 0 ≤ x ≤ π4 .
0
Ex. Find the length of the curve y =
Rx √
cos 2t dt, 0 ≤ x ≤ π4 .
0
Ex. Find the perimeter of the ellipse
x2
a2
+
y2
b2
= 1.
Ex. Find the length of the curve y =
Rx √
cos 2t dt, 0 ≤ x ≤ π4 .
0
Ex. Find the perimeter of the ellipse
x2
a2
+
y2
b2
= 1.
Ex. Find the length of the curve x = et sin t, y = et cos t,
0 ≤ t ≤ π2 .
Ex. Find the length of the curve y =
Rx √
cos 2t dt, 0 ≤ x ≤ π4 .
0
Ex. Find the perimeter of the ellipse
x2
a2
+
y2
b2
= 1.
Ex. Find the length of the curve x = et sin t, y = et cos t,
0 ≤ t ≤ π2 .
Ex. Find the length of the cardioid r = 1 − cos θ.
Ex. Find the length of the curve y =
Rx √
cos 2t dt, 0 ≤ x ≤ π4 .
0
Ex. Find the perimeter of the ellipse
x2
a2
+
y2
b2
= 1.
Ex. Find the length of the curve x = et sin t, y = et cos t,
0 ≤ t ≤ π2 .
Ex. Find the length of the cardioid r = 1 − cos θ.
Area between two curves: If f , g : [a, b] → R are continuous and
f (x) ≥ g(x) for all x ∈ [a, b], then we define the area between
Rb
y = f (x) and y = g(x) from a to b to be (f (x) − g(x)) dx.
a
Ex. Find the length of the curve y =
Rx √
cos 2t dt, 0 ≤ x ≤ π4 .
0
Ex. Find the perimeter of the ellipse
x2
a2
+
y2
b2
= 1.
Ex. Find the length of the curve x = et sin t, y = et cos t,
0 ≤ t ≤ π2 .
Ex. Find the length of the cardioid r = 1 − cos θ.
Area between two curves: If f , g : [a, b] → R are continuous and
f (x) ≥ g(x) for all x ∈ [a, b], then we define the area between
Rb
y = f (x) and y = g(x) from a to b to be (f (x) − g(x)) dx.
a
Ex. Find the area above the x-axis which is included between
the parabola y 2 = ax and the circle x 2 + y 2 = 2ax, where
a > 0.
Area in polar coordinates: Let f ; [α, β] → R be continuous. We
define the area bounded by r = f (θ) and the lines θ = α and
Rβ
θ = β to be 12 (f (θ))2 d θ.
α
Area in polar coordinates: Let f ; [α, β] → R be continuous. We
define the area bounded by r = f (θ) and the lines θ = α and
Rβ
θ = β to be 12 (f (θ))2 d θ.
α
Ex. Find the area of the region that is inside the cardioid
r = a(1 + cos θ) and also inside the circle r = 32 a.
Volume by slicing: V =
Rb
a
A(x) dx.
Volume by slicing: V =
Rb
A(x) dx.
a
Ex. A solid lies between planes perpendicular to the x-axis at
x = 0 and x = 4. The cross sections perpendicular to the axis
on the interval 0 ≤ x ≤ 4√are squares whose diagonals
run
√
from the parabola y = − x to the parabola y = x. Find the
volume of the solid.
Volume by slicing: V =
Rb
A(x) dx.
a
Ex. A solid lies between planes perpendicular to the x-axis at
x = 0 and x = 4. The cross sections perpendicular to the axis
on the interval 0 ≤ x ≤ 4√are squares whose diagonals
run
√
from the parabola y = − x to the parabola y = x. Find the
volume of the solid.
Rb
Volume of solid of revolution: V = π(f (x))2 dx.
a
Volume by slicing: V =
Rb
A(x) dx.
a
Ex. A solid lies between planes perpendicular to the x-axis at
x = 0 and x = 4. The cross sections perpendicular to the axis
on the interval 0 ≤ x ≤ 4√are squares whose diagonals
run
√
from the parabola y = − x to the parabola y = x. Find the
volume of the solid.
Rb
Volume of solid of revolution: V = π(f (x))2 dx.
a
Ex. Find the volume of a sphere of radius r .
Volume by slicing: V =
Rb
A(x) dx.
a
Ex. A solid lies between planes perpendicular to the x-axis at
x = 0 and x = 4. The cross sections perpendicular to the axis
on the interval 0 ≤ x ≤ 4√are squares whose diagonals
run
√
from the parabola y = − x to the parabola y = x. Find the
volume of the solid.
Rb
Volume of solid of revolution: V = π(f (x))2 dx.
a
Ex. Find the volume of a sphere of radius r .
Volume by washer method: V =
Rb
a
π((f (x))2 − (g(x))2 ) dx
Volume by slicing: V =
Rb
A(x) dx.
a
Ex. A solid lies between planes perpendicular to the x-axis at
x = 0 and x = 4. The cross sections perpendicular to the axis
on the interval 0 ≤ x ≤ 4√are squares whose diagonals
run
√
from the parabola y = − x to the parabola y = x. Find the
volume of the solid.
Rb
Volume of solid of revolution: V = π(f (x))2 dx.
a
Ex. Find the volume of a sphere of radius r .
Volume by washer method: V =
Rb
a
π((f (x))2 − (g(x))2 ) dx
√
Ex. A round hole of radius 3 is bored through the centre of a
solid sphere of radius 2. Find the volume of the portion bored
out.
Area of surface of revolution: S =
Rb
a
p
2πf (x) 1 + (f 0 (x))2 dx.
Area of surface of revolution: S =
Rb
a
p
2πf (x) 1 + (f 0 (x))2 dx.
Ex. Consider the funnel formed by revolving the curve y = x1
about the x-axis, between x = 1 and x = a, where a > 1. If Va
and Sa denote respectively the volume and the surface area of
the funnel, then show that lim Va = π and lim Sa = ∞.
a→∞
a→∞
Area of surface of revolution: S =
Rb
a
p
2πf (x) 1 + (f 0 (x))2 dx.
Ex. Consider the funnel formed by revolving the curve y = x1
about the x-axis, between x = 1 and x = a, where a > 1. If Va
and Sa denote respectively the volume and the surface area of
the funnel, then show that lim Va = π and lim Sa = ∞.
a→∞
a→∞
Ex. Find the volume and area of the curved surface of a
paraboloid of revolution formed by revolving the parabola
y 2 = 4ax about the x-axis, and bounded by the section x = x1 .