Calculus for Engineers (MAT1011)
Module 1 - Question Bank
1
Dr. T. Phaneendra
Department of Mathematics
Continuity and Differentiability
Definition 1.1 (Continuity). Let f : D → R, where D ⊂ R, and c ∈ D . Then f is said to
be continuous at c, if
(1.1)
lim f ( x) = f ( c).
x→ c
If f is continuous at every point of [a, b], then f is continuous on [a, b].
The following functions are known to be continuous on the real line:
• Exponential function e x
• Polynomial function a 0 x m + a 1 x m−1 + · · · + a m−1 x + a m
• Trigonometric sin x and cos x
(
• Modulus function f ( x) =
x,
x≥0
− x,
x<0
Definition 1.2 (Differentiability). Let f : D → R, where D ⊂ R, and c ∈ D . Then f is
said to be differentiable at c, if
l = lim
x→ c
f ( x) − f ( c )
x−c
(1.2)
exists, and the limit value l is called the derivative f 0 ( c) of f at c
Definition 1.3 (Critical Point). An interior point c of the domain D of a function f ( x)
is called a critical point of f , if
• f 0 ( c) is undefined or
• f 0 ( c) = 0 is zero
If f 0 ( c) = 0, then the tangent at the point ( c, f ( c)) is parallel to the x-axis
Example 1.1. Consider f ( x) = x(4 − x)3 for all x ∈ R. Since f ( x) is a polynomial function, it is differentiable at all real x, with
f 0 ( x) = x · 3(4 − x)2 (−1) + (4 − x)3 · 1
= (4 − x)2 (−3 x + 4 − x) = 4(4 − x)2 (1 − x).
Then f 0 ( x) = 0 only if (4 − x)2 (1 − x) = 0. Thus x = 1, 4 are the critical points of f .
Remark 1.1. For a rational polynomial function f ( x) = P ( x)/Q ( x), the zeros of the
denominator Q ( x), and the zeros of the numerator of f 0 ( x) are critical points.
Example 1.2. Consider f ( x) = x2 /( x − 2)for all x ∈ R. Then
f 0 ( x) =
SJT, 511 (A10)
( x−2)·(2 x)−( x2 )(1)
( x−2)2
=
x(2 x−4− x)(1)
( x−2)2
=
1
x( x−4)
( x−2)2
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Therefore, f 0 ( x) = 0 only if x = 0, 4. While, f 0 ( x) = ∞ at x = 2. Thus x = 0, 2, 4 are the
critical points of f .
p
Example 1.3. Consider f ( x) = x2 − 32 x for all x ∈ R. Then
p =2
f 0 ( x) = 2 x − 232
x
³
x3/2
p −8
x
´
Therefore, f 0 ( x) = 0 only if x3/2 − 8, that is only if x = 4. While, f 0 ( x) = ∞ at x = 0.
Thus x = 0, 4 are the critical points of f .
2
Absolute Extrema on a Finite Closed Interval
Let f : [a, b] → R and c ∈ [a, b].
Definition 2.1. We say that f ( c) is the absolute maximum value on [a, b], if
f ( x) ≤ f ( c) for all x ∈ [a, b].
(2.1)
Then c is called a point of absolute maximum of f .
Definition 2.2. We say that f (d ) is the absolute minimum value on [a, b], if
f ( d ) ≤ f ( x) for all x ∈ [a, b].
(2.2)
Then d is called a point of absolute minimum of f .
To find absolute extrema of a continuous function f on a finite closed interval [a, b]:
(a) Find all critical points of f in the open interval (a, b), and then evaluate f at
them
(b) Evaluate f at the end points a and b
(c) The largest of the computed values is the absolute maximum, while the smallest
is the absolute minimum for f on [a, b]
Example 2.1. Consider f ( x) = x2 for all x ∈ [−2, 1]. Then f is differentiable at every
real x, and hence on [−2, 1]. So the critical points are obtained from the condition
f 0 ( x) = 0, where f 0 ( x) = 2 x. Thus x = 0 is the only critical point.
Point
Critical Point x = 0
Left End Point x = −2
Right End Point x = 1
f ( x) = x2
0
4
1
Nature of f ( x)
The Smallest
The Largest
–
The function has the absolute maximum value of 4 at x = −2, and an absolute minimum value of 0 at x = 0.
Example 2.2. Consider f ( x) = (2 x/3) − 5 for all x ∈ [−2, 3]. Then f is differentiable
at every real x, and hence on [−2, 3]. So the critical points are obtained from the
SJT, 511 (A10)
2
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
condition f 0 ( x) = 0. But f 0 ( x) = 2/3 6= 0. Thus the critical points do not exist for f .
However, the functional values at the end points are f (−2) = −19/3 and f (3) = −3.
Therefore, f has the absolute maximum value of −3 at x = 3 and an absolute minimum
value of −19/3 at x = −2.
p
Example 2.3. Consider f ( x) = 5 − x2 for all x ∈ [−2, 1]. Then f 0 ( x) = −
x
5− x
−p
p1
(−2 x) =
2 5− x2
p
0
0
· Therefore, f ( x) = 0 at x = 0, and f ( x) = ∓∞ at x = ± 5 ∉ [−2, 1]. Thus f is
2
differentiable on [−2, 3], and 0 is the only critical point of f .
Point
Critical Point x = 0
Left End Point x = −2
Right End Point x = 1
f ( x)
p
5
1
2
Nature of f ( x)
The Largest
The Smallest
-
p
Therefore, f has the absolute maximum value of 5 at x = 0 and an absolute minimum
value of 1 at x = −2.
Example 2.4. Consider f ( x) = xa (1 − x)b , where a and b are positive real numbers.
To find the maximum value of f in [0, 1],
f 0 ( x) = xa · b(1 − x)b−1 (−1) + axa−1 (1 − x)b
= xa−1 (1 − x)b [− bx + a(1 − x)] for x ∈ (0, 1)·
Also f (0) = f (1) = 0. Therefore, f 0 ( x) = 0 only if −bx + a(1 − x) = 0 or x = a/(a + b).
Thus x = a/(a + b) is the only critical point of f , and
f
¡
a
a+ b
¢
=
¡
¢ ¡
¢b
a a
1 − a+a b
a+ b
=
aa b b
( a+ b ) a+ b
> 0.
Therefore, f has the absolute maximum value of aa b b /(a + b)a+b at x = 0 and an absolute minimum value of 0 at the end points x = 0, 1.
Exercise 2.1. Find the absolute maxima and absolute minima of the following functions on the given intervals:
( a) f ( x) =
2x
3
− 5, −2 ≤ x ≤ 3
( b) f ( x) = x2 − 1, −1 ≤ x ≤ 2
( c) f ( x) = − x12 , 21 ≤ x ≤ 2
p
( d ) f ( x) = 3 x, −1 ≤ x ≤ 8
p
( e) f ( x) = 5 − x2 , −2 ≤ x ≤ 1
Answers.
6= 0, critical points do not exist; f (−2) = − 19
3 , f (3) = −3; an absolute
maximum value at x = −3 and an absolute minimum value at x = −2
( a) f 0 ( x) =
SJT, 511 (A10)
2
3
3
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
( b) f 0 ( x) = 2 x = 0 if x = 0, the critical point is x = 0, f (0) = −1; f (−1) = 0, f (2) = 3;
an absolute maximum value at x = 2 and an absolute minimum value at x = 0
£
¤
( c) f 0 ( x) = x23 6= 0 for x 6= 0 in 21 , 2 . Also f 0 (0) = +∞ but x = 0 is not in the domain
[ 12 , ≤ 2] of f . Thus there is no critical point; f (1/2) = −4 and f (2) = −1/4; an
absolute maximum value at x = 2 and an absolute minimum value at x = 1/2
= ∞ at x = 0, f 0 ( x) 6= 0 for x 6= 0 in [−1, 8]. Thus x = 0 is the only
critical point; f (−1) = −1 and f (8) = 2; an absolute maximum value at x = 2 and
an absolute minimum value at x = −1
p
( e) f 0 ( x) = − p x 2 = 0 if x = 0, the critical point is 0; f (0) = 5, f (−2) = 1, f (1) = 2.
5− x
p
Thus the absolute maximum value 5 at x = 0, and an absolute minimum 1 at
x = −2
( d ) f 0 ( x) =
1
3 x2/3
Exercise 2.2 (Try Yourself). Find the absolute maxima and absolute minima of the
following functions on the given intervals:
( a) f ( x) =
p
3
x, −1 ≤ x ≤ 8
( b) f ( x) = x3 − 3 x2 + x, −1/2 ≤ x ≤ 4
( c) f ( x) = x/( x2 − x + 1), 0 ≤ x ≤ 3
3
Mean Value Theorems
Theorem 3.1 (Rolle’s Theorem). Consider f : [a, b] → R. Suppose that
(a) f is continuous at every point of the closed interval [a, b],
( b) f is differentiable on (a, b), and
( c) f (a) = f ( b).
Then there is at least one number c in (a, b) such that f 0 ( c) = 0.
Example 3.1. Verify Rolle’s theorem, and find an appropriate constant c of it for
f ( x) = ( x − a)m ( x − b)n on [a, b]. Note that f is continuous and differentiable at all
real values. Hence, f is continuous on [a, b], differentiable on (a, b) with
f 0 ( x) = ( x − a)m n( x − b)n−1 + m( x − a)m−1 ( x − b)n
= ( x − a)m−1 ( x − b)n−1 [ n( x − a) + m( x − b)] for x ∈ (a, b).
Also, f (a) = f (b) = 0.
Then by Rolle’s theorem, there exists at least one c ∈ (a, b) such that f 0 ( c) = 0. Now
an+ bm
solving f 0 ( x) = 0, we get x = anm++bm
n · Taking c = m+ n , we note that c ∈ (a, b), and is
the required constant c.
SJT, 511 (A10)
4
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Exercise 3.1 (Try Yourself). Verify Rolle’s theorem and and find an appropriate constant c of it for each of the following functions:
p
1 − x2 , [−1, 1]
i
h 2
( b) log xx(a++ab
b) , in [a, b], a > 0
( a)
( c)
p
x − x/3, [0, 9]
( d ) sin x, [0, π]
( e)
sin x
ex ,
[0, π]
Exercise 3.2 (Try Yourself). Give the reason why Rolle’s theorem is not applicable to
each of the following functions:
( a)
x2 −4 x
x−2 ,
[0, 4]
( b) 1 − ( x − 1)2/3 in [0, 2]
( c) sec x, [0, 2π]
( d ) | x|, [−1, 1]
( e) tan x, [0, π]
Theorem 3.2 (Lagrange’s Mean Value Theorem). Suppose that
(a) y = f ( x) is continuous on the closed interval [a, b],
( b) f is differentiable on (a, b).
Then there is at least one number c in (a, b) such that
f ( b ) − f ( a)
= f 0 ( c).
b−a
(3.1)
Example 3.2. Consider f ( x) = lx2 + mx + n for all x ∈ [a, b]. Then f is continuous
and differentiable at all real values. Hence, f is continuous on [a, b], differentiable on
(a, b) with f 0 ( x) = 2 lx + m. Then by the mean value theorem, there exists a c ∈ (a, b)
− f ( a)
such that f 0 ( c) = f (bb)−
a · That is,
2l c + m =
( lb2 + mb + n) − ( la2 + ma + n)
b−a
so that c = (a + b)/2. Since c, being the average of a and b, lies in between them. Thus
the mean value theorem is verified.
SJT, 511 (A10)
5
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Example 3.3. Now we find the values of a, m and b such that function f ( x) satisfy the
hypotheses of the mean value theorem on [0, 2], and hence an appropriate c, where


3,
f ( x) = − x2 + 3 x + a,


mx + b,
x=0
0<x<1
1 ≤ x ≤ 2.
From the conditions of the mean value theorem, f is continuous on [0, 2] and differentiable on (0, 2). From the continuity of f at x = 0, we have f (0) = f (0 + 0), that is
3 = limh→0 [−(0 + h)2 + 3(0 + h) + a] so that a = 3. While, by the differentiability of f at
x = 1, we see that R [ f 0 (1)] = L[ f 0 (1)] or m = |−2 x + 3| x=1 so that m = 1. Finally, by the
continuity of f at x = 1, we get f (1 − 0) = f (1 + 0) or limh→0 [−(1 − h)2 + 3(1 − h) + a] =
limh→0 [ m(1 + h) + b] so that 2 + a = m + b or b = a + 2 − m = 3 + 2 − 1 = 4. Thus with
a = 3, b = 4 and m = 1, we redefine the function as


3,
f ( x ) = − x 2 + 3 x + 3,


x + 4,
x=0
0<x<1
1 ≤ x ≤ 2.
By the mean value theorem, there exists a c ∈ (0, 2) such that
f 0 ( c) =
f (2)− f (0)
2−0
or f 0 ( c) = 23 ·
(3.2)
If c ∈ (0, 1), then f 0 ( c) = −2 c+3. Then from (3.2), we get −2 c+3 = 3/2 or c = 3/4. While,
if c ∈ (1, 2), then f 0 ( c) = 1. Then, from (3.2), we get 1 = 3/2, which is not possible. Thus
the constant c must lie in (0, 1) only.
Example 3.4. Verify the mean value theorem for the following functions f ( x) in the
given interval and find an appropriate constant c in each case:
(a) x + 1x , x ∈ [ 21 , 1]
p
( b) x − 1, x ∈ [1, 3]
( c) log x, x ∈ [1, e]
(d )
1
x,
x ∈ [a, b], a > 0
Answers. The constant c of the mean value theorem:
( a) c = 1
( b) c =
3
2
( c) c = e − 1
p
( d ) c = ab
SJT, 511 (A10)
6
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Example 3.5. Explain why for each of the following functions, the mean value theorem is not applicable:
(a) x2/3 in [−1, 8]
(
sin x
x , (−π ≤ x < 0)
( b)
0,
( x = 0)
(
x2 − x,
(−2 ≤ x ≤ −1)
( c)
2
2 x − 3 x − 3, (−1 < x ≤ 0)
4
Monotonic Functions
Definition 4.1. We say that f : [a, b] → R is
• increasing on [a, b] if f ( x1 ) ≤ f ( x2 ) for all x1 , x2 ∈ [a, b] such that x1 < x2 ,
• decreasing on [a, b] if f ( x1 ) ≥ f ( x2 ) for all x1 , x2 ∈ [a, b] such that x1 < x2 .
A function which is either increasing or decreasing is called a monotonic function.
Theorem 4.1 (Test of Monotonicity). Suppose that f ( x) is continuous on [a, b] and differentiable on (a, b).
• If f 0 ( x) > 0 at each point x ∈ (a, b), then f is increasing on [a, b],
• If f 0 ( x) < 0 at each point x ∈ (a, b), then f is decreasing on [a, b].
Example 4.1. Find the critical points of f ( x) = x4 − 8 x2 + 16 = ( x2 − 4)2 and separate
the intervals on which f is increasing and on which f is decreasing:
Solution. We see that f 0 ( x) = 2( x2 − 4)(2 x) = 4 x( x − 2)( x + 2). Then the critical points
of f are x = 0, ±2, and
x-values
x < −2
−2 < x < 0
0<x<2
x>2
4x
−
−
+
+
x−2
−
−
−
+
x+2
−
+
+
+
f 0 ( x)
−
+
−
+
f ( x)
&
%
&
%
Interval
(−∞, −2)
(−2, 0)
(0, 2)
(2, ∞)
Therefore, f is % on (−2, 0) ∪ (2, ∞) and & on (−∞, −2) ∪ (0, 2).
p
Example 4.2. Find the critical points of x − 6 x − 1 and separate the intervals on
which f is increasing and on which f is decreasing:
Solution. We see that
p
p
f 0 ( x) = 1 − 6/2 x − 1 = 1 − 3/ x − 1.
Then the critical points of f are x = 1, 10, and
SJT, 511 (A10)
7
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
x-values
x<1
1 < x < 10
x > 10
p
x−1
imaginary
<3
>3
Dr. T. Phaneendra
Department of Mathematics
p
3/ x − 1
f 0 ( x)
f ( x)
Interval
>1
<1
−
+
&
%
(1, 10)
(10, ∞)
Therefore, f is increasing on (10, ∞), decreasing on (1, 10)..
3
Example 4.3. Find the critical points of f ( x) = 3 xx2 +1 and separate the intervals on
which f is increasing and on which f is decreasing:
Solution. We see that
f 0 ( x) =
3 x2 ( x2 + 1)
≥ 0·
(3 x2 + 1)2
Then the only critical point of f is x = 0. Since f 0 ( x) > 0 when x 6= 0, we conclude that
f ( x) is always increasing and never decreasing on (−∞, ∞).
Exercise 4.1. Find the critical points of each of the following functions f ( x), and
separate the intervals on which f is increasing and on which f is decreasing:
(a) 2 x3 − 18 x
( b ) x4 − 4 x3 + 4 x2
p
( c ) x 8 − x2
( d ) x2/3 ( x + 5)
x2 −3
x−2 ,
x 6= 2
p
( f ) x2 5 − x
p
( g ) 4 x − x2 + 3
( e)
Answers
p
p
p
(a) f 0 (p
x) =p6 x2 − 18; Critical points are x = ± 3; % on (−∞, − 3) ∪ ( 3, ∞), & on
(− 3, 3)
( b) f 0 ( x) = 4 x3 − 12 x2 + 8 x; Critical points are x = 0, 1, 2; % on (0, 1) and (2, ∞), &
on (−∞, −0) ∪ (1, 2)
p
p
2
p − x ) ; Critical points are x = ±2, ±2 2; % on (−2, 2), & on (−2 2, 2) ∪
( c) f 0 ( x) = 2(4
2
p 8− x
(2, 2 2)
( d ) f 0 ( x) =
5( x+2)
p ; Critical points are
33x
( e ) f 0 ( x) =
(2, 3)
( x−3)( x−1)
; Critical points are
( x−2)2
SJT, 511 (A10)
x = −2, 0; % on (−∞, −2) ∪ (0, ∞), & on (−2, 0)
x = 1, 3; % on (−∞, 1) ∪ (3, ∞), & on (1, 2) ∪
8
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
( f ) f 0 ( x) = 1 − 5 xp(4− x) ; Critical points are x = 0, 4, 5; % increasing on (0, 4), & on
2 5− x
(−∞, 0) ∪ (4, 5)
( g ) f 0 ( x) =
2−p
2 x3/2
;
x
Critical points are x = 0, 1; % on (0, 1), & on (1, ∞)
Theorem 4.2 (First Derivative Test). If f 0 ( x) changes its sign from positive to negative on
passing through c from left to the right, that is
(
0
f ( x)
>0
<0
for x < c
for x > c
then f has a local maximum at c. If f 0 ( x) changes its sign from negative to positive on passing
through c from left to the right, that is
0
f ( x)
(
<0
>0
for x < c
for x > c
then f has a local minimum at c. If f 0 ( x) positive or negative on both sides of c, then f does
not have local extremum at c.
Example 4.4. Using the first derivative test, identify the local extreme values of
f ( x) = x2 − 4 x + 4 = ( x − 2)2 in [1, −∞) :
Solution. We have
Step 1. f ( x) is continuous for all x ≥ 1, and at the left-end point x = 1: f (1) = 1
Step 2. f 0 ( x) = 2( x − 2) = 0, Critical point is x = 2
Step 3. Note that
0
f ( x)
(
>0
<0
for x > 2
for x < 2.
Therefore, x = 2 is a point of local minimum and the local minimum value is
f (2) = 0
Step 4. Since f ( x) → ∞ as x → ∞, there is no absolute maximum. But the left-end
point x = 1 is a point of local maximum and the local maximum value is f (1) = 1
Example 4.5. Using the first derivative test, identify the local extreme values of
f ( x) =
p
25 − x2 in [−5, 5] :
Solution. We have
Step 1. f ( x) is continuous for all x ∈ [−5, 5], and f ( x) = 0 at the end points x = ±5
x
25− x2
Step 2. f 0 ( x) = − p
SJT, 511 (A10)
for x ∈ (−5, 5), the critical point is x = 0
9
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Step 3. Note that f 0 ( x) > 0 on (−5, 0), and f 0 ( x) < 0 on (0, 5). Thus, x = 0 is a point
of local maximum and the local maximum value is f (0) = 5. Also, at the end
points x = ±5, f has the absolute minimum value is f ( x) = 0
Example 4.6. Using the first derivative test, identify the local extreme values of
12 x − x3 in [−3, −∞) :
Solution. We have
Step 1. f ( x) is continuous for all x ≥ −3, at the left-end point x = −3: f ( x) = −9
Step 2. f 0 ( x) = 3(4 − x2 ), Critical points are x = ±2
Step 3. Now
0
f ( x)
(
>0
<0
for − 2 < x < 2
S
if x ∈ (−3, −2) (2, ∞)
f 0 ( x) changes its sign from negative to positive on passing through c = −2 from left to
right. So a local minimum value of −16 occurs at x = −2. Similarly, f 0 ( x) changes its
sign from negative to positive on passing through c = 2 from left to right. Therefore,
x = 2 is also a point of local minimum with the local minimum value is f (2) = 0
subinterval
sign of f 0
(−∞, −2)
(−2, 0)
(0, 2)
(2, ∞)
f 0 ( x) < 0
f 0 ( x) > 0
f 0 ( x) < 0
f 0 ( x) > 0
nature of f
decreasing
increasing
decreasing
increasing
Example 4.7. Find the intervals on which the function f ( x) = ax2 + bx + c with a 6= 0,
is increasing and decreasing. Describe the reasoning behind your answer.
³
Solution. We can write f ( x) = ax2 + bx + c = a x + 2ba
´2 ³ 2
´
− b 4−a42ac , which is a parabola
with vertex at x = − 2ba ;
³
´
³
´
• For a > 0, the parabola is oriented up and hence f is increasing on − 2ba , ∞
³
´
and decreasing on ∞, − 2ba .
• If a < 0, the parabola is oriented down and hence f is decreasing on − 2ba , ∞
³
´
and increasing on ∞, − 2ba .
³
´
Note that f 0 ( x) = 2a x + 2ba = 0 so that x = − 2ba is a critical point of f .
Example 4.8. Determine the constants a and b so that f ( x) = ax2 + bx has an absolute
maximum at the point (1, 2).
SJT, 511 (A10)
10
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Solution. We see that f (1) = 2 implies a + b = 2. while f 0 ( x) = 2ax + b. Since x = 1
is a critical point of f , f 0 (1) = 0 gives 2a + b = 0. Solving these conditions, we get
a = −2, b = 4. Hence f ( x) = 2 x2 + 4 x.
Exercise 4.2. Using the first derivative test, identify the local extreme values of each
of the following functions f ( x) in the given domain:
(a) x3 + 3 x2 + 3 x + 1, x ≤ 0
( b)
x2
,
4− x2
−2 < x ≤ 1
Answer (a) f 0 ( x) = 3( x + 1)2 ; Critical point is x = −1;
f (−1) = 0, f (0) = 1
local maximum 1 at x = 0 No local minimum
Answer (b) f 0 ( x) = (4−8xx2 )2 ; Critical point is x = 0; f (0) = 0, f (1) = 1/3;
local maximum 1/3 at x = 1, a local minimum 0 at x = 0
Definition 4.2 (Concavity). If the graph of f ( x) lies above the tangents at its points
on an interval I , we say that C is concave up in I . While, If the graph of f ( x) lies
below the tangents at its points on an interval I , we say that C is concave down in I .
Theorem 4.3. Let y = f ( x) be a plane curve C . Then the graph of f ( x) is concave up or
down in I according as f 00 ( x) > 0 or f 00 ( x) < 0 for all x ∈ I respectively.
Definition 4.3. A point P on a curve y = f ( x) is called a point of inflection, if f
is continuous at P and the concavity of the curve reverses on passing through P .
P ( c, f ( c)) is a point of inflection on the curve y = f ( x) if the sign of f 00 ( x) is different
on either side of the ordinate x = c
Theorem 4.4 (Second Derivative Test). Suppose that f "( x) is continuous near the point
c.
• If f 0 ( c) = 0 and f 00 ( c) > 0, then f ( x) has a local minimum at c
• If f 0 ( c) = 0 and f 00 ( c) < 0, then f ( x) has a local maximum at c
Example 4.9. Determine the critical points, points of local maxima and local minima
3
2
of f ( x) = x3 − x2 − 2 x + 13 , and then identify the intervals on which f is concave up and
concave down. Also find the points of inflection of f .
Solution. We have
• f 0 ( x) = x2 − x − 2, f 00 ( x) = 2 x − 1; Critical points are x = −1, 2;
• f is % on (−∞, −1) ∪ (2, ∞), f is & on (−1, 2) and (0, 2);
• local maximum
3
2
at x = −1, local minimum −3 at x = 2;
• concave up for x > 1/2, concave down for x < 1/2;
SJT, 511 (A10)
11
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
• x = 1/2 is point of inflection
Example 4.10. Determine the critical points, points of local maxima and local minima of f ( x) = p x2 , and then identify the intervals on which f is concave up and
x +1
concave down. Also find the points of inflection of f .
Solution. We have
• f 0 ( x) =
1
,
x2 +1)3/2
x
f 00 ( x) = − x2 +31)
5/2 ; No critical points;
• f is % on (−∞, ∞),
• No local maxima and No local minima;
• concave up on (−∞, 0) and concave down on (0, ∞);
• x = 0 is a point of inflection
Example 4.11. Determine the critical points, points of local maxima and local minx1/3 2
( x − 7), and then identify the intervals on which f is concave up and conima of 914
cave down. Also find the points of inflection of f .
Solution. We have
−2/3
• f 0 ( x) = 3 x 2 ( x2 − 1), f 00 ( x) = x−5/3 (2 x2 + 1); The critical points are x = 0, ±1;
• f is % on (−∞, −1) ∪ (1, ∞), & on (−1, 1);
• local maximum
27
7
at x = −1, local minimum − 27
7 at x = −1;
• concave up on (0, ∞), concave down on (−∞, 0);
• x = 0 is a point of inflection
Exercise 4.3 (Try Yourself). Determine the critical points, points of local maxima and
local minima of each of the following functions f ( x), and then identify the intervals
on which f is concave up and concave down. Also find the points of inflection in each
case:
( a)
3 2
2/3
4 ( x − 1)
( b ) 6 − 2 x − x2
( c) −2 x3 + 6 x2 − 3
( d ) x4 ( x − 5)
( e) x2/5
( f ) (2 − x2 )3/2
p
3
( g ) x3 + 1
SJT, 511 (A10)
12
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Hints and Answers
(a) f 0 ( x) = x( x2 − 1)−1/3 , f 00 ( x) =
x2 −3
; Critical points are
3( x2 −1)4/3
x = 0, ±1; f is increas-
ing on (−1, 0)∪(1, ∞), decreasing on (−∞, −1)∪(0, 1); local maximum 43 at x = 0,
p
p
local minima
= ±1; concave up on (−∞, − 3) ∪ ( 3, ∞), concave down
p p 0 at x p
on (− 3, 3); x = ± 3 are points of inflection
( b) f 0 ( x) = −2(1 + x), f 00 ( x) = −2; Critical point is x = −1; increasing on (−∞, −1),
decreasing on (−1, ∞); local maximum 7 at x = −1, No minimum; concave down
for all x; No point of inflection
( c) f 0 ( x) = 6 x(2 − x), f 00 ( x) = 12(1 − x); Critical points are x = 0, 2; increasing on
(0, 2), decreasing on (−∞, 0) ∪ (2, ∞); local maximum at x = 2, local minimum
at x = 0; concave up on (−∞, 1) and concave down on (1, ∞); x = 1 is a point of
inflection
( d ) f 0 ( x) = 5 x3 ( x − 4), f 00 ( x) = 20 x2 ( x − 3); Critical points are x = 0, 4; increasing on
(−∞, 0) ∪ (4, ∞), decreasing on (0, 4); local maximum at x = 0, local minimum
at x = 4; concave up on (3, ∞) and concave down on (−∞, 3); x = 3 is a point of
inflection
f 00 ( x) = − 5 x49/5 ; x = 0 is a critical point; increasing on (−∞, ∞); No
maxima, No minima; concave up on (−∞, 0) and concave down on (0, ∞); x = 0
( e ) f 0 ( x) =
1
,
5 x4/5
is a point of inflection
p
p
2
p − x ) ; Critical points are x = ± 2; f is increasing on (− 2, 0), de( f ) f 0 ( x) = − 6(1
2− x2p
p
creasing on (0p, 2); Local maximum
at x = 0, Local minima at x = ± 2; conp
cave up on (− 2, −1) ∪ (1, 2) and concave down on (−1, 1); x = ±1 are points
of inflection
2
x
( g) f 0 ( x) = ( x3 +x1)2/3 , f 00 ( x) = ( x3 +21)
5/3 ; Critical points are x = 0, −1; f is increasing on
(−∞, −1) ∪ (−1, 0) ∪ (0, ∞); No maxima, No minima; concave up on (−∞, −1) ∪
(0, ∞) and concave down on (−1, 0); x = 0, −1 are points of inflection
5
Area of the Region bounded by two Plane Curves
The area A of a region D enclosed by the plane curves y = f ( x) and y = g( x) between
the ordinates x = a and x = b, where f ( x) and g( x) are continuous and f ( x) ≥ g( x) for
x ∈ [a, b], is given by the definite integral of f ( x) − g( x) from x = a and x = b. That is
Zb
A=
[ f ( x) − g( x)] d x
x= a
Remark 5.1. If f ( x) ≥ g( x) for x ∈ [a, c] and f ( x) ≤ g( x) for x ∈ [ c, b], then
Zb
A=
x= a
SJT, 511 (A10)
| f ( x) − g( x)| d x =
Zc
[ f ( x) − g( x)] d x +
x= a
Zb
[ g( x) − f ( x)] d x
x= c
13
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Example 5.1. Find the area of the region enclosed by the parabolas y = 2 x − x2 and
y = x2 .
Solution. Let f ( x) = 2 x − x2 and g( x) = x2 . From y = 2 x − x2 , we see that y − 1 =
−(1 − 2 x + x2 ) or y − 1 = −( x − 1)2 , which is a negatively oriented parabola, with vertex
at (1, 1), passing through the origin (0, 0). This intersects the parabola y = x2 in two
points (0, 0) and (1, 1). So the x-limits are x = 0 to x = 1. Also, f ( x) − g( x) = 2 x2 − x2 −
x2 = 2( x − x2 ) ≥ 0 for all 0 ≤ x ≤ 1. Therefore,
Z1
A=
[ f ( x) − g( x)] d x = 2
x=0
Z1
¯
¯ 2
3 ¯1
¯
( x − x2 ) d x = 2 ¯ x2 − x3 ¯
x=0
=2
¡1
2
¢
− 13 =
1
3
x=0
Example 5.2. Find the area of the region enclosed by the cubical parabola y = x3 and
the straight line y = x.
Solution. The curve y = x3 and the straight line y = x intersect in the points (0, 0) and
(1, 1). So the x-limits are x = 0 to x = 1. Also, x ≥ x3 for all 0 ≤ x ≤ 1. Therefore,
Z1
A=
¯ 2
¯
4 ¯1
¯
( x − x3 ) d x = ¯ x2 − x4 ¯
x=0
=
¡1
2
¢
− 14 =
1
4
x=0
Example 5.3. Find the area of the region enclosed by the curves y = sin x and y =
cos x between the ordinates x = 0 and x = π/2.
Solution. The curves y = sin x and y = cos x intersect in the point where x = π4 . Also,
cos x ≥ sin x for 0 ≤ x ≤ π4 , while sin x ≥ cos x for π4 ≤ x ≤ π2 . Therefore,
Zπ/4
Zπ/2
A=
(cos x − sin x) d x +
(sin x − cos x) d x
x=π/4
π/4
= |sin x + cos x| x=0 − |sin x + cos x|πx=/2π/4
x=0
=
³
p1
2
´ ³
´
p
+ p1 − 0 − 1 − 0 + 1 − p1 − p1 = 2( 2 − 1)
2
2
2
Example 5.4. Find the area enclosed by the curves y = sin x and y = sin 2 x between
the ordinates x = 0 and x = π
Solution. The curves y = sin x and y = sin 2 x intersect in the point where x = π3 . Also,
sin 2 x ≥ sin x for 0 ≤ x ≤ π3 , while sin x ≥ sin 2 x for π3 ≤ x ≤ π.
Therefore,
A=
Zπ/3
Zπ
(sin 2 x − sin x) d x +
(sin x − sin 2 x) d x
x=0
x=π/3
¯
¯π/3 ¯
¯π
= ¯− cos22 x + cos x¯ x=0 + ¯− cos x + cos22 x ¯ x=π/3 =
SJT, 511 (A10)
14
5
2
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Example 5.5. Find the area enclosed by the parabolas y2 = 4ax and x2 = 4a y in the
first quadrant.
Solution. The parabolas intersect in the points, given by 4ax = ( x2 /4a)2 , that is at x = 0
p
and x = 4a. Also, 2 ax ≥ x2 /4 for all 0 ≤ x ≤ 4a. Therefore,
A=
4
Ra
x=0
¯ p
¯
3
p
p
3/2
3 ¯4 a
¯
= 43 · a · (4a)3/2 − (412aa) = 16a2 /3.
(2 ax − x2 /4a) d x = ¯2 a · x3/2 − 3x.4a ¯
x=0
Example 5.6. Find the area enclosed by the grapha y = 2 x and y = x2 − 8
Solution. The graphs intersect in the points, given by 2 x = x2 − 8, that is or x = −2 and
x = 4. Also 2 x ≥ x2 − 8 for all −2 ≤ x ≤ 4. Therefore,
Z4
A=
¯4
¯ 2
¯
¯
[2 x − ( x2 − 8)]) d x = ¯− x3 + x2 + 8 x¯
x=−2
= 56
x=−2
Example 5.7. Find the area enclosed by the graphs y = | x| and y = 1 − | x|
Solution. The graphs intersect in the points, given by 1 − | x| = | x|, that is x = ±1/2.
Also, 1 − | x| ≥ | x| for all − 21 ≤ x ≤ 12 .
Therefore,
Z1/2
A=
[1 − | x| − | x|] d x =
x=−1/2
Z0
=
Z1/2
(1 − 2 | x|) d x
x=−1/2
Z1/2
(1 + 2 x) d x +
(1 − 2 x) d x = 1/2
x=0
x=−1/2
Exercise 5.1 (Try Yourself). Find the area of the region enclosed by
(a) the curves y = sin x and y = cos x between the ordinates x = 0 and x = π
(b) the parabola y = x2 and the line y = x
(c) the parabola y2 = x and the line y = x
(d) the astroid
p
x+
p
y = 1 and the line x + y = 1 in the first quadrant
(e) the graphs of y = cos x and y = cos 2 x from x = 0 to x = 2π/3
(f) the graphs of y = 3 x2 + 12 and y = 4( x + 1) from x = −3 to x = 3
(g) the parabolas y = 20 + x − x2 and y = x2 − 5 x
(h) the parabola y = 8 − 3 x2 and the curve y = x3 − 6 x (from x = −1 to x = 2)
SJT, 511 (A10)
15
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
6
Dr. T. Phaneendra
Department of Mathematics
Volumes of Solids of Revolution
Disk Method
Let f ( x) be a continuous function on [a, b]. The volume of solid of revolution obtained
by revolving the arc of the plane curve y = f ( x) from x = a to x = b about the x-axis,
is
b
Z
V=
a
b
Z
π y2 dx =
(6.1)
π[ f ( x)]2 dx.
a
Let g( y) be a continuous function on [ c, d ]. The volume of solid of revolution obtained
by revolving the arc of the plane curve x = g( y) from y = c to y = d about the y-axis,
is
d
Z
V=
c
π x2 d y =
d
Z
(6.2)
π[ g( y)]2 d y.
c
Example 6.1. Find the volume of the solid of revolution of the arc of the parabola
p
y = x from x = 0 to x = 1 about the x-axis.
Solution. The solid of revolution of the arc of the parabola is a paraboloid with x-axis
R1
as its axis, and its volume is V =
π y2 dx =
x=0
R1
π xdx = π/2.
x=0
Example 6.2. Regarding a sphere of radius a, as a solid of revolution of the semicircular arc x2 + y2 = a2 from x = −a to x = a about the x-axis, find its volume.
Solution. The volume of the sphere of radius a is
Za
2
Za
π y dx =
V=
x=−a
2
Za
2
π(a − x )dx = 2π
x=−a
(a2 − x2 ) dx
x=0
¯a
¯
µ
¶
¯
a3
x3 ¯
= 2π a3 −
= 4πa3 /3.
= 2π ¯¯a2 x − ¯¯
3 x=0
3
Exercise 6.1 (Try Yourself). Find the volume of the hyperboloid, obtained by revolving the hyperbola y2 − x2 = 1 from x = −a to x = a about the x-axis.
Example 6.3. Find the volume of the solid of revolution of the arc of the curve y = x3
from y = 0 to y = 8 about the y-axis.
Solution. The required volume is
Z8
2
Z8
πx d y =
V=
y=0
¯ 5/3 ¯8
¯ 3y ¯
¯
5 ¯
π y2/3 d y = 4π ¯¯
y=0
= 96π/5.
y=0
Example 6.4. Find the volume of the solid generated by revolving the region enclosed by the hyperbola x y = 2 about the y-axis, between the limits y = 1 to y = 8.
SJT, 511 (A10)
16
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Solution. The required volume is
Z4
V=
π x2 d y =
y=1
Z1
¯4
¯
¯
¯
1
π(2/ y)2 d y = 4π ¯¯− ¯¯
y
x=0
y=1
= 4π ·
3
= 3π.
4
Exercise 6.2 (Try Yourself). Find the volume of the solid of revolution of the graph
of each of following functions about the x-axis, between the given limits
(a) x2 − 4 x + 5 from x = 1 to x = 4
(b) x2 − 3 x from x = 0 to x = 3
(c) x5/3 from x = 1 to x = 8
(d)
p
x + 1 from x = 1 to x = 4
(e) 2/( x + 1) from x = 1 to x = 3
(f)
p
cos x + 1 from x = 0 to x = π
Exercise 6.3 (Try Yourself). Regarding a cone of height h and radius a, as a solid of
revolution of the straight line segment joining the vertex (0, 0) to the point (a, h) from
y = 0 to y = h about the y-axis, find its volume.
Solution. V = πa2 h/3
Exercise 6.4 (Try Yourself). Regarding a cylinder of height h and radius a, as a solid
of revolution of the rectangle with edges x = 0, x = a, y = 0 and y = h about the y-axis,
find its volume.
Solution. V = πa2 h
Washer Method
The volume of solid of revolution of the region enclosed by the plane curves y = f ( x)
and y = g( x) with f ( x) ≥ g( x) from x = a to x = b about the x-axis, is
b
Z
V=
a
(6.3)
π[ f ( x)2 − g( x)2 ]dx.
The volume of solid of revolution of the region enclosed by the plane curves x = f ( y)
and x = g( y) with f ( y) ≥ g( y) from y = c to y = d about the y-axis, is
d
Z
V=
c
(6.4)
π[ f ( y)2 − g( y)2 ]d y.
Example 6.5. Find the volume of the solid of revolution of the region enclosed by
the parabola y = x2 and the line y = x about the x-axis.
SJT, 511 (A10)
17
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Solution. The two curves intersect in the points (0, 0) and (1, 1), and x2 ≤ x for all
0 ≤ x ≤ 1. Therefore, the volume of the solid of revolution is
Z1
2
Z1
2 2
¯ 3
¯x
π( x2 − x4 )dx = π ¯¯
π[ x − ( x ) ] dx =
V=
x=0
x=0
3
−
¯1
x5 ¯¯
= 2π/5.
5 ¯ x=0
Example 6.6. Find the volume of the solid of revolution of the region enclosed by
the parabola y = x2 + 1 and the line x + y = 3 about the x-axis.
Solution. The two curves intersect in the points, where x = −2 and x = 1. Also, x2 + 1 ≤
3 − x for all −2 ≤ x ≤ 1. Hence
Z1
2
2
2
Z1
π[(3 − x) − ( x + 1) ]dx =
V=
x=−2
π(8 − 6 x − x2 − x4 )dx = 117π/5.
x=−2
Exercise 6.5 (Try Yourself). Find the volume of the solid of revolution of each of the
following regions enclosed by the given curves about the x-axis (between the given
limits):
(a) y = x3 and y = x2
(b) y2 = 4( x − 1) and y = x − 1
(c) y = x2 + 2 and y = 10 − x2
(d) y = 1/ x and 2 y = 5 − 2 x
Exercise 6.6 (Try Yourself). Find the volume of the solid of revolution of each of the
following regions enclosed by the given curves about the y-axis:
(a) y = x1/3 and x = 4 y, x, y ≥ 0
(b) x2 − 2 x and y = x
(c) y = 16 − x and y = 3 x + 2
(d) y = x3 and y = x1/3
Exercise 6.7 (Try Yourself). Imagine the region D, which gives a cylindrical shell
with height h, in-radius r and out-radius R , as a solid of revolution about x-axis. Then
find its volume.
7
Average Value of a Function
Let y = f ( x) be continuous on [a, b]. The average value of f on the interval is given by
f ave =
SJT, 511 (A10)
1
b−a
Z
b
(7.1)
f ( x) dx.
x= a
18
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Example 7.1. A ball is thrown in the air vertically from ground level with an initial
velocity of 64 feet per second. Find the average height of the ball over the time interval
extending from the time of ball’s release to its return to ground level.
Solution. Given that the initial velocity is u = 64 feet per second, and the acceleration
due to gravity is g = 42 feet per second squared. Therefore, the height of the ball at
any time t is given by h ≡ h( t) = ut − gt2 /2. The minus sign is due to the fact that
the object is thrown against the gravity. Thus h( t) = 64 t − 16 t2 . The ball will be at
the ground when h( t) = 64 t − 16 t2 = 0, that is t = 0 and t = 4. Therefor, the average
height of the ball over 0 ≤ t ≤ 4 is
Z 4
Z 4
1
1
h( t) dt =
[64 t − 16 t2 ] dt
4 − 0 t=0
4 − 0 t=0
¯4
1¯
= ¯32 t2 − 16 t3 /3¯ t=0 = 42.667 ft.
4
h ave =
Exercise 7.1 (Try Yourself). Find the average value f ave of each of the following functions on the indicated interval:
(a) x2 + 10, [0, 2]
(b) 2 x3 − 3 x2 + 4 x − 1, [−1, 1]
p
5 x + 1, [0, 3]
p
(d) x x2 + 16, [0, 3]
(c)
(e) 2/( x + 1)2 , [3, 5]
(f) cos 2 x, [3, π/4]
(g) x2/3 − x−2/3 , [1, 4]
(h) | x| − 1, [−1, 3]
Exercise 7.2 (Try Yourself). A retarding force slows the motion of the weighted
spring so that the position of the mass at time t is y = e− t cos t, t ≥ 0. Then find the
average value of y over 0 ≤ t ≤ 2π.
Exercise 7.3 (Try Yourself). In the absence of damping, the position of a mass m
on a freely vibrating spring is given by x( t) = A cos(ω t + φ), where A , ω and φ are
constants. The period of oscillation is 2π/ω. The potential energy of the system is
U ( x) = κ x2 /2, where κ is the spring constant. The kinetic energy of the system is
K = mv2 /2, where v = dx/ dt. If ω2 = κ/ m, show that the average potential energy
and the average kinetic energy over one period are the same, and that each equals
κ A 2 /4.
Exercise 7.4 (Think about It). If y = f ( x) is a continuous odd function, what is its
average value on any interval [−a, a]?
Exercise 7.5 (Think about It). If y = f ( x) is a differentiable function, what is the
average value of f 0 on the interval [ x, x + h], where h > 0?
SJT, 511 (A10)
19
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
8
Dr. T. Phaneendra
Department of Mathematics
Gamma Function
(a) Definition
Γ( p) =
Z∞
e− x x p−1 dx,
(8.1)
p>0
0
(b) Recurrence Relation
Γ(1) = 1 and Γ( p) = ( p − 1)Γ( p − 1) for
(8.2)
p>1
(c) Gamma for Positive Integers Γ(n) = (n − 1)! for n = 1, 2, 3, . . .
Example 8.1.
Γ(6)
5!
5×4×3×2×1
=
=
= 30
2Γ(6) 2 × 2!
2×2
¡ ¢ p
Example 8.2. Γ 21 = π
Example 8.3. For n = 1, 2, 3, ..., we have
¶ µ
¶µ
¶ µ ¶µ ¶
1
1
3
3 1 p
Γ n+
= n−
n − ···
π
2
2
2
2 2
µ
Thus
µ ¶
3
=
2
µ ¶
5
Γ
=
2
µ ¶
7
Γ
=
2
µ ¶
9
Γ
=
2
Γ
( a)
( b)
( c)
(d )
µ ¶
1
1
·Γ
=
2
2
µ ¶
3
3
=
·Γ
2
2
µ ¶
5
5
=
·Γ
2
2
µ ¶
7
7
·Γ
=
2
2
p
π
2
p
p
3
π 3 π
·
=
2 2
4
p
p
5 3 π 15 π
·
=
2
4
8
p
p
7 15 π 105 π
·
=
2
8
16
Example 8.4.
Γ
¡5¢
2
Γ(3)
7 · Γ(1/2)
=
3
2
· 12 Γ
¡1¢
3 1 2
3
2 · 2!
¡1¢ = · · =
2
2
7
14
7·Γ
2
(c) Gamma for Negative Real Numbers For p < 0, p 6= 0, −1, −2, ..., we have
Γ( p) =
1
1
1
1
· Γ( p + 1) = ·
···
· Γ( p + k + 1),
p
p p+1
p+k
where k is the least positive integer such that p + k + 1 > 0
SJT, 511 (A10)
20
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
Dr. T. Phaneendra
Department of Mathematics
Example 8.5.
( a)
( b)
( c)
(d )
9
µ
¶
1
Γ − =
2
µ
¶
3
Γ − =
2
¶
µ
5
Γ − =
2
µ
¶
7
Γ − =
2
µ ¶
p
1
1
= −2 π
·Γ
−1/2
2
µ
¶
p
p
1
1
· Γ − = −2(−2 π) = 4 π
−1/2
2
¶
µ
1
2 p
8 p
3
π
·Γ − = − ·4 π = −
−3/2
2
3
15
p
¶ ½
¾½
¾
µ
1
2
8
16 π
5
−
=
·Γ − = −
−5/2
2
5
15
105
Beta Function
(a) Definition
Z1
B( p, q) =
x p−1 (1 − x) q−1 dx,
(9.1)
p > 0, q > 0
0
(b) Beta in terms of Trigonometric Sine and Cosine
π
Z2
B( p, q) = 2
sin2 p−1 θ cos2 q−1 θ d θ ,
(9.2)
p > 0, q > 0.
0
(c) Trigonometric integral in terms of Beta
π
Z2
µ
¶
1
r+1 s+1
sin θ cos θ d θ = · B
,
for r > 0, s > 0.
2
2
2
r
s
(9.3)
0
(d) Beta in terms of Gamma If p, q and r are positive real numbers, then
B( p, q) =
Γ ( p) Γ ( q)
Γ ( p + q)
(9.4)
Example 9.1.
Γ (2) Γ (5) 1 · 4!
1
=
=
Γ (2 + 5)
6!
30
Γ (4) Γ (7) 3! · 6!
1
( b) B(4, 7) =
=
=
Γ (4 + 7)
10!
1680
Γ (2) Γ (3/2) 1 · Γ (3/2)
( c) B(2, 3/2) =
=
=
Γ (2 + 3/2)
Γ (7/2)
( a)
B(2, 5) =
Γ (3/2)
4
¡3¢ =
3
15
· 2 ·Γ 2
·
¸2
2
Γ (9/2) Γ (7/2) 7 [Γ (7/2)]
1 5 3 1
5π
( d ) B(9/2, 7/2) =
= ·
=
· · Γ (1/2) =
Γ ((9 + 7)/2)
2
7!
2.6! 2 2 2
2048
SJT, 511 (A10)
21
5
2
phaneendra.t@vit.ac.in
Calculus for Engineers (MAT1011)
Module 1 - Question Bank
10
Dr. T. Phaneendra
Department of Mathematics
Evaluation of Integrals Using the Gamma
Example 10.1.
( a)
Zπ/2
sin2 θ cos3 θ d θ =
( b)
Zπ/2
sin10 θ d θ =
1
2
³ ´
3
Γ 2 Γ(2)
· Γ((3/2)+2) =
1
2
.1
· ΓΓ(3/2)
(7/2) =
1
2
Γ(3/2)
· (5/2)(3/2)
Γ(3/2) = 2/15
0
1
2
·
³ ´
11
Γ 2 Γ(1/2)
Γ((12/2)+2)
=
1
2
63π
9.7.5.3
· 15 · 92 · 72 · 52 · 32 · 12 · Γ (1/2) = 2π
.5! · 25 = 512
0
( c)
Zπ/2
128
cos9 θ d θ =
315
Zπ/2
16
sin6 θ cos7 θ d θ =
3003
(d )
0
0
An Important Result: Γ( p)Γ(1 − p) = sinπpπ for 0 < p < 1
Example 10.2.
( a)
Zπ/2p
Zπ/2
tan θ d θ = sin1/2 θ cos−1/2 θ d θ =
0
0
=
( b)
1
2
1
2
Γ(1/4)
· Γ(3/4)
=
Γ(1)
1
2
·
·
µ
¶µ
¶
(1/2)+1 (−1/2)+1
Γ
2
2
µ
¶
(1/2)−(1/2)+1
Γ
2
p
π
= π/2 2
sin(π/4)
Zπ/2p
Zπ/2q
Zπ/2p
p
¡π
¢
cot θ d θ =
cot 2 − θ d θ =
tan θ d θ = π/2 2
0
0
0
Example 10.3.
 π/2
  π/2

¾½
¾
 ½
Z p
Z
Γ(1/4)
1 Γ(1/2)
1
p
¡ 3 Γ(1/2)
¡
¢
¢
sin θ d θ
·
d θ = 12 · Γ(3/4)
2
Γ 4 + 12
Γ 12 + 41
sin θ



0
0
np
onp
o
π
π Γ(1/4)
= 14 2 · ΓΓ(3/4)
·
(5/4)
2
Γ(3/4)
Γ(1/4)
=π
4 · Γ(1/4)/4 = π
R1
Exercise 10.1 (Try Yourself). Find x3/2 (1 − x2 )5/2 dx, with the substitution x = sin θ .
0
Exercise 10.2 (Try Yourself). Evaluate
R1
0
SJT, 511 (A10)
p x
dx, using the substitution x = (sin θ )2/5 .
1− x5
22
phaneendra.t@vit.ac.in