Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
12.4 The Fundamental Theorem Calculus (5.4)
Fundamental Theorem of Calculus provides a connection between differentiable and
integral calculus.
Consider we interesting in computing the area that lies between a positive function
f(x) and x-axis on [a,b], i.e. A =
b
∫ f (t ) dt . In additional let the upper limit b vary, i.e.
a
we consider a function A ( x ) =
x
∫ f (t ) dt .
a
x
Ex 1.
∫ cos (t ) dt = sin t
x
a
= sin x − sin a
a
x
x
1
x3 − 0
∫ t dt = 4 t 3 = 4
0
0
Ex 2.
3
Let F be antiderivative of f , calculate the derivative of A ( x ) =
x
∫ f (t ) dt :
a
d
d
A ( x) =
dx
dx
Thus
d
dx
x
d
∫ f (t ) dt = dx ( F ( x ) − F (a)) = F ' ( x ) = f ( x )
a
x
∫ f (t ) dt = f ( x ) .
a
Ex 3.
⎧0
⎩1
Does f ( x ) = ⎨
0 ≤ x <1
has antiderivative?
1≤ x ≤ 2
Suppose F(x) is an antiderivative of f(x) on [0,2], then F(x) is differentiable
function, furthermore F’(x)=f(x). Define g(x)=F(x)-Cx for an arbitrary constant
C ∈ ( f ( 0 ), f ( 2 )) = ( 0,1) . The function g is clearly continuous function on [0,2],
because it differentiable, and therefore there is c, s.t. g’(c)=0. But g’(c)=f(c)-C,
i.e. f(c)=C for an arbitrary C ∈ ( 0,1) which is impossible. Note that
f ( 0 ) < C < f ( 2 ) so c ≠ 0 and c ≠ 2 . Thus f has no antiderivative.
Let calculate
existence:
d
dx
x
∫ f (t ) dt
a
for continuous f without assumptions of antiderivative
Course: Accelerated Engineering Calculus I
x+h
d
dx
x
∫ f (t ) dt = lim
∫
a
∫ f (t ) dt − ∫ f (t ) dt
a
x
f (t ) dt −
a
⇒
x
a
h
h→0
x+h
∫
x
f (t ) dt =
a
d
dx
Instructor: Michael Medvinsky
∫
x+h
∫
f (t ) dt +
a
x
x
f (t ) dt −
x
∫
x+h
f (t ) dt =
a
∫ f (t ) dt
x
x+h
1
∫ f (t ) dt = lim h ∫ f (t ) dt
h→0
a
x
From the continuity of f(x) on [x,x+h] we have
x+h
f ( xm ) = m < f ( x ) < M = f ( xM ) ⇒ mh <
∫
f (t ) dt < Mh ⇒ m <
x
1 x+h
∫ f (t ) dt < M .
h x
Since xm , xM ∈ [ x, x + h ] and h → 0 we have xm , xM → x , so f ( xm ) = m, f ( xM ) = M → f ( x ) ,
therefore by intermediate value theorem
d
dx
x
∫ f (t ) dt = f ( x )
a
The Fundamental Theorem of Calculus:
Suppose f(x) is continuous on [a,b].
1) If g ( x ) =
x
∫ f (t ) dt
then g' ( x ) = f ( x )
a
2)
b
∫ f (t ) dt = F (b) − F (a) for any F antiderivative of f,
that is F ' = f
a
Ex 4.
Ex 5.
Ex 6.
x
d
dx
d
dx
∫ cos
2
t dt = cos2 x = cos2 x
0
1/x
∫
1 $ 1'
⋅ &− )
x % x2 (
x dx =
3
2
2
x2
d x 1
d" c 1
1 % " d sin x 1
d
dx = $$ ∫ dt + ∫ dt '' = $$ −
dt +
∫
∫
dx sin2 x t
dt # sin2 x t
dx
c t
& # dx c t
=−
"1
%
1
1
⋅ ( 2sin x cos x ) + 2 ⋅ 2x = 2 $ − cot x '
2
#x
&
sin x
x
12.5 The Substitution Rule (5.5)
b
Indefinite Integral: ∫
a
( ( )) ( )
f g x g ' x dx
=
t=g ( x )
dt=g '( x )dx
∫ f (t )dt
1 %
∫ t dx '' =
c
&
x2
Course: Accelerated Engineering Calculus I
Ex 1.
∫
(2x + 7) dx =
2
1
1
1
49
t dt = t 2 = 2x + 7 + C = x 2 + 7x + + C
2
4
4
4
(
∫
t=2 x+7
dt=2dx
Instructor: Michael Medvinsky
)
=C
Ex 2.
∫
Ex 3.
∫
Ex 4.
∫
(3x + 5) 2 dx =
3
t=3x+5
dt=3dx
x
3x + 5
dx =
dx
=
1+ cos x
∫
1
t=x+5/3
dt=dx
∫
3
3
1 2
1t
1
t dt =
+ C = 3x + 5 + C =3x 3 +15x 2 + 25x + C
3
33
9
∫
5
3 dt = 1
t
3
t−
dx / 2
cos 2 x / 2
(
)
∫
=
t=
x
2
∫
)
3
5
1 #2 2 5 &
%% t −
t − t −1/2 =
t(
3
6 ('
3$3
"x%
dt
=
tan
t
+
C
=
tan
$ '+C
cos 2 t
#2&
dt=dx/2
u n+1
Ex 5.
∫ u ( x ) u!( x ) dx = n +1 + C
Ex 6.
∫ cos
Ex 7.
∫
Ex 8.
Show that if
n
(
n ≠ −1
tan 2 x
1
1
dx =
u 2 du = u 3 = tan 3 x
∫
2
u=tan
x
3
3
x du=dx/cos2 x
dx
x
( ) dx = ln u + C
( x)
u! x
= ln x + C
∫u
f '( x)
∫ f ( x ) + 1 dx = − x
3
then f ( x ) = 0, ∀x ≥ 0
Rewrite it as ln ( f ( x ) + 1) = − x3 and learn that f(x)+1>0. Next, implicitly differentiate
∫
()
()
f' x
f '( x)
= −2x 2 which gives f ' = −2x 2 f x +1 < 0 , i.e. f(x) is
dx = − x 3 to get
f ( x) +1
f x +1
( () )
decreasing function. However x = 0 ⇒ ln ( f (0) +1) = −03 = 0 , thus f (0) +1 = 1 and we get
f 0 = 0 . Since f decreasing starting from 0 we get f ( x ) = 0, ∀x ≥ 0 .
()
Ex 9.
Ex 10.
x4
1 5x 4
1
∫ x5 − 9 dx = 5 ∫ x5 − 9 dx = 5 ∫
sin x
∫ tan x dx = ∫ cos x dx u=cos=x ∫
du=−sin xdx
Inverse Substitution
dx 5
1
= ln x 5 − 9 + C
5
x −9 5
−du
= −ln u + C = −ln cos x + C
u
Course: Accelerated Engineering Calculus I
x>0
Ex 11.
dx
∫
(
x 1+ 3 x
(
)
=6
x=t
dx=6t 5dt
)
= 6 t − arctan t + C = 6
1− x 2 dx
∫
=
∫
=
x=sin t
dx=costdt
#
t 2 dx
1 &
=
6
1−
dx =
%
∫
2(
1+ t 2
$ 1+ t '
3
(
x − arctan 6 x + C
(1+ t )
2
)
6
cos 2 t costdt =
∫
( ) dt = t + sin (2t ) + C = 1
1+ cos 2t
2
=6∫
∫t
1− sin 2 t cost dt =
∫
6t 5dx
Instructor: Michael Medvinsky
2
∫ cos
2
t dt =
arcsin x + x 1− x ) + C
2(
4
2
sin 2t = 2sin t cost = 2sin t 1− sin 2 t
Definite Integral Substitution:
b
g ( b)
a
t=g ( x )
g a
dt=g '( x )dx ( )
∫ f ( g ( x )) g ' ( x ) dx
1
π
t2
∫ cos x sin x dx t=cos=x − ∫ t dt = − 2
0
dt=−sin x dx −1
x=0→t=1
x=π →t=−1
Ex 12.
∫ cos x sin x dx = ∫ cos x sin x dx
0
Ex 13.
−1
∫ f (t )dt
1 2
2
1 − (−1) = 0
2
(
)
x=π
=
t=cos x
x=0 dt=−sin x dx
− ∫ t dt
x=0
t2
=−
2
1x=π
−1
x=0
(
1
6 − 5x =
∫
t=6−5x
dt=−5dx 6
0
)
6
dt 1 6
1 2 3/2
2 3/2 3/2
t
= ∫ t dt =
t
=
6 −1
−5 5 1
5 3 1 15
(
)
Integral of Symmetric Function:
If f is even
0
∫
0
f ( x ) dx
−a
a
⇒
t=− x
dt=−dx
(−a,0)→(a,0)
0
−
∫
a
f (−x ) dx =
a
∫
a
f (−x ) dx
0
a
=
f ( x )= f (− x )
a
∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx = 2 ∫ f ( x ) dx
−a
If f is odd then
=
−a
0
0
∫ f ( x ) dx
0
1π
cos2 x
=−
=
2 0x=0
1
1
2
cos2 π − cos2 0 ) = − 12 − (−1) = 0
(
2
2
1
∫
=−
x=π
π
=−
1
=
Course: Accelerated Engineering Calculus I
0
0
∫ f ( x ) dx
−a
=
−
t=− x
dt=−dx
Instructor: Michael Medvinsky
a
∫ f (−x ) dx = ∫ f (−x ) dx
a
0
a
=
− f ( x )= f (− x )
−
∫ f ( x ) dx
0
(−a,0)→(a,0)
a
⇒
0
a
∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx = 0
−a
0
−a
12.6 Integration be parts (5.6)
Recall the Chain Rule
d
df
dg
( fg) = g + f , rewrite in a different way d ( fg) = gdf + fdg
dx
dx
dx
and then integrate (both sides)
Ex 1.
Ex 2.
∫ d ( fg ) = fg = ∫ g df + ∫
f dg . Thus
∫
f dg = fg −
∫ x sin xdx = − ∫ x cos' x dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C
8 ∫ x sin x cos x = 4 ∫ x sin 2xdx = ∫ t sin tdt = −t cost + ∫ cost dt
t=2 x
dt=2dx
= −t cost + sin t + C = −2x cos 2x + sin 2x + C
Ex 3.
Ex 4.
∫ 1⋅ ln x dx = x ⋅ ln x − ∫ dx = x ⋅ ln x − x + C
∫ 1⋅ arctan x dx = x ⋅ arctan x − ∫ dx = x ⋅ arctan x − ∫
x
x 2 +1
= x ⋅ arctan x − 12
Definite Integral:
b
∫
Ex 5.
π
2x
x 2 +1
dx =
( x +1)$ dx = x ⋅ arctan x − 1 ln 1+ x 2 + C
∫ x +1
( )
2
2
2
b
f dg = fg a −
a
1
2
b
∫ g df
a
π
π
π
π
xsin x 0 − ∫ sin x dx = cos x 0 = cos π − cos0 = −1−1 = −2
∫ x cos x dx = ∫ xsin' x dx = 
 
 0
0
0
=0
∫ g df