More Related Content Similar to Integral Calculus Similar to Integral Calculus (20) Integral Calculus2. Integral calculus
Integration is the inverse process of differentiation. Instead
of differentiating a function, we are given the derivative of a
function and asked to find its primitive, i.e., the original
function. Such a process is called integration or anti
differentiation.
Let us consider the following examples:
We know that,
2
3
3
cos
sin
x
x
dx
d
e
dx
de
x
dx
xd
x
x
=
=
=
We observe that, the function cos x is the derived function of sinx.
We say that sinx is an anti derivative (or an integral) of cos x.
Similarly , in (2) and (3), and ex
are the anti derivatives of x2
and ex
--------------- (1)
--------------- (2)
--------------- (3)
3
3
x
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3. we note that for any real number C, treated as constant function,
its derivative is zero and hence, we can write (1), (2) and (3) as
follows :
( ) 2
3
3
,cos)(sin xC
x
dx
d
eCe
dx
d
xCx
dx
d xx
=
+=+=+
Thus, anti derivatives (or integrals) of the above cited
functions are not unique. Actually , there exist infinitely many
anti derivatives of each of these functions which can be obtained
by choosing C arbitrarily from the set of real numbers. For this
reason C is customarily referred to as arbitrary constant.
We introduce a new symbol, namely , which will
represent the entire class of anti derivatives read as the
integral of f with respect to x
∫ dxxf )(
An antiderivative of f `(x) = f(x)
The indefinite integral: `( ) ( )f x dx f x c= +∫
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5. You need to remember all the integral identities from higher.
1
1
1
cos( ) sin( )
1
sin( ) cos( )
n na
ax dx x c
n
ax b dx ax b c
a
ax b dx ax b c
a
+
= +
+
+ = + +
+ = − + +
∫
∫
∫
A definite integral is where limits are given.
This gives the area under the curve of f `(x) between these
limits.
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6. Standard forms
From the differentiation exercise we know:
2
( )
1
(ln )
(tan ) sec
x xd
e e
dx
d
x
dx x
d
x x
dx
=
=
=
This gives us three new anti derivatives.
2
1
ln
sec tan
x x
e dx e c
dx x c
x
x dx x c
= +
= +
= +
∫
∫
∫
Note: when 0 but when 0x x x x x x= ≥ = − ≤
Example
1 2
3 3
0 1
1. Find x x
e dx e dx+∫ ∫
1 2 2
3 3 3
0 1 0
x x x
e dx e dx e dx+ =∫ ∫ ∫
2
3
0
1
3
x
e
=
61 1
3 3
e= −
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7. 5
2. Find dx
x∫
5 1
5dx dx
x x
=∫ ∫ 5lnx c= + 5
lnx c= +
2
3. Find tan x dx∫
(We need to use a little trig here and our knowledge of integrals.)
From a few pages ago we know 2
sec tanx dx x c= +∫
2 2
sin cos 1x x+ =
2 2
2 2 2
sin cos 1
cos cos cos
x x
x x x
⇔ + = 2 2
tan 1 secx x⇔ + =
2 2
tan sec 1x dx x dx= −∫ ∫
2 2
tan sec 1 tanx dx x dx x x c= − = − +∫ ∫
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8. Integration by Substitution
When differentiating a composite function we made use of the
chain rule.
3
(2 3)y x= + Let 2 3u x= + 3
y u⇒ =
2
3
dy
u
du
= and 2
du
dx
=
dy dy du
dx du dx
= ×
2
3 2u= × 2
6u= 2
6(2 3)x= +
When integrating, we must reduce the function to a standard
form – one for which we know the antiderivative.
This can be awkward, but under certain conditions, we can use
the chain rule in reverse.
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9. If we wish to perform we can proceed as follows( ( )). `( )g f x f x dx∫
Let ( )u f x= then `( )du f x dx=
The integral then becomes which it is hoped will be a
standard form.
( )g u du∫
2 3
1. Find ( 3)x x dx+∫
2
Let 3u x= + 2du x dx=
2 3 31
( 3)
2
x x dx u du+ =∫ ∫
41
8
u c= + Substituting back gives,
2 41
( 3)
8
x c= + +
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10. 2 3 5
2. Find 3x ( 4)x dx−∫
3
Let 4u x= − 2
then 3du x dx=
2 3 5 5
3x ( 4)x dx u du− =∫ ∫
61
6
u c= +
3 6
( 4)
6
x
c
−
= +
Putting the value of u
3
3. Find 8cosxsin x dx∫
Let sinu x= then cosdu x dx=
3 3
8cosxsin 8x dx u du=∫ ∫
4
2u c= +
4
2sin x c= +
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11. For many questions the choice of substitution will not
always be obvious.
You may even be given the substitution and in that
case you must use it.
4ln
1. Find . Let ln
x
dx u x
x
=∫
1
du dx
x
=
4ln
4
x
dx udu
x
=∫ ∫ 2
2u c= + 2
2(ln )x c= +
2 1
2. Find (1 ) . Let sinx dx u x−
− =∫
cosdx udu=
2 2
(1 ) 1 sin cosx dx u udu− = −∫ ∫
2
cos u du= ∫
sinx u= Substituting gives,
We can not integrate this yet. Let us use trig.
2 1
cos (1 cos2 )
2
u u= +
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12. 1 1
cos2
2 2
u du= +∫
1
sin2
2 4
u
u c= + +
1
.2sin cos
2 4
u
u u c= + + ( )
1
sin cos
2
u u u c= + +
( )1 21
sin 1
2
x x x c−
= + − +
Now for some trig play…..
sin , but what does cos equal?x u u=
( )
1
sin cos
2
u u u c= + +
2 2
sin cos 1u u+ =
2 2
cos 1 sinu = −
2
cos 1 sinu u= −
( )2 1
(1 ) sin cos
2
x dx u u u c− = + +∫
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13. 2
2
3. Find . Let x 2sin
4
x
dx
x
θ=
−
∫
2cosdx dθ θ=
2 2
2 2
4sin
.2cos
4 4 4sin
x
dx d
x
θ
θ θ
θ
=
− −
∫ ∫
2
2
4sin
.2cos
2 1 sin
d
θ
θ θ
θ
=
−
∫
2
2
2sin
.2cos
cos
d
θ
θ θ
θ
= ∫
2
4sin dθ θ= ∫
2 1 1
sin cos2
2 2
θ θ
= − ÷
2 2cos2 dθ θ= −∫
2 sin2 cθ θ= − +
We now need to substitute theta
in terms of x.
2 2sin cos cθ θ θ= − +
14. 2sinx θ=
sin
2
x
θ
= ÷
1
sin
2
x
θ −
= ÷
2
2
2 2sin cos
4
x
dx c
x
θ θ θ= − +
−
∫
2 2
4sinx θ=
2
4 4cos θ= −
2 2
4cos 4 xθ = −
2
2cos 4 xθ = −
21
cos 4
2
xθ = −
1 21
2sin 2. . 4
2 2 2
x x
x c−
= − − + ÷
1 2
2sin 4
2 2
x x
x c−
= − − + ÷
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15. Now for some not very obvious substitutions……………….
5
1. Find sin x dx∫
5 4
sin sin sinx x x= 2 2
sin (sin )x x= 2 2
sin (1 cos )x x= −
5 2 2
sin sin (1 cos )x dx x x dx= −∫ ∫ Let cosu x= sindu x dx= −
5 2 2
sin (1 )x dx u du= − −∫ ∫
2 4
(1 2 )u u du= − − +∫
2 4
1 2u u du= − + −∫
3 52 1
3 5
u u u c= − − +
3 52 1
cos cos cos
3 5
x x x c= − − +
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16. 1
2. Find
1
dx
x−
∫ Let 1u x= −
1
2
1
2
du x dx
−
= −
1
2
2dx x du⇒ = − ( )2 1dx u du⇒ = − −
1 2( 1)
1
u
dx du
ux
−
=
−
∫ ∫
2
2 du
u
= −∫
2 2lnu u c= − +
2 2 2ln 1x x c= − − − +
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17. Substitution and definite integrals
Assuming the function is continuous over the interval, then
exchanging the limits for x by the corresponding limits for u
will save you having to substitute back after the integration
process.
2
2 3
1
1. Evaluate (2x+4)(x 4 )x dx+∫
2
Let 4u x x= + 2 4du x dx= + When 2, 12; 1, 5x u x u= = = =
2 12
2 3 3
1 5
(2x+4)(x 4 )x dx u du+ =∫ ∫
12
4
5
1
4
u
=
5027.75=
18. Special (common) forms
Some substitutions are so common that they can be treated
as standards and, when their form is established, their
integrals can be written down without further ado.
1
( ) ( )f ax b dx F ax b c
a
+ = + +∫
`( )
ln ( )
( )
f x
dx f x c
f x
= +∫
21
`( ) ( ) ( ( ))
2
f x f x dx f x c= +∫
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19. Area under a curve
a b
y = f(x)
( )
b
a
A f x dx= ∫ a b
y = f(x)
( )
b
a
A f x dx= −∫
Area between the curve and y - axis
b
a
y = f(x)
( )
b
a
A f y dy= ∫
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20. 1. Calculate the area shown in the diagram below.
5
2
y = x2
+ 1
2
1y x= +
2
1x y= −
1x y= −
1
5 2
2
( 1)A y dy= −∫
53
2
2
2
( 1)
3
y
= −
14
units squared.
3
=
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21. Volumes of revolution
Volumes of revolution are formed when a curve is rotated
about the x or y axis.
2
b
a
V y dxπ= ∫ π= ∫
2
d
c
V x dy
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22. 1. Find the volume of revolution obtained between x = 1 and
x = 2 when the curve y = x2
+ 2 is rotated about
(i) the x – axis (ii) the y – axis.
2
2
1
( )i V y dxπ= ∫
2
4 2
1
( 4 1)x x dxπ= + +∫
25 3
1
4
5 3
x x
xπ
= + +
32 32 1 1
2 1
5 3 5 3
π π
= + + − + + ÷ ÷
3263
units
15
π=
( ) when 1, 3 and when 2, 6ii x y x y= = = =
6
3
( 2)y dyπ= −∫
62
3
2
2
y
yπ
= −
( )
9
18 12 6
2
π π
= − − − ÷
315
units
2
π=
6
2
3
V x dyπ= ∫
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23. The End
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