Presentation about integration - mathematics - University of Zakho ... College of Engineering - Petroleum Department

1. Integration indefinite
Prepared by:
Safeen Dlshad
Bafreen Amjad
Darya Mahdi
Harikar Mikail
Supervisor:
Dawoud Sadullah

2. 2
1.Integration ……………………………………………………… 3
2.Integration Rules
Examples ……………………………………………………… 3-9
3.Integration by Parts
Examples ……………………………………………………… 10-14
4.Integration by Substitution
Examples ……………………………………………………… 15-18
5.Reference ……………………………………………………… 19
Subject Page

3. 3
Integration
Integration can be used to find areas, volumes, central
points and many useful things. But it is often used to find
the area underneath the graph of a function like this:
Integration Rules
Rules Function Integral
Constant ∫a dx ax + C
Power Rule (n≠-1) ∫xn
dx Xn+1
/n+1 + C
Reciprocal ∫(1/x) dx ln|x| + C
Exponential ∫ex
dx ex
+ C
∫eg(x)
dx
1
𝑔′(𝑥)
eg(x)
+C
∫ax
dx ax
/ln(a) + C
∫ 𝑓′(𝑥)
𝑓(𝑥)
dx ln (f(x)) + C
Multiplication by constant ∫cf(x) dx c∫f(x) dx
Sum Rule ∫(f + g) dx ∫f dx + ∫g dx
Difference Rule ∫(f - g) dx ∫f dx - ∫g dx

4. 4
Examples:
Example 1: What is ∫x3
dx ?
The question is asking "what is the integral of x3
?"
We can use the Power Rule, where n=3:
∫xn
dx = xn+1
/n+1 + C
∫x3
dx = x4
/4 + C
Example 2: What is ∫√x dx ?
√x is also x0.5
We can use the Power Rule, where n=½:
∫xn
dx = xn+1
/n+1 + C
∫x0.5
dx = x1.5
/1.5 + C
Example 3: What is ∫e3x
dx ?
=
1
3
e3x
+ C

5. 5
Example 4: What is ∫6x2
dx ?
We can move the 6 outside the integral:
∫6x2
dx = 6∫x2
dx
And now use the Power Rule on x2
:
= 6 x3
/3 + C
Simplify:
= 2x3
+ C
Example 5: What is ∫2x4
3x5
 dx ?
=∫2x4
dx +∫3x5
dx
= 2∫ x4
dx + 3∫ x5
dx
=
2x5
5
+
𝑥6
2
+ 𝑐
Example 6: What is ∫8z + 4z3
− 6z2
dz ?
Use the Sum and Difference Rule:
∫8z + 4z3
− 6z2
dz =∫8z dz + ∫4z3
dz − ∫6z2
dz
Constant Multiplication:
= 8∫z dz + 4∫z3
dz − 6∫z2
dz
Power Rule:
= 8z2
/2 + 4z4
/4 − 6z3
/3 + C
Simplify:
= 4z2
+ z4
− 2z3
+ C

6. 6
Example 7: What is ∫
5
3+5𝑥
 dx ?
The derivative of the denominator is 5 which is the same as the numerator, hence
= ln (3 + 5x) + C
Example 8: What is ∫
𝑥
1+𝑥2
 dx ?
The derivative of the denominator is 2x. This is not the same as the numerator but we can make
it the same by re-writing the function
𝑥
1+𝑥2
as
1
2
∗
2𝑥
1+𝑥2
, therefore
∫ (
𝑥
1+𝑥2
) dx =
1
2
∫
2𝑥
1+𝑥2
dx
=
1
2
ln (1 + x2) + C

7. 7
Function Integral
∫ln(x) dx x ln(x) − x + C
∫cos(x) dx sin(x) + C
∫cos(gx) dx
1
𝑔′(𝑥)
Sin( 𝑔𝑥)+ C
∫sin(x) dx -cos(x) + C
∫sin(gx) dx
−1
𝑔′(𝑥)
Cos( 𝑔𝑥) + C
∫1/cos2
(x) dx
∫1+tan2
(x) dx
Tan(x) + C
∫sec2
(x) dx tan(x) + C
∫tan(x) dx -ln|cos(x)| + C
∫csc2
(x) dx -cot(x) + C
∫cot(x) dx ln|sin(x)| + C
∫sec(x) dx ln|sec(x)+tan(x)| + C
∫csc(x) dx -ln|csc(x)+cot(x)| + C
∫sec(x)tan(x) dx Sec(x) + C

8. 8
Examples
Example 1: What is ∫cos (3x2
) dx ?
=
1
6
Sin(3x2) + C
Example 2: What is ∫ sin(- 5x) dx?
=
1
5
Cos(−5x) + C
Example 3: What is ∫cos x + x dx ?
Use the Sum Rule:
∫cos x + x dx = ∫cos x dx + ∫x dx
Work out the integral of each (using table above):
= sin x + x2
/2 + C
Example 4: What is ∫5cos3x3e7x
 dx ?
∫5cos3xdx - ∫3e7x
dx
5∫cos3xdx - 3∫ e7x
dx
=
5
3
sin(3𝑥) −
3
7
e7x

9. 9
Example 5: What is ∫
cos 2𝑥
sin2 𝑥 cos2 𝑥
 dx ?
=∫cos2 𝑥−sin2 𝑥
sin2 𝑥 cos2 𝑥
𝑑𝑥
=∫
cos2 𝑥
sin2 𝑥 cos2 𝑥
−
sin2 𝑥
sin2 𝑥 cos2 𝑥
)𝑑𝑥
=∫(
1
sin2 𝑥
−
1
cos2 𝑥
) 𝑑𝑥
=∫
dx
sin2 𝑥
− ∫
dx
cos2 𝑥
= - cot(x) – tan(x) + C
Example 6: What is ∫sec x(cos x + tan x dx ?
=∫(sec x cos x + sec x tan x) dx
=∫(1 + sec x tan x) dx
= x + sec x + C

10. 10
Integration by Parts
Integration by Parts is a special method of integration that is often useful when
two functions are multiplied together, but is also helpful in other ways.
∫u v dx = u∫v dx −∫u' (∫v dx) dx
 u is the function u(x)
 v is the function v(x)
As a diagram:

11. 11
Examples
Example 1: What is ∫x cos(x) dx ?
OK, we have x multiplied by cos(x), so integration by parts is a good choice.
First choose which functions for u and v:
 u = x
 v = cos(x)
So now it is in the format ∫u v
dx we can proceed:
Differentiate u: u' = x' = 1
Integrate v: ∫v dx = ∫cos(x) dx = sin(x)
Now we can put it together:
Simplify and solve: x sin(x) − ∫sin(x) dx
x sin(x) + cos(x) + C
Example 2: What is ∫ln(x)/x2
dx ?
First choose u and v:
 u = ln(x)
 v = 1/x2
Differentiate u: ln(x)' = 1/x
Integrate v: ∫1/x2
dx = ∫x-2
dx = −x-1
= -1/x
Now put it together:
Simplify:
−ln(x)/x − ∫−1/x2
dx
= −ln(x)/x − 1/x + C
=− (ln(x) + 1)/x + C

12. 12
Example 3: What is ∫ln(x) dx ?
But there is only one function! How do we choose u and v ?
Hey! We can just choose v as being "1":
 u = ln(x)
 v = 1
Differentiate u: ln(x)' = 1/x
Integrate v: ∫1 dx = x
Now put it together:
Simplify:
x ln(x) − ∫1 dx = x ln(x) − x + C

13. 13
Example 4: What is ∫ex
x dx ?
Choose u and v:
 u = ex
 v = x
Differentiate u: (ex
)' = ex
Integrate v: ∫x dx = x2
/2
Now put it together:
Well, that was a spectacular disaster! It just got more complicated.
Maybe we could choose a different u and v?
Example 5: ∫ex
x dx (continued)
Choose u and v differently:
 u = x
 v = ex
Differentiate u: (x)' = 1
Integrate v: ∫ex
dx = ex
Now put it together:
Simplify:
x ex
− ex
+ C
ex
(x−1) + C

14. 14
Example 6: ∫ex
sin(x) dx
Choose u and v:
 u = sin(x)
 v = ex
Differentiate u: sin(x)' = cos(x)
Integrate v: ∫ex
dx = ex
Now put it together:
∫ex
sin(x) dx = sin(x) ex
-∫cos(x) ex
dx
Looks worse, but let us persist! We can use integration by parts again:
Choose u and v:
 u = cos(x)
 v = ex
Differentiate u: cos(x)' = -sin(x)
Integrate v: ∫ex
dx = ex
Now put it together:
∫ex
sin(x) dx = sin(x) ex
- (cos(x) ex
−∫−sin(x) ex
dx)
Simplify:
∫ex
sin(x) dx = ex
sin(x) - ex
cos(x) −∫ ex
sin(x)dx
Now we have the same integral on both sides (except one is subtracted) ...
... so bring the right hand one over to the left and we get:
2∫ex
sin(x) dx = ex
sin(x) − ex
cos(x)
Simplify:
∫ex
sin(x) dx = ex
(sin(x) - cos(x)) / 2 + C

15. 15
Integration by Substitution
Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain
Rule") is a method to find an integral, but only when it can be set up in a special
way.
Note that we have g(x) and its derivative g'(x)
Like in this example:
Here f=cos, and we have g=x2
and its derivative 2x
This integral is good to go!
When our integral is set up like that, we can do this substitution:
Then we can integrate f(u), and finish by putting g(x) back as u.
Like this:

16. 16
Examples
Example 1: ∫cos(x2
) 2x dx
We know (from above) that it is in the right form to do the substitution:
Now integrate:
∫cos(u) du = sin(u) + C
And finally put u=x2 back again:
sin(x2
) + C
So ∫cos(x2) 2x dx = sin(x2) + C
That worked out really nicely! (Well, I knew it would.)
But this method only works on some integrals of course, and it may need rearranging:
Example 2: ∫cos(x2
) 6x dx
Oh no! It is 6x, not 2x like before. Our perfect setup is gone.
Never fear! Just rearrange the integral like this:
∫cos(x2
) 6x dx = 3∫cos(x2
) 2x dx
(We can pull constant multipliers outside the integration, see Rules of Integration .)
Then go ahead as before:
3∫cos(u) du = 3 sin(u) + C
Now put u=x2
back again:
3 sin(x2
) + C
Done!

17. 17
Example 3: ∫x/(x2
+1) dx
Let me see ... the derivative of x2
+1 is 2x ... so how about we rearrange it like this:
∫x/(x2
+1) dx = ½∫2x/(x2
+1) dx
Then we have:
Then integrate:
½∫1/u du = ½ ln(u) + C
Now put u=x2
+1 back again:
½ ln(x2
+1) + C
Example 4: ∫(x+1)3
dx
Let me see ... the derivative of x+1 is ... well it is simply 1.
So we can have this:
∫(x+1)3
dx = ∫(x+1)3
· 1 dx
Then we have:
Then integrate:
∫u3
du = (u4
)/4 + C
Now put u=x+1 back again:
(x+1)4
/4 + C

18. 18
Example 5: ∫(5x+2)7
dx
If it was in THIS form we could do it:
∫(5x+2)7
/5 dx
So let's make it so by doing this:
1/5 ∫(5x+2)7
/5 dx
The 1/5 and 5 cancel out so all is fine.
And now we can have u=5x+2
And then integrate:
1/5 ∫u7
du = 1/5 u8
/8 + C
Now put u=5x+2 back again, and simplify:
(5x+2)8
/ 40 + C
Example 6: ∫−𝟐𝒙√ 𝟖 − 𝒙2 dx
-2x = -2x
∫−2𝑥√8 − 𝑥2 dx =
(8−𝑥2)3/2
3
2
=
2
3
(8 − 𝑥2)√8 − 𝑥2 + C

19. 19
Reference:
 https://www.mathsisfun.com/calculus/integration-rules.html
 DIFFERENTIAL AND INTEGRAL CALCULUS (WITH EXAMPLES AND
APPLICATIONS) BY GEORGE A. OSBORNE, S.B.
 http://ncert.nic.in/ncerts/l/lemh201.pdf
 Integral Calculus – Exercises
http://www.buders.com/UNIVERSITE/Universite_Dersleri/Math101/Arsi
v/integral_sorulari_ve_cozumleri.pdf
 Calculas - by Gilbert Strang
 Guide to Integration Mathematics 101 - Mark MacLean and Andrew
Rechnitzer