Definite Integrals
 Link between sigma notation and integration
n
b
i 1
a
lim  f (c)(x)   f ( x)dx
x 

Fundamental Theorem of Calculus
b
 f ( x)dx  F (a)  F (b)
a

Properties of Definite Integrals
b
1.
 f x dx  0
a
b
2.
a
 f xdx   f xdx
a
b
3.

a
4.
5.
b
c
b
f x dx   f x dx   f x dx (where c is any constant on the interval [a,b])
a
c
b
b
a
a
 kf x dx  k  f x dx
b
b
a
a
  f x  g xdx   f xdx  g xdx
Example: f x   x 2
Evaluating Definite Integrals
 Evaluation Theorem
o If f is continuous on the interval [a,b], then
b
so F  x  
1 3
x
3
The evaluation theorem tell us that
1
 x dx  F 1  F 0  3 1
1
2
3
0
1
1
  03 
3
3
If vt  is the velocity of an object and s t 
 f x dx  F b  F a 
a
Where F is any antiderivative of f , that is F ’= f
is its position at time t , then vt  = s' t  ,
so s is an antiderivative of v .
b
 vt dt  sb  sa 
a

Indefinite integrals
 f x dx  F x 
Means
F ' x   f x 
You should distinguish carefully between definite and indefinite integrals. A definite integral
b
 f x dx is a number, whereas an indefinite integral is  f x dx is a function
a

Integral rules:
Power Rule:
Constant Multiple Rule:
Sum and Difference Rule:
n 1
 cf x dx  c  f x dx
 x dx  n  1  C n  1
 f x dx  g x dx   f x dx   g x dx
x
n
e u Rule:
 e dx  e
x
a u Rule:
x
C
Antiderivatives and the Natural Log
a
x
 x dx  ln x   C
1
 a dx  ln a  C
x
Trig Integrals:
 sin x dx   cosx   C
 cosx dx  sin x   C
 cscx dx   ln cscx   cot x   C  secx dx  ln secx   tan x   C
Inverse Trig Antiderivatives:
a
2
1
1
u
dx  arctan  C
a
a
u

1
a u
2
2
dx  arcsin
u
C
a
 tan x dx   ln coxx   C
 cot x dx  ln sin x   C
U Substitution:

Group Definition: when you cannot force the derivative of the inner by any other method you
can utilize the u substitution method by changing the variable x to the variable u, allowing for
integration/evaluation in places that might otherwise have been impossible to find using antidifferentiation formulas or the reverse chain rule.

Book Definition: If ∫ 𝑓 (𝑔(𝑥))𝑔′ (𝑥) 𝑑𝑥 is present and u = g(x) is a differentiable function whose
range is an interval I and f is continuous on I then…
∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥) 𝑑𝑥 = ∫ 𝑓(𝑢) 𝑑𝑢
Example) ∫ 𝑥 3 cos(𝑥 4 + 2) 𝑑𝑥
*Notice how you cannot integrate this expression through reverse chain or anti-derivative formulas
*Make the cos the u because it is differentiable; differentiate then isolate the dx so you can substitute
u = x4+2
du = 4x3 dx
dx =
𝑑𝑢
4𝑥 3
*Now that you have dx isolated, plug it back into the original equation so you can integrate in terms of
u; when you have this completed, cancel out like terms
𝑑𝑢
∫(𝑥 3 cos(𝑢))(4𝑥 3 )
∫
∫
𝑥 3 cos(𝑢)𝑑𝑢
4𝑥 3
𝑥 3 cos(𝑢)𝑑𝑢
4𝑥 3
1
=> ∫ cos(𝑢) 𝑑𝑢
4
*At this point you may choose to move the constant outside of the integrand; either way, you are now
able to fully integrate in terms of u
1
∫ 4 cos(𝑢) 𝑑𝑢 =>
1
4
sin(𝑢) + 𝐶
1
4
∫ cos(𝑢) 𝑑𝑢
*At this point you will substitute your original value that you set equal to u back into your integrated
expression
u = x4+2
1
4

sin(𝑢) + 𝐶
1
4
sin(𝑥4 + 2) + 𝐶
Some other helpful tips from the textbook:
o Make u some function in the integrand whose differential can occur except that of a constant
 This means you want to choose a u who can be differentiated
o Choose the most complicated part of the integrand to be the u
o If your first u value still doesn’t allow you to integrate, choose another u value
Example)
4𝑥+5
∫ (4𝑥 2 +10𝑥)3 dx
*This is an example where you would want to set u as the most complicated inner function
u = (4x2 +10x)
du = 8x+10 dx
dx =
dx =
∫
∫
𝑑𝑢
1
8𝑥+10
2
𝑑𝑢
2(4𝑥+5)
(4𝑥+5)𝑢−3 𝑑𝑢
2(4𝑥+5)
(4𝑥+5)𝑢−3 𝑑𝑢
2(4𝑥+5)
∫ 𝑢−3 𝑑𝑢
1
1
4
4
− 𝑢−2 + 𝑐  − (4𝑥 2 + 10𝑥)−2 + 𝑐
Example) ∫ cos 2 (4𝑥) sin(4𝑥) 𝑑𝑥
*This example demonstrates how important it is to choose a viable u value. Because you are working with two
different trig expressions that are multiplied together, reverse chain is almost impossible in this case. In this case,
you want to choose a u value that will allow you to cancel out one of the trig expressions.
𝑐𝑜𝑠 2 (𝑢) sin(4𝑥)𝑑𝑢
u = cos (4x)
∫
du = -4sin(4x) dx
− ∫ 𝑢2 𝑑𝑢
4
dx =
𝑑𝑢
−4sin(4𝑥)
−4sin(4𝑥)
 ∫
𝑐𝑜𝑠 2 (𝑢) sin(4𝑥)𝑑𝑢
−4sin(4𝑥)
1
1 1
1
4 3
12
− ( ) 𝑢3 + 𝑐  −
𝑐𝑜𝑠 3 (4𝑥) + 𝑐
Example) ∫(3𝑒
2𝑥−5
+ 7) 𝑑𝑥
*This is an example that consists of a u substitution and ea, a being some value or expression. Usually
in these cases, you will make your u value whatever value/expression a exists as
u = 2x-5
du = 2 dx
dx =
∫
𝑑𝑢
2
(3𝑒 𝑢 +7)𝑑𝑢
2
*This is also an example of an integration problem where you need to split the integral into two
separate integral fractions order to take its derivative
∫
3
2
(3𝑒 𝑢 )
2
7
∫ 𝑒 𝑢 𝑑𝑢 + ∫ 2 𝑑𝑢
3 𝑢
𝑒
2

7
𝑑𝑢 + ∫ 𝑑𝑢
2
7
2
+ 𝑥+𝑐 
3 (2𝑥−5)
𝑒
2
7
2
+ 𝑥+𝑐
Sometimes u can be more than one value; in these cases you are lucky and do not need to retest separate u
values
Example) ∫ √2𝑥 + 1 𝑑𝑥
*You can make u = 2x+1
𝑑𝑢 = 2 𝑑𝑥
dx =
∫ √𝑢
1
𝑑𝑢
𝑑𝑢
2
3
du =
𝑑𝑥
√2𝑥+1
∫ √2𝑥 + 1 𝑑𝑥
1
2
*You can make u=√2𝑥 + 1
du = √2𝑥 + 1 𝑑𝑢 = 𝑢 𝑑𝑢
2
∫ 𝑢2 𝑑𝑢
2
0.5 *
OR
3
𝑢2 + c
∫ 𝑢 ∗ 𝑢 𝑑𝑢 𝑂𝑅 ∫ 𝑢2
𝑢3
3
+ 𝑐
1
3
3
1
(2𝑥
3
(2𝑥 + 1)2 + 𝐶
3
+ 1)2 + 𝐶
U substitution and Definite Integrals

There are two possible methods when evaluating definite integrals and utilizing u substitution
o Evaluate the integral, then theorem
o Change the limits of integration when the variable is changed, evaluate, then re-substitute

Book definition: if g’ is continuous on [a,b] and f is continuous on the range of u = g(x) then
𝑏
𝑔(𝑏)
∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥) 𝑑𝑥 = ∫
𝑎
𝑓(𝑢) 𝑑𝑢
𝑔(𝑎)
2
Example) ∫0 𝑥√3𝑥 + 2 𝑑𝑥
*Start by solving like an indefinite integral, pick a u value and take its derivative and isolate
u = 3x + 2
du = 3 dx
dx =
𝑑𝑢
3
*Here’s where things change; because you are introducing the variable u, you must change the bounds
on the integral!
(You must change the bounds on the integral because the original bounds given define the expression
when it is in terms of x; because we are now working with a different variable, u, we must use different
bounds that will define the expression in terms of u).
*To do this, take the lower bound and plug it into your u; then take your upper bound and plug it into
your original u
Lower Bound: u = 3x + 2  u = 3(0) + 2 = 2
Upper Bound: u = 3x + 2  u = 3(2) + 2 = 8
*Substitute these new bounds into the integral; substitute your dx value into the integral as well
1
8
𝑑𝑢
∫2 𝑥𝑢2 3
1
OR
8 𝑥𝑢2 𝑑𝑢
∫2
3
*Because the variable x still exists in our integral, we can rewrite our original u in terms of x and
substitute that value into our integral in order to continue integrating
1
8 𝑢−2 𝑢2 𝑑𝑢
∫2 ( 3 ) 3
u = 3x + 2
u – 2 = 3x
𝑢−2
3
=𝑥
*From here we simplify and pull out any constants
1
81
∫2 9 ( 𝑢 − 2) (𝑢2 ) 𝑑𝑢
8 3
∫ 𝑢2
9 2
1
1
2
− 2𝑢 𝑑𝑢
*Integrate in terms of u
1
9
[
2
5
5
𝑢2 −
4
3
3
8
𝑢2 | ]
2
*From this point on there are two options for evaluation, we can leave the expression the way it is (in
terms of u) and substitute in the bounds (2,8) for u1 or we can substitute in our original value for u,
change the bounds back to the originals given (because we will be working in terms of x, again), and
evaluate.2
*Method 1:
1
9
2
5
[ ( (8)2 −
5
4
3
3
5
2
(8)2 ) − ( (2)2 −
5
4
3
3
(2)2 )] = 4.8607
*Method 2:
1
9
1
9
[(
2
5
[
2
5
5
2
(3𝑥 + 2) −
5
(3(2) + 2)2 −
4
3
4
3
3
2
2
(3𝑥 + 2) | ]
0
3
2
5
(3(2) + 2)2 ) − ( (3(0) + 2)2 −
5
4
3
3
(3(0) + 2)2 ) = 4.8607
*If you would like to see an additional example, or would like to see an online tutorial, please visit:
http://www.math.hmc.edu/calculus/tutorials/substitution
http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleDefinite.aspx
1
2
Method 1
Method 2
Integration by Parts
Group Definition: If u and v are functions for x and have continuous derivatives, then…
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
Proof Provided by Textbook:


The rule that corresponds to the Product Rule for differentiation is the called integration by parts
Start with the product rule which states that if f and g are differentiable functions, then…
𝑑
[𝑓(𝑥)𝑔(𝑥)] = 𝑓(𝑥)𝑔′ (𝑥) + 𝑔(𝑥)𝑓 ′ (𝑥)
𝑑𝑥

If you then take the indefinite integrals the equation becomes…
∫[𝑓(𝑥)𝑔′ (𝑥) + 𝑔(𝑥)𝑓 ′ (𝑥)]𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥)

Simplified this equation becomes…
∫ 𝑓(𝑥)𝑔′ (𝑥) 𝑑𝑥 + ∫ 𝑔(𝑥)𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥)

If you simplify once more you have this equation…
∫ 𝑓(𝑥)𝑔′ (𝑥) 𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − ∫ 𝑔(𝑥)𝑓 ′ (𝑥) 𝑑𝑥 (Or in simpler terms) ∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
Guidelines:

The purpose of using Integration by parts is to obtain a simpler integral than the original so
below are some helpful tips to utilize when integrating by parts
1. Try letting u be the most difficult function
o For integrals in the forms below…let u=xn and dv= eaxdx, sin(ax)dx, cos(ax)dx
 ∫ 𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥, ∫ 𝑥 𝑛 sin(𝑎𝑥) 𝑑𝑥, ∫ 𝑥 𝑛 cos(𝑎𝑥) 𝑑𝑥
o For integrals in the forms below…let u=ln(x), arcsin(ax), arctan(ax) and let dv=xndx
 ∫ 𝑥 𝑛 𝑑𝑥, ∫ 𝑥 𝑛 arcsin(𝑎𝑥) 𝑑𝑥, ∫ 𝑥 𝑛 arctan(𝑎𝑥) 𝑑𝑥
o For any integrals given in the forms below…let u=sin(bx), cos(bx) and let dv=eaxdx
 ∫ 𝑒 𝑎𝑥 sin(𝑏𝑥) 𝑑𝑥, ∫ 𝑒 𝑎𝑥 cos(𝑏𝑥) 𝑑𝑥
Example) ∫ 𝑥 cos(𝑥) 𝑑𝑥
*This example deals with integration by parts and with an integral that has a trig expression
*The first step is to choose a u value and then choose a dv value (most logically, it would make sense to
make the x your u value because you would simply get dx as your derivative, therefore allowing for
simple integration; for a vast majority of problems in a similar form, you should try making x=u)
u=x
dv=cos(x)dx
du=dx
v=sin(x)
𝑥𝑠𝑖𝑛(𝑥) − ∫ sin(𝑥) 𝑑𝑥
𝑥𝑠𝑖𝑛(𝑥) − (− cos(𝑥)) + 𝑐
𝑥𝑠𝑖𝑛(𝑥) + cos(𝑥) + 𝑐
Example) ∫ 𝑥 3 ln(𝑥) 𝑑𝑥
*This is an example of an integral expression that includes a natural log as well as an x. In these cases
you always want to make your u value the natural log, this is due to the fact that you cannot take the
derivative of ln(x). Therefore, you are only left with this option.
dv=x3dx
u=ln(x)
du=
1
1
v= 𝑥 4
𝑥
4
1
1
1
4
4
𝑥
(ln(𝑥 )) ( 𝑥 4 ) − ∫ 𝑥 4 ( ) 𝑑𝑥
1
1
1
𝑥 4 ln(𝑥 ) − ( 𝑥 4 ) + 𝑐
𝑥 4 ln(𝑥 ) − ∫ 𝑥 3 𝑑𝑥
4
4
4
1
4
1 1
4 4
𝑥 4 ln(𝑥 ) −
1
16
𝑥4 + 𝑐
Example) ∫ 𝑥𝑒 𝑥 𝑑𝑥
*This is an example of a problem where you have ex in the integral expression
*The procedure for this specific problem is similar to all the others; first, choose a value for u and dv
that will simplify your integral expression, then substitute and integrate
u=x
dv=exdx
du=dx
𝑥𝑒 𝑥 − ∫ 𝑒 𝑥 𝑑𝑥
𝑥𝑒 𝑥 − 𝑒 𝑥 + 𝑐
v=ex
Approximate Integrals
Used to approximate the area under a function with domain [a,b].
Left End Point Approximation
1. Divide the interval into n equal subintervals by the means of n + 1 points
ba
2. Each of the subintervals has a width of ∆x =
n
3. The left end point of the rectangles constructed should touch the function.
To find the left end points: a + ∆x(i – 1), where a is the first value of your domain and
i=interval you are in
4. The height of each rectangle is: f(a+ ∆x(i – 1) )
n
5. To approximate the area, do:  width  height or
i 1
n
 x  f (a  x(i  1))
i 1
For easier calculations, the ∆x can be pulled out front and multiplied by the summation of
n
the function:
x  f (a  x(i  1))
i 1
EXAMPLE:
f(x)=2x2+5 on the interval [0,2] with n=5
Work
Explanation
∆x =
20 2
5 =5
LEP=a+ ∆x(i -1)
=
2
5 (i – 1)
plug in 2 for b, 0 for a, and 5 for n
formula
2
substitute 0 in for a in 5 for ∆x
2
2
= 5i- 5
distribute
2
2
Height= f( i - )
5
5
f(a + ∆x(i -1))
2
2
=2( 5 i - 5 )2 +5
A=
2 5
2 2
(2( i  )  5)

5 i 1
5 5
A= 13.84
2
2
substitute i in for x in the original function 2x2 + 5
5
5
n
x  f (a  x(i  1))
i 1
solve in calculator
Right End Point Approximation
1. Divide the interval into n equal subintervals by the means of n + 1 points
ba
2. Each of the subintervals has a width of ∆x =
n
3. The right end point of the rectangles constructed should touch the function.
To find the right end points: a + (∆x)(i), where a=the first value in your domain and
i=interval you are in
4. The height of each rectangle is: f(a + ∆x(i) )
n
n
5. To approximate the area, do:
 width  height
i 1
or
 x  f (a  x(i))
i 1
For easier calculations, the ∆x can be pulled out front and multiplied by the summation of
n
the function:
x  f (a  x(i ))
i 1
EXAMPLE:
f(x)=
on the interval [1,5] with n=8
Work
∆x =
Explanation
5 1 4 1

8 =8 2
REP= a + ∆x(i)
1
= 1+ i
2
1
Height= f(1 + i)
2
=
A=
1
1
1 i
2
1 8
1
(
)

1
2 i 1
1 i
2
A= 1.429
plug in 5 for b, 1 for a, and 8 for n
formula
1
substitute in 1 for a and 2 for ∆x
f(a + ∆x(i ))
1
1
substitute (1 + i) in for x in the original function
2
x
n
x  f (a  x(i ))
i 1
solve in calculator
Method Number Two
A second method to solve for both right and left end point approximations are to find the actual
x-values of the subintervals (xo, x1, x2,…xn-2,xn-1,xn), and then:
A  x [ f ( xo)  f ( x1)  f ( x2)  ...  f ( xn2)  f ( xn1)  f ( xn)]
EXAMPLE:
Right end point approximation of f(x)=5x-x3 on the interval [0,1] with n =3
Work
∆x =
Explanation
0 1 1
3 =3
plug in 1 for b, 0 for a, and 3 for n
1 2
REP= 3 , 3 , 1
Use the formula ∆x(i). In interval 1, the REP is
REP is
A=
1
, in interval 2, the
3
2
, and in interval 3, the REP is 1.
3
1
1
2
[ f ( )  f ( )  f (1)]
3
3
3
formula
1
1
2
= [1.6296 + 3.037+ 4] Find the values of f ( ) , f ( ), f (1) by using calculator or plugging
3
3
3
1 2
, , 1 into f(x)
3 3
=2.8889
Add and multiply!
Midpoint Approximation
1. Divide the interval into n equal subintervals by the means of n + 1 points
ba
2. Each of the subintervals has a width of ∆x =
n
3. The midpoint of each rectangle constructed should touch the function.
To find the midpoints:
 divide ∆x in half
x
 add
to a (the first x value in the domain of the function)
2
 continue to add ∆x to this number until you have the values of all n midpoints. The
x
last number should be
below b (the last x value in the domain of the function)
2
4. The height of each rectangle is: f( midpoint)
5. The area is approximately:
∆x [f(midpoint1)+f(midpoint2)+f(midpoint3)+…+ f(midpointn)]
EXAMPLE:
f(x)= x2+3 on the interval [2,6] with 3 subintervals
Work
∆x =
Explanation
62 4
3 =3
8 16
midpoints= 3 ,4, 3
plug in 6 for b, 2 for a, and 3 for n
Divide ∆x in half 
2
3
8
Add this to a, the 1st value in the domain, which is 2, to get 3 . Add
∆x to this value until you have three midpoints.
4
8
16
[ f ( )  f (4)  f ( )]
3
3 formula
A= 3
A=
4
[10.1111+19+31.4444] Find values by using calculator or substituting into f(x)
3
A=80.7407
Add and multiply!
Trapezoidal Approximation
1. Divide the interval into n equal subintervals by the means of n + 1 points
ba
2. Each of the subintervals has a width of ∆x =
n
3. Construct trapezoids out of the subintervals. Both ends of the trapezoid should touch the
function.
4. To find the endpoints of the trapezoids:
 Start with a (the first x-value in the domain.)
 Add ∆x until you reach b (the last x-value)
5. The area is approximately:
1
x  [ f ( first endpo int)  2 f (sec ond endpo int)  2 f (third )....  f (last endpo int)]
2
EXAMPLE:
f(x)= x2+2 on the interval [0,6] with n=4.
Work
∆x =
Explanation
60 3
4 =2
3
9
endpoints= 0, 2 , 3, 2 ,6
plug in 6 for b, 0 for a, and 4 for n
Start with a (0) and add ∆x until you get to b (6)
3 1
3
9
 [ f (o)  2 f ( )  2 f (3)  2 f ( )  f (6)
2
2
A= 2 2
formula
A=
3
[2  8.5  22  44.5  38] Find values by using calculator or substituting into f(x)
4
A=86.25
Add and multiply!
Fundamental Theorem of Calculus
Part 1
If f is a continuous function on [a,b] then the function g, which is defined by
𝑔(𝑥)
𝑑
[∫
𝑓(𝑡)𝑑𝑡] = 𝑓(𝑔(𝑥 ))𝑔′ (𝑥 )
𝑑𝑥 𝑎
Part 2
If f is a continuous function on [a,b], and F is any antiderivative of f, then
b
 f ( x)dx  F (b)  F (a)
a
Using the Fundamental Theorem of Calculus
Part 1 allows you to take the derivative of an integral with variable bounds.
EXAMPLE:
x
Find the derivative of the function g(x)=  1  t 2 dt
0
Since f(t)= 1  t 2 dt is continuous, Part 1 of the Fundamental Theorem of Calculus gives that
g’(x)=f(x). To evaluate f(x), substitute x in for t in 1  t 2 . You find that
g’(x)=
1  x2
Part 2 allows you to evaluate definite integrals.
EXAMPLE:
1
Evaluate
x
2
dx
0
x3
Take the indefinite integral of
 x dx = 3
Part 2 of the Fundamental Theorem of Calculus gives that to find the definite integral from 0 to
1 we evaluate the integral at x=1 and subtract from that the integral at x=0. Or, F(1)-F(0).
x2.
This can be written as:
2
x3 1
3
0
Which can be evaluated as
13 0 3

3 3
=
1
0
3
=
1
3
Both parts together allow you to understand the inverse relationship between integrals and
derivatives.
Area of a Curve
1.) Find the bounds
2.) Take the integral of function
3.) Solve
Example 1: Find the area under the curve y=
x + 2 between the values x = 4 and x = 9
9
Area  


x  2 dx
4
9
3
  2 x 2  2 x 
 3
 4
4
9
 3 9
 2
 36
3
2
  3 4
 29  2
3
2

 24
3
Example 2: Find the area between y = cos x and the x-axis between the values x = 0 and x = 2π
2
Area 
 cosxdx
0
4
π
OR
1.5708
Area 
 cosxdx 
4.7124
0
1.5708
 sin x
1.5708
0
 cosxdx 
  sin x  
2
 cosxdx
4.7124
4.7124
1.5708
 sin x
2π
4.7124
4
Area Between Two Curves
y = f(x)
y = g(x)
b
Area   g ( x)  f ( x)dx
a
1.) Find the bounds
2.) Figure out which function is the upper and which
function is the lower OR which function is left and
which function is right
3.) Plug in the functions (upper – lower) OR ( right-left)
4.) Solve
Example 1: Find the area bound by y= x +1 and y=x+1
 
1
Bounds
y = x+1


x  1  x  1 dx
0
y=
1
x +1
 x 12  x dx
0 

x 1  x 1
1
xx
2 x
3
2 3 1
X2-x=0
3
2
3
2
1 x
2
2
0

3
2
2
 1 1  2 0 2  1 0
2
3
2

X = x2
2  1  0  1 6u2
3
2
X(x-1) =0
X = 0, 1
Example 2: Find the area bound by y= x3-5x2+x-1 and y= 2x-6
Bounds
X3-5x2+x-1= 2x-6
X=-1, 1, 5
 x
1
3


 5 x 2  x  1  2 x  6  dx 
1
 x
3

 5 x 2  x  1 dx
1
1
1
 2 x  6  x
5
3

5
 5 x 2  x  5 dx    x 3  5 x 2  x  5 dx
1
1 x 4  5 x 3  1 x 2  5x
4
3
2
1
-1
+  1 x 4  5 x 3  1 x 2  5x
4
3
2
5
1
4
3
2
4
3
2
( 1 1  5 1  1 1  51)  ( 1  1  5  1  1  1  5 1)  (  1 54  5 53  1 52  55
4
3
2
4
3
2
4
3
2
 14 1
4

3
2
 5 1  1 1  51
3
2
= 32 u 2
3
Y= x3-5x2+x-1
Y= 2x-6