[<< wikibooks] A-level Mathematics/OCR/C2/Integration
Integration, is the process of anti-differention. In the last segment we saw that differentiating a function gives the gradient (or rate of change) of that function. Integrating a function, on the other hand, gives the area underneath the curve of that function.

== Basic integration ==
Integration is the reverse of differentiation.

=== Notation ===
The integral sign is

∫
f

(
x
)

d
x

{\displaystyle \int f\left(x\right)dx}
. This symbol was chosen by Leibniz because it looks like an elongated S and an integral is a limit of sums. The f(x) is known as the integrand. The dx means that we integrate with respect to x.

=== Integration Rules ===
An arbitrary constant must be created every time a function is integrated. This is because when a constant is differentiated it becomes zero, so there is no way of telling if such a constant used to exist.

∫

x

n

d
x
=

1

n
+
1

x

n
+
1

+
C
,

(
n
≠
−
1
)

{\displaystyle \int x^{n}\,dx={\frac {1}{n+1}}x^{n+1}+C,\ (n\neq -1)}

∫
k

x

n

d
x
=
k
∫

x

n

d
x

{\displaystyle \int kx^{n}\,dx=k\int x^{n}\,dx}

∫

{

f

′

(
x
)
+

g

′

(
x
)

}

d
x
=
f
(
x
)
+
g
(
x
)
+
C

{\displaystyle \int \left\{f^{'}(x)+g^{'}(x)\right\}\,dx=f(x)+g(x)+C}

∫

{

f

′

(
x
)
−

g

′

(
x
)

}

d
x
=
f
(
x
)
−
g
(
x
)
+
C

{\displaystyle \int \left\{f^{'}(x)-g^{'}(x)\right\}\,dx=f(x)-g(x)+C}

==== Examples ====

∫

x

3

=

1

3
+
1

x

3
+
1

+
c
=

1
4

x

4

+
c

{\displaystyle \int x^{3}={\frac {1}{3+1}}x^{3+1}+c={\frac {1}{4}}x^{4}+c}

∫
4

x

3

=
4
∫

x

3

=
4

(

1

3
+
1

x

3
+
1

)

+
c
=
4

(

1
4

x

4

)

+
c
=

x

4

+
c

{\displaystyle \int 4x^{3}=4\int x^{3}=4\left({\frac {1}{3+1}}x^{3+1}\right)+c=4\left({\frac {1}{4}}x^{4}\right)+c=x^{4}+c}

∫

x

4

−
3

x

3

+
4

x

2

−
x
+
3
=
∫

x

4

−
∫
3

x

3

+
∫
4

x

2

−
∫
x
+
∫
3
=

1
5

x

5

−

3
4

x

4

+

4
3

x

3

−

1
2

x

2

+
3
x
+
c

{\displaystyle \int x^{4}-3x^{3}+4x^{2}-x+3=\int x^{4}-\int 3x^{3}+\int 4x^{2}-\int x+\int 3={\frac {1}{5}}x^{5}-{\frac {3}{4}}x^{4}+{\frac {4}{3}}x^{3}-{\frac {1}{2}}x^{2}+3x+c}

=== Fundamental Theorem of Calculus ===
This is the most important rule in calculus it establishes that differentiation and integration are inverse processes. Here is the theorem:
Let f be continuous on [a,b].

If

g
(
x
)
=

∫

a

x

f

(
t
)

d
t

{\displaystyle g(x)=\int _{a}^{x}f\left(t\right)\ dt}
then

g
′

(
x
)

=
f

(
x
)

{\displaystyle g'\left(x\right)=f\left(x\right)}

∫

a

b

f

(
x
)

d
x
=
F

(
b
)

−
F

(
a
)

{\displaystyle \int _{a}^{b}f\left(x\right)\ dx=F\left(b\right)-F\left(a\right)}
, F is the antiderivative of f such that F' = fThe first rule states that if we integrate a function and then we differentiate the resultant then we will arrive at the same function. The second part says that we can find a definite integral by subtracting the value of F at the endpoints.

== Indefinite Integrals ==
The first part of the Fundamental theorem of calculus establishes that if we differentiate a function and then integrate a function we get a indefinite integral. When we evaluate an indefinite integral the resultant will be a function. For an indefinite integral we use the appropriate rule to get the general antiderivative. If a point on the graph is given we solve for C, to get the complete antiderivative.

=== Example ===
The rate of a function is given by the equation

4

x

3

+
3

x

2

−
4
x
+
2

{\displaystyle 4x^{3}+3x^{2}-4x+2}
, the point (0, -7) is on the curve. Find the equation of the curve.

We need to find the general indefinite integral of

4

x

3

+
3

x

2

−
4
x
+
2

{\displaystyle 4x^{3}+3x^{2}-4x+2}
.

∫
(
4

x

3

+
3

x

2

−
4
x
+
2
)
d
x
=

x

4

+

x

3

−
2

x

2

+
2
x
+
C

{\displaystyle \int (4x^{3}+3x^{2}-4x+2)dx=x^{4}+x^{3}-2x^{2}+2x+C}

Now we substitute the point (0, -7) into the general antiderivative to get the value of C.

−
7
=

(
0
)

4

+

(
0
)

3

−
2

(
0
)

2

+
2

(
0
)

+
C

{\displaystyle -7=\left(0\right)^{4}+\left(0\right)^{3}-2\left(0\right)^{2}+2\left(0\right)+C}

C
=
−
7

{\displaystyle C=-7}

Now we can write the complete antiderivative.

F

(
x
)

=

x

4

+

x

3

−
2

x

2

+
2
x
−
7

{\displaystyle F\left(x\right)=x^{4}+x^{3}-2x^{2}+2x-7}

== Definite Integrals ==
The definite integral is used to find the area underneath a curve. The second part of the fundamental theorem of calculus allows to evaluate these integrals, the resultant will be a number. The Definite Integral is denoted as

∫

a

b

f

(
x
)

d
x

{\displaystyle \int _{a}^{b}f\left(x\right)\,dx}
. In the definite integral a is the lower limit and b is the upper limit together they are known as the limits of integration. The limits of integration on which interval to find the area. When we evaluate a definite integral we don't write the +c because they will always cancel out and are liable to cause confusion. When we write

F
(
x
)

]

a

b

{\displaystyle F(x){\Big ]}_{a}^{b}}
this means that we have found the indefinite integral and are going to find the definite integral from b to a.

=== Rules of Definite Integrals ===

∫

a

b

f

(
x
)

d
x
=
F

(
b
)

−
F

(
a
)

{\displaystyle \int _{a}^{b}f\left(x\right)\ dx=F\left(b\right)-F\left(a\right)}
, F is the antiderivative of f such that F' = f

∫

a

b

f

(
x
)

d
x
=
−

∫

b

a

f

(
x
)

d
x

{\displaystyle \int _{a}^{b}f\left(x\right)\ dx=-\int _{b}^{a}f\left(x\right)\ dx}

∫

a

a

f

(
x
)

d
x
=
0

{\displaystyle \int _{a}^{a}f\left(x\right)\ dx=0}

Area between a curve and the x-axis is

∫

a

b

y

d
x

(

for

y
≥
0
)

{\displaystyle \int _{a}^{b}y\,dx\ ({\mbox{for}}\ y\geq 0)}
Area between a curve and the y-axis is

∫

a

b

x

d
y

(

for

x
≥
0
)

{\displaystyle \int _{a}^{b}x\,dy\ ({\mbox{for}}\ x\geq 0)}

=== Area of a Region Bounded By A Curve ===
When we evaluate area underneath curve we need to make sure that the interval over which we are finding the area is not in part or in full below the x-axis. In order to determine this fact we need to find x-intercepts of the function. Then we see if a x-intercept is in the interval. If so we need to determine which part of the interval is below the x-axis. If the whole interval is below the x-axis we take the absolute value of the area

∫

a

b

|

f

(
x
)

|

d
x

{\displaystyle \int _{a}^{b}{\begin{vmatrix}f\left(x\right)\end{vmatrix}}dx}
.If only part of the interval is under the x-axis we need to break up the integration function into positive and negative parts. For the negative parts we need take the absolute value of the area. For example: If we need to find the area of

∫

a

b

{\displaystyle \int _{a}^{b}}
and on the interval's

a
,

a

1

{\displaystyle a,a_{1}}
and

b

1

,
b

{\displaystyle b_{1},b}
the curve is above the x-axis and on the interval

a

1

,

b

1

{\displaystyle a_{1},b_{1}}
the curve is below the x-axis, we would break the integral into these parts:

∫

a

a

1

f

(
x
)

d
x
+

∫

a

1

b

1

|

f

(
x
)

|

d
x
+

∫

b

1

b

f

(
x
)

d
x

{\displaystyle \int _{a}^{a_{1}}f\left(x\right)dx+\int _{a_{1}}^{b_{1}}{\begin{vmatrix}f\left(x\right)\end{vmatrix}}dx+\int _{b_{1}}^{b}f\left(x\right)dx}
. If you are unable to evaluate the definite integral you will need to use a numerical method to estimate the area.

==== Example ====
Evaluate

∫

0

3

x

3

−
6
x

d
x

{\displaystyle \int _{0}^{3}x^{3}-6x\ dx}
.

First we need to find the x-intercepts. In this case they will be at 0 and 2.
Then we determine where x is positive or negative on the interval.0__f(1)=-5__2__f(3)=9.

Now we break the integral into parts.

∫

0

2

|

x

3

−
6
x

|

d
x
+

∫

2

3

x

3

−
6
x

d
x

{\displaystyle \int _{0}^{2}{\begin{vmatrix}x^{3}-6x\end{vmatrix}}\ dx+\int _{2}^{3}x^{3}-6x\ dx}
.
Now we evaluate the two definite integrals.

(

(

1
4

(
3
)

4

−
3

(
3
)

2

)

−

(

1
4

(
2
)

4

−
3

(
2
)

2

)

)

{\displaystyle \left(\left({\frac {1}{4}}\left(3\right)^{4}-3\left(3\right)^{2}\right)-\left({\frac {1}{4}}\left(2\right)^{4}-3\left(2\right)^{2}\right)\right)}
+

(

|

(

1
4

(
2
)

4

−
3

(
2
)

2

)

−

(

1
4

(
0
)

4

−
3

(
0
)

2

)

|

)

{\displaystyle \left({\begin{vmatrix}\left({\frac {1}{4}}\left(2\right)^{4}-3\left(2\right)^{2}\right)-\left({\frac {1}{4}}\left(0\right)^{4}-3\left(0\right)^{2}\right)\end{vmatrix}}\right)}
.

(

−
6.75
−
−
8

)

{\displaystyle \left(-6.75--8\right)}
+

(

|

−
8
−
0

|

)

{\displaystyle \left({\begin{vmatrix}-8-0\end{vmatrix}}\right)}
= 9.25

The area under the curve of

x

3

−
6
x

{\displaystyle x^{3}-6x}
between 0 and 3 is 9.25.

== Areas Involving Two Curves ==

=== Area Bounded By Two Curves ===
In order to calculate the area bounded curves we need to:

Find the places where the 2 curves intercept.

f

(
x
)

=
g

(
x
)

{\displaystyle f\left(x\right)=g\left(x\right)}
. The lower point of intersection will become the lower limit and the upper point of intersection will become the upper limit of integration.
Decide whether to integrate with respect to x or y. This is easy to do in most cases, see if the function is more easily integrated as y = f(x) or x = f(y).
Determine which curve is above the other curve.
Subtract the bottom curve from the top curve.

A
=

∫

a

b

|

f

(
x
)

−
g

(
x
)

|

d
x

{\displaystyle A=\int _{a}^{b}{\begin{vmatrix}f\left(x\right)-g\left(x\right)\end{vmatrix}}dx}

Evaluate the definite integral

==== Example ====
Find the area bounded by

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
and

y
=
x

{\displaystyle y=x}
.

x

2

+
3
x
−
3
=
x

{\displaystyle x^{2}+3x-3=x\,}

x

2

+
2
x
−
3
=
0

{\displaystyle x^{2}+2x-3=0\,}

(

x
+
3

)

(

x
−
1

)

=
0

{\displaystyle \left(x+3\right)\left(x-1\right)=0}

x
=
−
3

o
r

1

{\displaystyle x=-3\ or\ 1\,}

We will integrate with respect to x because the equation is in the form f(x).
We now have to determine which line is above the other. For this we just test a point. I will use f(x) = 0

y
=

0

2

+
3
×
0
−
3

{\displaystyle y=0^{2}+3\times 0-3}
and

y
=
0

{\displaystyle y=0\,}

y
=
−
3

{\displaystyle y=-3}
and

y
=
0

{\displaystyle y=0\,}

So

y
=
x

{\displaystyle y=x}
will be above

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
.

Now we write out the integral.

∫

−
3

1

|

(
x
)

−

(

x

2

+
3
x
−
3

)

|

d
x

{\displaystyle \int _{-3}^{1}{\begin{vmatrix}\left(x\right)-\left(x^{2}+3x-3\right)\end{vmatrix}}dx}

∫

−
3

1

|

−

x

2

−
2
x
+
3

|

d
x

{\displaystyle \int _{-3}^{1}{\begin{vmatrix}-x^{2}-2x+3\end{vmatrix}}dx}

Finally we evaluate the definite integral.

|

−

1
3

x

3

−

x

2

+
3
x

|

]

−
3

1

{\displaystyle {\begin{vmatrix}-{\frac {1}{3}}x^{3}-x^{2}+3x\end{vmatrix}}{\Big ]}_{-3}^{1}}

|

[

−

1
3

(
1
)

3

−

(
1
)

2

+
3

(
1
)

]

−

[

−

1
3

(

−
3

)

3

−

(

−
3

)

2

+
3

(

−
3

)

]

|

{\displaystyle {\begin{vmatrix}{\begin{bmatrix}-{\frac {1}{3}}\left(1\right)^{3}-\left(1\right)^{2}+3\left(1\right)\end{bmatrix}}-{\begin{bmatrix}-{\frac {1}{3}}\left(-3\right)^{3}-\left(-3\right)^{2}+3\left(-3\right)\end{bmatrix}}\end{vmatrix}}}

|

1

2
3

−
(
−
9
)

|

{\displaystyle {\begin{vmatrix}1{\frac {2}{3}}-(-9)\end{vmatrix}}}

A
r
e
a
=
10

2
3

{\displaystyle Area=10{\frac {2}{3}}}

=== Area Between Two Curves ===
When we calculate the area between two curves the procedure is very similar as to when we calculate the area bounded by two curves. In the problem the limits of integration will be given. The major difference is that in some cases we need to split the integral up.

Find the places where the 2 curves intercept.

f

(
x
)

=
g

(
x
)

{\displaystyle f\left(x\right)=g\left(x\right)}
.
If the two graphs intercept at a point within the limit of integration, we will need to split the integral into parts. We will use the points of interception as break points.
Decide whether to integrate with respect to x or y. This is easy to do in most cases, see if the function is more easily integrated as y = f(x) or x = f(y).
Determine which curve is above the other curve. If the integral has been split up do this on every interval.
Subtract the top curve from the bottom curve.

A
=

∫

a

b

|

f

(
x
)

−
g

(
x
)

|

d
x

{\displaystyle A=\int _{a}^{b}{\begin{vmatrix}f\left(x\right)-g\left(x\right)\end{vmatrix}}dx}
. If the integral has been split up we add up the parts.
Evaluate the definite integral(s).

==== Example ====
Find the area between by

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
and

y
=
x

{\displaystyle y=x}
between x = -4 and x = 4.

x

2

+
3
x
−
3
=
x

{\displaystyle x^{2}+3x-3=x\,}

x

2

+
2
x
−
3
=
0

{\displaystyle x^{2}+2x-3=0\,}

(

x
+
3

)

(

x
−
1

)

=
0

{\displaystyle \left(x+3\right)\left(x-1\right)=0}

x
=
−
3

o
r

1

{\displaystyle x=-3\ or\ 1\,}

Since the graph have points of interception on the interval we need to split the integral into parts. The parts will be (-4, -3), (-3, 1), and (1,4)
We will integrate with respect to x because the equation is in the form f(x).
We now have to determine which line is above the other on each interval. For this we just test a point on each interval.For the line

y
=
x

{\displaystyle y=x}
__f(-4)=-4__f(-2)=-2___f(3)=3__.
For the curve

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
__f(-4)=-1__f(-2)=-5___f(3)=15__.
Now we will know that on the interval (-4, -3)

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
is above

y
=
x

{\displaystyle y=x}
, (-3, 1)

y
=
x

{\displaystyle y=x}
is above

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
, and (1,4)

y
=

x

2

+
3
x
−
3

{\displaystyle y=x^{2}+3x-3}
is above

y
=
x

{\displaystyle y=x}
.

Now we write out the integral.

∫

−
4

−
3

|

(

x

2

+
3
x
−
3

)

−

(
x
)

|

d
x

{\displaystyle \int _{-4}^{-3}{\begin{vmatrix}\left(x^{2}+3x-3\right)-\left(x\right)\end{vmatrix}}dx}
+

∫

−
3

1

|

(
x
)

−

(

x

2

+
3
x
−
3

)

|

d
x

{\displaystyle \int _{-3}^{1}{\begin{vmatrix}\left(x\right)-\left(x^{2}+3x-3\right)\end{vmatrix}}dx}
+

∫

1

4

|

(

x

2

+
3
x
−
3

)

−

(
x
)

|

d
x

{\displaystyle \int _{1}^{4}{\begin{vmatrix}\left(x^{2}+3x-3\right)-\left(x\right)\end{vmatrix}}dx}

∫

−
4

−
3

|

(

x

2

+
2
x
−
3

)

|

d
x

{\displaystyle \int _{-4}^{-3}{\begin{vmatrix}\left(x^{2}+2x-3\right)\end{vmatrix}}dx}
+

∫

−
3

1

|

−

x

2

−
2
x
−
3

|

d
x

{\displaystyle \int _{-3}^{1}{\begin{vmatrix}-x^{2}-2x-3\end{vmatrix}}dx}
+

∫

1

4

|

(

x

2

+
2
x
−
3

)

|

d
x

{\displaystyle \int _{1}^{4}{\begin{vmatrix}\left(x^{2}+2x-3\right)\end{vmatrix}}dx}

Now we evaluate the definite integrals.

|

1
3

x

3

+

x

2

−
3
x

|

]

−
4

−
3

{\displaystyle {\begin{vmatrix}{\frac {1}{3}}x^{3}+x^{2}-3x\end{vmatrix}}{\Big ]}_{-4}^{-3}}
+

|

−

1
3

x

3

−

x

2

+
3
x

|

]

−
3

1

{\displaystyle {\begin{vmatrix}-{\frac {1}{3}}x^{3}-x^{2}+3x\end{vmatrix}}{\Big ]}_{-3}^{1}}
+

|

1
3

x

3

+

x

2

−
3
x

|

]

1

4

{\displaystyle {\begin{vmatrix}{\frac {1}{3}}x^{3}+x^{2}-3x\end{vmatrix}}{\Big ]}_{1}^{4}}

|

−
9
−

−
76

3

|

{\displaystyle {\begin{vmatrix}-9-{\frac {-76}{3}}\end{vmatrix}}}
+

|

2
3

−
18

|

{\displaystyle {\begin{vmatrix}{\frac {2}{3}}-18\end{vmatrix}}}
+

|

76
3

−

−
5

3

|

{\displaystyle {\begin{vmatrix}{\frac {76}{3}}-{\frac {-5}{3}}\end{vmatrix}}}

A
r
e
a
=
60

2
3

{\displaystyle Area=60{\frac {2}{3}}}

== Estimating Area Underneath A Curve ==
It becomes necessary to estimate the area underneath a curve when a function is very difficult or impossible to integrate or when we obtain our curve from a set of values.

=== Trapezium Rule ===
The Trapezium Rule estimates the area under a curve between limits by turning the curve into a set of trapeziums (or strips) and each strip is made out of two ordinates, so there is always one more ordinates than there are strips. The formulae is:

∫

a

b

y

d
x
≈

1
2

h

{

(

y

0

+

y

n

)

+
2

(

y

1

+

y

2

+
…
+

y

n
−
1

)

}

{\displaystyle \int _{a}^{b}y\,dx\approx {\frac {1}{2}}h\left\{\left(y_{0}+y_{n}\right)+2\left(y_{1}+y_{2}+\ldots +y_{n-1}\right)\right\}}

Where:

h
=

b
−
a

n

{\displaystyle h={\frac {b-a}{n}}}

==== Example ====
Use the Trapezium Rule to evaluate

∫

0

1

(
2
x
+
1

)

1

/

2

d
x

{\displaystyle \int _{0}^{1}(2x+1)^{1}/2\ dx}
using 4 strips.
Firstly, we work out h.

h
=

1
−
0

4

=

1
4

=
0.25

{\displaystyle h={\frac {1-0}{4}}={\frac {1}{4}}=0.25}

Now we start setting up the Trapezium Rule.

∫

0

1

(
2
x
+
1

)

1

/

2

d
x
≈

1
8

[

f

(
0
)

+
f

(
1
)

+
2

{

f

(
0.25
)

+
f

(
0.5
)

+
f

(
0.75
)

}

]

{\displaystyle \int _{0}^{1}(2x+1)^{1}/2\ dx\approx {\frac {1}{8}}\left[f\left(0\right)+f\left(1\right)+2\left\{f\left(0.25\right)+f\left(0.5\right)+f\left(0.75\right)\right\}\right]}

Solving for the f(n) we get:

∫

0

1

(
2
x
+
1

)

1

/

2

d
x
≈

1
8

[

1
+
1.73
+
2

{

1.22
+
1.41
+
1.42

}

]

{\displaystyle \int _{0}^{1}(2x+1)^{1}/2\ dx\approx {\frac {1}{8}}\left[1+1.73+2\left\{1.22+1.41+1.42\right\}\right]}

∫

0

1

(
2
x
+
1

)

1

/

2

d
x
≈

1.39375

{\displaystyle \int _{0}^{1}(2x+1)^{1}/2\ dx\approx \ 1.39375}

== Integrals to Infinity ==
An integral that has either it upper limit going to

∞

{\displaystyle \infty }
or is lower limit going to

−
∞

{\displaystyle -\infty }
is an improper integral and cannot be computed directly instead we need to find the limit of the function. The improper integrals

∫

a

∞

f

(
x
)

d
x

{\displaystyle \int _{a}^{\infty }f\left(x\right)\ dx}
and

∫

−
∞

b

f

(
x
)

d
x

{\displaystyle \int _{-\infty }^{b}f\left(x\right)\ dx}
is convergent if the limit exists and divergent if the limit does not exist. The rules to calculate a integral going to infinity are:
1) If

∫

a

∞

f

(
n
)

d
x

{\displaystyle \int _{a}^{\infty }f\left(n\right)\ dx}
exist for all number

n
≥
a

{\displaystyle n\geq a}
, then

∫

a

∞

f

(
x
)

d
x
=

lim

n
→
∞

∫

a

∞

f

(
x
)

d
x

{\displaystyle \int _{a}^{\infty }f\left(x\right)\ dx=\lim _{n\to \infty }\int _{a}^{\infty }f\left(x\right)\ dx}
The limit has to be a finite number.2) If

∫

n

b

f

(
x
)

d
x

{\displaystyle \int _{n}^{b}f\left(x\right)\ dx}
exist for all number

n
≤
b

{\displaystyle n\leq b}
, then

∫

−
∞

b

f

(
x
)

d
x
=

lim

n
→
−
∞

∫

−
∞

b

f

(
x
)

d
x

{\displaystyle \int _{-\infty }^{b}f\left(x\right)\ dx=\lim _{n\to -\infty }\int _{-\infty }^{b}f\left(x\right)\ dx}
The limit has to be a finite number.3) If

∫

a

∞

f

(
n
)

d
x

{\displaystyle \int _{a}^{\infty }f\left(n\right)\ dx}
and

∫

n

b

f

(
x
)

d
x

{\displaystyle \int _{n}^{b}f\left(x\right)\ dx}
are convergent, then

∫

−
∞

∞

f

(
n
)

d
x
=

∫

n

a

f

(
x
)

d
x
+

∫

a

∞

f

(
n
)

d
x

{\displaystyle \int _{-\infty }^{\infty }f\left(n\right)\ dx=\int _{n}^{a}f\left(x\right)\ dx+\int _{a}^{\infty }f\left(n\right)\ dx}
a is any real number.You can now attempt the Integration practice problems.