This document provides an overview of key concepts related to indefinite and definite integrals including: 1) How to bring variables to the numerator when solving indefinite integrals. 2) The rule for antideriving powers and the importance of the constant C. 3) How to perform u-substitutions when integrals contain functions within functions. 4) The difference between indefinite and definite integrals and how to set up definite integrals. 5) How to calculate upper and lower sums to estimate the area under a curve. 6) How the Fundamental Theorems of Calculus relate definite and indefinite integrals.

1. Integration FTW
Ronnie Trinh
Jacob Flores
Cyrus Byramji

2. Indefinite Integrals
• When solving an indefinite integral, the first thing to do is bring all
variables to the numerator (if any variables are on the denominator)
• For Example, when approaching the problem:
the integral of x2 + 8x + (3/x2) dx, where dx is saying what to antiderive in
terms of, the first thing that should be noticed is that there is an x2 on the
bottom. In order to bring the x2 to the numerator simply move the variable
to the top, BUT remember to put a negative sign, as in x-2. Once that is
taken care of, antiderive which means instead of deriving, ask yourself
what did you derive to get the problem given in the integral.
• The rule for antideriving is simple - Xn+1 / n + 1
• Remember when antideriving, the only possible way to antiderive is if there
is ONE variable to a power.

3. Indefinite Integrals Continued
• Let‟s put everything together!
• Given the integral of x2 + 8x + (3/x2) dx the first
thing is to move the x2 to the top.
• Now we have the ingtegral of x2 + 8x + (3x-2) dx
• Now Antiderive.
• x3 / 3 + 4x2 - 3x + C
• Since we are going backwards it is important to
add the variable C because we don‟t know if there
could be a number present.

4. Indefinite Integrals With U
Substitution
• When given an indefinite integral with a parenthesis, square root, or a
function within an trigonomic function such as tan (2x), where 2x is the
function within the tangent function, the U substitution comes into play
• Given the integral of 3x(1-2x2)1/2 dx, the first thing to notice is that 1-2x2 is
inside of a square root.
• Now let 1-2x2 be equal to u.
• Since we are now using a different variable, it is important to understand
now we must derive in terms of that variable…du!
• du = the derivative of u which is -4x dx.
• Although we do not have a -4x to substitute for we can isolate x dx because
we can substitute x dx into the equation.

5. Indefinite Integrals With U
Substitution Continued
• Now we carefully substitute.
• Step One – Draw the integral
• Step Two – substitute u for 1-2x2 since they are equal to each
other and then substitute du for x dx. Remember that -3/4 is a
constant so that can be outside of the integral.
• Step Three - antiderive the integral of (u) ½ multiplied by -3/4
• -1/2 (u)3/2 + C
• Step Four – plug back the equation u equals for u.
• In other words… -1/2(1-2x2) 3/2 + C

6. Definite Integrals
• The difference between an indefinite and
definite integral is that an integral is bounded
by either two x or y values.
• The top number on the integral is “b” and the
bottom number on the integral is “a”.
• Common problems include write the integral
from 1 to 4 of x2

7. Definite Integrals – Upper Sum, Lower Sum
• If a problem ask to find the upper and lower sum of the curves the problem is
really asking to find the area under the curve.
• Say a problem states find the upper and lower sum of the integral from 2 to 5
of the function x, where there are 3 subintervals of equal length.
• Remember the area of the curve is essentially the area of multiple rectangles
or in other words length multiplied by height.
• When solving for the upper sum look between the first subinterval between 2
and 3 and see which of the two points has the greatest height and multiply
that by 1 (width). Take that height and add it to the greatest height of the
second subinterval between 3 and 4 multiplied by 1 (width) and so on and so
forth.
• Same for the lower sum…look between the first subinterval between 2 and 3
and see which of the two points has the lowest height and multiply the height
by 1 (width). Take that height and add it to the lowest height of the second
subinterval between 3 and 4 multiplied by 1 (width) and so on and so forth.

8. 1st Fundamental Theorem of Calculus
• F(b) – F(a)
• Given the problem the integral from 0 to pie/2 of cos(2x/3) dx,
the first thing to note that we are dealing with a definite
integral, but will be using u substitution.
• In this case u = 2x/3 3/2 du = x
• Substitute and antiderive….3/2 sin (u) from 0 to pie/3 because
those values are now in terms of u instead of x.
• Now solve using F(b) – F(a)
• (3 radical 3)/4

9. 2nd Fundamental Theorem of Calculus
• Although different than the 1st fundamental theorem of
Calculus, the 2nd fundamental theorem of calculus is a piece of
cake.
• A typical problem will give a function equal to the integral of
a value to a variable of a function.
• For example, the integral from 0 to x of sin(t) dt.
• It is important to understand the 2nd fundamental theorem of
Calculus allows for the top value which is “x” in this case to
be substituted for every t value in the integral.
• Thus we have F „(x) = sin (x).