E1 - Antiderivatives and
Indefinite Integrals
IB Math HL/SL
(A) Review - Simple Antiderivatives
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If v(t) = t4 – 6t2 + 4, find d(t)
So we need to “undifferentiate” or “antidifferentiate” 3 functions (t4, 6t2, 3)
So what equation did we differentiate to get t4  recalling the power rule, it must
have been a t5 expression  but the derivative of t5 is actually 5t4  so if we have
only t4 as our “derivative”, then we are “off” by 1/5  so our “antiderivative” must
be 1/5t5 (test by differentiating  which does in fact give us 1t4)
Likewise, to have 6t2 as a derivative, we must have started with a t3 function  but
which one?  if we differentiate t3, we get 3t2, so seeing that we have a 6t2, we
clearly have twice what we need  so we must have started with 2t3 or one-third of
6t3 (test by differentiating  which does in fact give us 6t2)
And then consider the 4 (or 4x0)  we must have started with an x1 function  but
which one?  if we differentiate x1, then we get 1x0  we have 4x0, so we have
four times what we need  so we must have started with 4x1 or simply 4x (test by
differentiating  which does in fact give us 3)
So we have simply looked at power functions and seen a method for “undoing” a
power rule  add one to the exponent, then adjust the coefficient by dividing by
the new exponent
(B) Review - Simple Antiderivatives
& Initial Conditions
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Another point needs to be addressed  so let’s
work with an easier function
Say v(t) = 4t (or rather d`(t) = 4t)  then what
is d(t)??
If we anti-differentiate, we would predict d(t)
= 2t2, but we need to realize that there is more
to it than that  Why?  lets consider some
derivatives and some graphs
(B) Review - Simple Antiderivatives
& Initial Conditions
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So our derivative of y` = 4x could
have come from any of the three
“antiderivatives” of y = 2x2 or y = 2x2
+ 4 or y = 2x2 – 6 or any other
combination of y = 2x2 + C, where C
is simply a constant that changes the
y-intercept of the original parabola
2x2
(B) Review - Simple Antiderivatives
& Initial Conditions
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So back to our original problem  if v(t) = t4 – 6t2 + 4, find d(t)
We have worked out that d(t) = 1/5t5 – 2t3 + 4t
Having just seen the last graphic discussion about “families of curves”, we
realize that we need to add some constant to the “antiderivative” equation
(the distance equation) so our equation for d(t) becomes d(t) = 1/5t5 – 2t3 +
4t + C  which we will call our most general solution
Now, if we knew some information about the distance function, we could
solve for C and complete our equation  say at t = 0 sec, then d(0) = 2 m
(the initial position of the object was 2 meters beyond some reference
point)
Now d(0) = 1/5(0)5 – 2(0)3 + 4(0) + C = 2  C = 2
Thus d(t) = 1/5t5 – 2t3 + 4t + 2 is our final specific solution
(C) Review - Common Antiderivatives
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The functions:
dy/dx = xn
dy/dx = 0
dy/dx = 1
dy/dx = 1/x
dy/dx = ekx
dy/dx = coskx
dy/dx = sinkx
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The antiderivatives:
y = xn+1 / (n+1) + C
y=1
y=x+C
y = ln|x| + C
y = 1/k ekx + C
y = 1/k sin(kx) + C
y = -1/k cos(kx) + C
(D) Indefinite Integrals - Definitions
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We have worked with motion applications and velocity functions  so we
recognize that v(t) = s`(t) (i.e. that the velocity equation is simply the derivative of
the position/displacement equation
So we will generalize the anti-derivative idea by saying that the function we are
trying to anti-differentiate simply represents a derivative equation (i.e. the velocity
function is the derivative of the displacement function)
Definitions: an anti-derivative of f(x) is any function F(x) such that F`(x) = f(x) 
If F(x) is any anti-derivative of f(x) then the most general anti-derivative of f(x) is
called an indefinite integral and denoted  f(x)dx = F(x) + C where C is any
constant
In this definition the  is called the integral symbol, f(x) is called the integrand, x is
called the integration variable and the “C” is called the constant of integration
The process of finding an indefinite integral (or simply an integral) is called
integration
(E) Properties of Indefinite Integrals
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[c  f(x)]dx = c  f(x)dx
-f(x)dx = - f(x)dx
[f(x) + g(x)]dx = f(x)dx + g(x)dx
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which is similar to rules we have seen for derivatives
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And two other interesting “properties” need to be
highlighted:
 g`(x)dx = g(x) + C
d/dx f(x)dx = f(x)
(F) Examples
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(x4 + 3x – 9)dx = x4dx + 3
xdx - 9 dx
(x4 + 3x – 9)dx = 1/5 x5 +
3/2 x2 – 9x + C
e2xdx =
sin(2x)dx =
(x2x)dx =
(cos + 2sin3)d =
(8x + sec2x)dx =
(2 - x)2dx =
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Continue now with
these questions on line
Problems & Solutions
with Antiderivatives /
Indefinite Integrals from
Visual Calculus
(G) Indefinite Integrals with Initial
Conditions
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Given that f(x)dx = F(x) + C, we can determine a
specific function if we knew what C was equal to 
so if we knew something about the function F(x),
then we could solve for C
Ex. Evaluate (x3 – 3x + 1)dx if F(0) = -2
F(x) = x3dx - 3 xdx + dx = ¼x4 – 3/2x2 + x + C
Since F(0) = -2 = ¼(0)4 – 3/2(0)2 + (0) + C
So C = -2 and
F(x) = ¼x4 – 3/2x2 + x - 2
(H) Examples – Indefinite Integrals
with Initial Conditions
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Problems & Solutions with Antiderivatives /
Indefinite Integrals and Initial Conditions from
Visual Calculus
Motion Problem #1 with Antiderivatives /
Indefinite Integrals from Visual Calculus
Motion Problem #2 with Antiderivatives /
Indefinite Integrals from Visual Calculus
(I) Internet Links
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Calculus I (Math 2413) - Integrals from Paul
Dawkins
Tutorial: The Indefinite Integral from Stefan
Waner's site "Everything for Calculus”
The Indefinite Integral from PK Ving's
Problems & Solutions for Calculus 1
Karl's Calculus Tutor - Integration Using Your
Rear View Mirror
(J) Homework
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Stewart, 1989, Chap 11.2, p505, Q1,2
Handout