An antiderivative of a function f is a function F such that F'= f.

Antiderivative in words:

1.	An antiderivative of 4x3 is x4     Because the derivative of x4 is 4x3
2.	Another antiderivative of 4x3 is x4 + 7     Because the derivative of x4 + 7 is 4x3
3.	An antiderivative of 2x is x2 + 12.     Because the derivative of x2+12 is 2x

4.	An antiderivative of 5 is
5.	An antiderivative of x is

The expression

f(x)

dx

is read "the indefinite integral of f(x) with respect to x," and stands for the set of all antiderivatives of f.

Thus, ∫ f(x) dx is a collection of functions; it is not a single function, nor a number. The function f that is being integrated is called the integrand, and the variable x is called the variable of integration. (The expression dx is short for "with respect to x.")

1.		4x3 dx  =	x4 + C	C is an arbitrary constant
2.		2x dx  =	x2 + C

The constant of integration, C, reminds us that we can substitute any number for C and get a different antiderivative.

3.		5	dx	=
4.		x	dx	=

	xn dx	=	xn+1 n + 1	+ C	if n ≠ -1
	x-1 dx	=	ln|x|+ C

In Words:

1.	x55 dx	=	x56 56 + C
2.	1 x55 dx	=	x-55 dx = - x-54 54 + C
3.	dx	=	x0 dx = x1 1 + C = x + C

4.		x3	dx	=
5.		1 x2	dx	=

ex dx = ex + C	Because d dx ex = ex
cos x dx = sin x + C	Because d dx sin x = cos x
sin x dx = -cos x + C	Because d dx (-cos x) = sin x
sec2x dx = tan x + C	Because d dx tan x = sec2x

(a) Sum and Difference Rules

[f(x) ± g(x)] dx

=

f(x) dx

±

g(x) dx

In Words:

(b) Constant Multiple Rule

kf(x) dx

=

k

f(x) dx

(k constant)

In Words:

Why are these rules true? Because the derivative of a sum is the sum of the derivatives, and similarly for differences and constant multiples.

Sum:	(x3 + 1) dx	=	x3 dx + 1 dx
		=	x4 4 + x + C

Constant Multiple:	5x3 dx	=	5 x3 dx
		=	5x4 4 + C

Constant Multiple:	4 dx	=	4 1 dx
		=	4x + C

Both Rules:	(6x2 + 4) dx	=	6 x dx + 4 1 dx
		=	6x3 3 + 4x + C
		=	2x3 + 4x + C

Want some practice? Try the tutorial or exercises.

If u is a function of x, we can use the following formula to evaluate an integral.

f dx

=

f du/dx

du

Using the Formula

Use of the formula is equivalent to the following procedure:

1. Write u as a function of x.
2. Take the derivative du/dx, and solve for the quantity dx in terms of du.
3. Use the expression you obtain in part 2 to substitute for dx in the given integral.

Deciding What to Use for u

There is no hard and fast rule, but some guidelines that sometimes work are the following.

• Take u to be an expression that is being raised to a power.
• Take u to be an expression whose derivative appears as a factor of the integrand.
• Take u to be the denominator is a rational expression.
• If the variable x cannot be eliminated using a substitution, try another substitution.
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To evaluate (x²+1)(x³ + 3x - 2)² dx, proceed as follows.

u = x3 + 3x - 2     Decide on a choice of u du dx = 3x2 + 3 = 3(x2 + 1)     Take du/dx dx = du 3(x2 + 1)     Solve for dx

Now substitute in the integral to obtain the solution, as follows:

(x2+1)(x3 + 3x - 2)2

dx

=	(x2+1)u2 du 3(x2 + 1)	Substitute for u and dx
=	u2 du 3	Cancel to eliminate x
=	1 3 u2 du	Constant multiple rule
=	1 3 u3 3 + C	Do the integration
=	(x3+3x-2)3 9 + C	Substitute back for x

Want some practice? Try the tutorial or exercises.

If s(t) represents position at time t, then velocity is given by v(t) = s'(t) and acceleration by a(t) = v'(t). This means that

v(t) = a(t) dt and

s(t) = v(t) dt.

Moreover, for motion due to gravity close to the earth's surface, ignoring air resistance, a(t) ≈ -32 ft/s² is constant. Integrating this twice gives the equations

v(t) = v₀ - 32t ft/sec     and

s(t) = s₀ + v₀t -16t²

where v₀ is the initial velocity and s₀ is the initial position.

If u is a function of x, we can use the following formula to evaluate an integral.

Riemann Sum

If f is a continuous function, the left Riemann sum with n equal subdivisions for f over the interval [a, b] is defined follows.

First, partition the interval [a, b] into n equal parts:

Δx = (b-a)/n,
x₀ = a,
x₁ = a + Δx,
x₂ = a + 2Δx,
...
x_n = a + nΔx = b

Next, add up the n products f(x₀)Δx, f(x₁)Δx, f(x₂)Δx, ..., f(x_{n -1})Δx, to get the Riemann sum.

Thus,

(Left) Riemann sum	=	n-1 n = 0 f(xk)Δx
	=	f(x0)Δx + f(x1)Δx + ... + f(xn -1)Δx
	=	[f(x0) + f(x1) + ... + f(xn -1)]Δx

The left Riemann sum gives the area shown below.

The Definite Integral

If f is a continuous function, the definite integral of f from a to b is defined as

b     a

f(x) dx

=

lim n

n-1 n = 0

f(xk)Δx

In Words:

The function f is called the integrand, the numbers a and b are the limits of integration, and the variable x is the variable of integration.

Approximating the Definite Integral To approximate the definite integral, we use a Riemann sum with a large number of subdivisions.

Let us compute the Riemann sum for the integral_-1¹(x²+1) dx using n = 5 subdivisions.

First, to compute the subdivisions:

Δx = (b-a)/n = (1-(-1)/4 = 0.4.
x₀ = a = -1
x₁ = a + Δx = -1 + 0.4 = 0.6
x₂ = a + 2Δx = -1 + 2(0.4) = 0.2
x₃ = a + 3Δx = -1 + 3(0.4) = 0.2
x₄ = a + 4Δx = -1 + 4(0.4) = 0.6
x₅ = b = 1

The Riemann sum we want is

f(x₀)Δx + f(x₁)Δx + ... + f(x₄)Δx
  = [f(-1) + f(-0.6) + f(-0.2) + f(0.2) + f(0.6)]0.4

We can organize this calculation in a table as follows.

The Riemann sum is therefore

6.8Δx = 6.80.4 = 2.72.
 

If you have Excel and want to see a visual representation of Riemann sums like the picture on the left, download the Excel Riemann Sum Grapher.

Geometric Interpretation of the Definite Integral (Non-Negative Functions)

If f(x) ≥ 0 for all x in [a, b], then _a^bf(x) dx is the area under the graph of f over the interval [a, b], as shaded in the figure.

Geometric Interpretation of the Definite Integral (All Functions)

For general functions, _a^bf(x) dx is the area between x = a and x = b that is above the x-axis and below the graph of f, minus the area that is below the x-axis and above the graph of f.

1.	2     0 x dx = 2	Area Shown = 2
2.	1     -1 x dx = 0	Green - Red = 0

Relationship between Riemann Sum Definition and Area Definition

The following figure illustrates the relationship between the (left) Riemann sum and the area for the integral ₀¹ (1-x²) dx.

If Δx is the width of each rectangle, then:

Area of the first (leftmost) rectangle = height × width = f(0)Δx = f(x₀)Δx
Area of the second rectangle = height × width = f(x₁)Δx
....
Area of the last rectangle = height × width = f(x_{n -1})Δx

Adding the areas of all the rectangles together gives the Riemann Sum. As the number n of rectangles gets larger (so that each of them has width approaching zero) the area represented by the Reimann sum gets closer and closer to the actual area. Thus,

Limit of Riemann Sum = Integral = Area

The Fundamental Theorem of Calculus (FTC)

Let f be a continuous function defined on the interval [a, b]. Then:

(a) If A(x) = _a^x f(t) dt, then A'(x) = f(x), i.e., A is an antiderivative of f, and

(b) If f is any continuous antiderivative of f, and is defined on [a, b], then

b     a

f(x) dx

=

F(b) - F(a).

Part (b) in words:

Example of (a)

If A(x) =

x     0

et2 dt,  then  A'(x) = ex2.

Example of (b)
Since F(x) = x² is an antiderivative of f(x) = 2x,

1     0

2x dx   =   F(1) - F(0) = 12 - 02 = 1.

Another Example of (b)

1     -1

(1-x2) dx

=

x - x3/3

1     -1

=

2 3

- (-

2 3

) =

4 3