Intro To Integration

Lets pick up where we left off in Kinematics.

From the kinematic concepts of position, velocity and acceleration we saw that the area under the curve of a derivative function gives us a connection back to the original function. In the kinematics example we used numbers, now we're going to go a bit more abstract.

### Area Under the Curve

Given the function $f(x)=2x$ the derivative is $f'(x)=2$. The graph of the derivative it looks something like: Now the area under the curve is simply a rectangle. The height of the rectangle is 2 and the width… With a length of x. So the area under the curve is $2x$ which is the original function!

Lets try a slightly more complicated example. Starting with the function $g(x)=x^2$ then the derivative is $g'(x)=2x$. The graph of the derivative looks like:

The area under the curve is in the shape of a triangle. Now at any point the height of the triangle is $g'(x)$ and the length is x. So…

(1)
\begin{align} Area=\frac{1}{2}(2x)(x)=x^2 \end{align}

Which is once again equal to the original function. So far we have a "proof by suggestion," that is, two examples that worked. They showed that the area under the curve of the derivative is equal to the original function, but is this always true?

Turns out it is.

### The Anti-Derivative

Calculating the area under the curve essentially "undoes" the derivative. In math lingo this is sometimes called the "Anti-Derivative." It is an inverse operation! Just like addition and subtraction are inverse operations.

For less nice functions say $f(x)=\frac{1}{3}x^3 + x$ whose derivatives $f '(x)=x^2+1$ don't have areas under them that make nice shapes (perfect rectangles or triangles) we have to be a bit more clever. One approach is to break up the area under the curve into rectangles as shown below:

Notice that the rectangles are only an approximation, that either we over or underestimate the area. However, if the number of rectangles increases (move the slider) the approximation gets better and better. As the number of rectangles increases the width of each rectangle decreases. Imagine that the width of the rectangle gets infinitely small (but not zero) then the area of all the rectangles will in fact exactly equal the area under the curve. Hmm. Clever!

The area of each rectangle will be the height of the rectangle (the value of the function at that point, $f'(x)$) multiplied by the width of the rectangle which we'll call $dx$. More formally the area of one rectangle is:

(2)
\begin{align} f '(x) \cdot dx \end{align}

To find the total area under the curve we must add up all of the rectangles. Since the width of each rectangle is infinitely small we then have an infinite number of rectangles to add up… In symbols we represent this addition or sum of the rectangles by:

(3)
\begin{align} \int f'(x) dx = f(x) \end{align}

Since the area under the curve is equal to the function then the sum of the areas of all the rectangles must equal the original function. This is the famous "Integral."

### The Integral

The integral $\int f'(x) \cdot dx = f(x)$ can be viewed to be asking the questions:

• What function $f(x)$ has the derivative $f'(x)$ and/or
• What is the sum of all the infinitely small areas $f '(x) \cdot dx$

Both of these questions get asked in a math class. It is easy to lose perspective and focus and see these questions as fundamentally different. Sometimes it is easier to think of one versus the other, but they are in fact the same question…

Still with me? Lets move on to calculating indefinite integrals.