The area beneath a curve

In differential calculus the starting question was how to find the slope of a tangent to a curve. Later we saw that this "slope" can have many different interpretations, depending on the context and the problem at hand. The same will be the case for integral calculus. We start with the question of how to calculate the area beneath a curve, and we will see later that there are many different interpretations of this "area".

Another parallel between differential calculus and integral calculus concerns the technique. In differential calculus we first found the slope of the tangent by approximating the tangent by secants. This was difficult to do, and involved many calculation steps. Only later did we find short cuts in form of differential rules for calculating the slope of the tangent.

To find the area beneath a curve, we first have to do it the hard way as well. Only later we will learn about short cuts in form of of integration rules.

So, let's start. We already know how to find the area of geometrical objects like rectangles and triangles. These areas are simple to find because their borders are straight. In fact, the circle is the only geometrical object with curved borders whose area you can actually calculate.

To understand how to calculate the area with curved borders, we first have to find a way to describe this curved border. And the simplest way to do this is to use the graph of a function, as is shown in the figure below:

The graph of the function (shown in green) has the function equation

f(x)=14x2+12f(x)=\frac{1}{4} x^2+\frac{1}{2}

Of course we can take any other function, but let's stick with this one for the moment. We want to find the area of the shaded region, that is, the region that is enclosed by the graph of ff, the xx-axis, and the vertical lines at x=ax=a and x=bx=b. Here we used a=0.1a=0.1 and b=1.5b=1.5.

For a couple of functions we can already find the area under it. See the exercise below.

Exercise 1

Determine the area of the region from aa to bb beneath the graph of ff:

  1. f(x)=3f(x)=3 for all xx (constant function), from a=0.5a=0.5 to b=4b=4.

  2. f(x)=0.5x+1f(x)=0.5x+1, from a=−1a=-1 to b=3b=3.

  3. f(x)=1−x2f(x)=\sqrt{1-x^2}, from a=−1a=-1 to b=1b=1.

Solution

  1. The region is a rectangle with side lengths 3.53.5 and 33. The area is A=3⋅3.5=10.5‾A=3\cdot 3.5=\underline{10.5}.

  2. The region is a trapezoid, that is, a triangle with the top cut off (the top is to the left). Let's calculate the area of the triangle with the top, and then subtract from it the area of the top that was cut off

    A=Awith top−AtopA=A_{with\, top}-A_{top}

    The triangle with the top has height 55 and base

    f(3)=0.5⋅3+1=2.5f(3)=0.5\cdot 3+1=2.5

    thus its area is

    Awith top=2.5⋅2.5=6.25A_{with\, top}=2.5\cdot 2.5=6.25

    The triangle forming the top has height 11 and base

    f(−1)=0.5⋅(−1)+1=0.5f(-1)=0.5\cdot (-1)+1=0.5

    Thus its area is

    Atop=0.5⋅0.5=0.25A_{top}=0.5\cdot 0.5=0.25

    The area of the trapezoid is therefore

    A=6.25−0.25=6‾A=6.25-0.25=\underline{6}

  3. The graph forms the upper half of a circle of radius r=1r=1, because from y=1−x2y=\sqrt{1-x^2} follows y2=1−x2y^2=1-x^2 and therefore x2+y2=1x^2+y^2=1, which means that the distance between the origin and the point with the coordinates (x∣y)(x\vert y) has always length 11. Thus, the shaded area is

    A=r2π2=π2‾A=\frac{r^2\pi}{2}=\underline{\frac{\pi}{2}}