From a Definite Integral to a Function
A definite integral usually has two fixed endpoints and produces a number. A different perspective is to keep the starting point fixed while allowing the ending point to vary. Each choice of endpoint then gives a definite integral, and therefore a number. In this way, one integral expression defines a function.
This construction is useful whenever \(f\) describes a rate or a density and we want to record how much has accumulated from a chosen starting point up to each later input. For now, the central task is to define this function precisely and calculate its values from signed areas.
Definition and Domain
Let \(a<b\), and suppose \(f\) is Riemann integrable on \([a,b]\). The basepoint \(a\) is fixed. For every \(x\) in the interval \([a,b]\), define an accumulation function by integrating \(f\) from \(a\) to \(x\).
The integrability condition ensures that the definite integral exists for every endpoint in the stated domain: the restriction of an integrable function to any subinterval is integrable. Continuity of \(f\) on \([a,b]\) is one familiar sufficient condition, but continuity is not required if integrability is already known.
The domain matters. This definition assigns values only to \(x\in[a,b]\), so each integral runs from the fixed basepoint to an endpoint at or to the right of it. The endpoint \(x\) varies, but the basepoint does not. At the basepoint itself, the interval has length zero, giving \(A(a)=0\).
Why This Definition Gives a Function
A function must assign exactly one output to each input in its domain. The integral in the definition does that: once \(x\) is chosen, the lower endpoint \(a\), upper endpoint \(x\), and integrand \(f\) are all specified. The definite integral has one real value. Thus every \(x\in[a,b]\) receives exactly one real output.
More explicitly, for any chosen \(x\in[a,b]\), the restricted function \(f\) is integrable on \([a,x]\). Hence \(\int_a^x f(t)\,dt\) exists and is a unique real number. This works for every input in the domain, so the rule defines a function \(A:[a,b]\to\mathbb{R}\). In particular, when \(x=a\), the integral is over an interval of zero width, so \(A(a)=0\).
Calculating Values from Signed Areas
When the graph of \(f\) is made of simple geometric pieces, the integral can be evaluated as signed area. A region above the horizontal axis contributes its ordinary positive area; a region below the axis contributes the negative of its ordinary area. To find \(A(x)\), account for the signed regions between the basepoint and the chosen endpoint.
This interpretation follows the sign convention of the definite integral. In Riemann sums, a rectangle over a point where \(f\) is positive contributes a positive height times a positive width. A rectangle where \(f\) is negative contributes a negative height times a positive width. The limit of these signed sums is the integral. For a graph made of straight or horizontal segments, these contributions can often be computed exactly using rectangles and triangles.
Worked Example: A Constant Positive Integrand
Let \(f(t)=2\) on \([0,3]\), and define \(A(x)=\int_0^x 2\,dt\). Find \(A(0)\), \(A(1)\), and \(A(5/2)\).
The graph is a horizontal line at height \(2\), so the region from \(0\) to any endpoint \(x\) is a rectangle of height \(2\) and width \(x\). Its area is \(2x\).
At \(x=0\), the width is \(0\), so \(A(0)=2\cdot 0=0\). At \(x=1\), the width is \(1\), so \(A(1)=2\cdot 1=2\). At \(x=5/2\), the width is \(5/2\), so \(A(5/2)=2\cdot(5/2)=5\).
Check: The respective rectangle areas are \(2\cdot 0=0\), \(2\cdot 1=2\), and \(2\cdot(5/2)=5\), matching all three values.
Worked Example: A Graph That Crosses the Axis
Let \(f(t)=t-2\) on \([0,4]\), with basepoint \(0\). Calculate \(A(1)\), \(A(2)\), \(A(3)\), and \(A(4)\), where \(A(x)=\int_0^x(t-2)\,dt\).
The graph is a line that crosses the horizontal axis at \(t=2\). It lies below the axis from \(0\) to \(2\), and above it from \(2\) to \(4\). We therefore keep the below-axis areas negative and the above-axis areas positive.
From \(0\) to \(1\), the graph is below the axis, with endpoint heights \(2\) and \(1\) in magnitude. The region is a trapezoid of width \(1\), so its ordinary area is \(1(2+1)/2=3/2\). It is below the axis, giving \(A(1)=-3/2\).
From \(0\) to \(2\), the region below the axis is a triangle with base \(2\) and height \(2\). Its ordinary area is \(2\cdot2/2=2\), so \(A(2)=-2\).
From \(2\) to \(3\), there is a positive triangle with base \(1\) and height \(1\), of area \(1\cdot1/2=1/2\). The accumulation from \(0\) to \(3\) includes both regions: \(A(3)=-2+1/2=-3/2\).
From \(2\) to \(4\), the positive triangle has base \(2\) and height \(2\), so its area is \(2\cdot2/2=2\). Thus \(A(4)=-2+2=0\).
Check: At \(x=4\), the negative triangle on \([0,2]\) has signed area \(-2\), and the positive triangle on \([2,4]\) has signed area \(2\). Their sum is \(-2+2=0\), as found for \(A(4)\).
Accumulation Across Several Pieces
If the graph changes shape or sign along the interval, the value at an endpoint includes the signed contributions of all the pieces between the basepoint and that endpoint. Splitting the calculation at the points where the graph changes makes the bookkeeping clear.
Worked Example: Adding Signed Rectangles
Suppose \(f\) is defined on \([1,6]\) by \(f(t)=3\) for \(1\le t<2\), \(f(t)=-2\) for \(2\le t<5\), and \(f(t)=1\) for \(5\le t\le6\). Let \(A(x)=\int_1^x f(t)\,dt\). Find \(A(2)\), \(A(4)\), \(A(5)\), and \(A(6)\).
On \([1,2]\), the graph makes a rectangle of width \(1\) and height \(3\), with signed area \(3\cdot1=3\). Therefore \(A(2)=3\).
To reach \(x=4\), include that first rectangle and the below-axis rectangle from \(2\) to \(4\). The second rectangle has width \(4-2=2\) and signed height \(-2\), so its signed area is \((-2)\cdot2=-4\). Hence \(A(4)=3+(-4)=-1\).
To reach \(x=5\), the below-axis rectangle extends from \(2\) to \(5\), with width \(3\). Its signed area is \((-2)\cdot3=-6\), so \(A(5)=3+(-6)=-3\).
Finally, the interval from \(5\) to \(6\) adds a positive rectangle of width \(1\) and height \(1\), with area \(1\cdot1=1\). Thus \(A(6)=3+(-6)+1=-2\).
Check: At \(x=6\), the three signed rectangle areas are \(3\), \(-2\cdot3=-6\), and \(1\). Their sum is \(3-6+1=-2\), confirming the result.
A Common Sign Mistake
A frequent error is to add the ordinary areas of all regions as positive numbers, even when part of the graph is below the axis. The integral records signed contributions, not just geometric sizes.
For instance, in the second example, the region from \(0\) to \(2\) has area \(2\) below the axis, and the region from \(2\) to \(3\) has area \(1/2\) above it. Treating both as positive gives the incorrect calculation \(2+1/2=5/2\) for \(A(3)\). The correction is to assign the below-axis region a negative sign: \(A(3)=-2+1/2=-3/2\).
Why the Variable Endpoint Matters
For one fixed integrand and one fixed basepoint, each endpoint gives a potentially different accumulated value. The examples show that these values can be positive, negative, or zero. A value of zero does not necessarily mean that there were no regions: in the second example, nonzero regions cancel because one is below the axis and the other is above it.
The main discipline is to identify the ingredients before calculating: the basepoint sets where accumulation starts, the chosen input sets where it ends, and the integrand determines the signed contributions in between. With those roles clear, an expression of the form \(\int_a^x f(t)\,dt\) can be read as a rule assigning a number to each allowed endpoint.
Check Your Understanding
Use the definition and signed-area method to answer these questions.
- If \(f(t)=4\) on \([0,2]\) and \(A(x)=\int_0^x f(t)\,dt\), what are \(A(0)\) and \(A(3/2)\)?
- For an accumulation function defined on \([a,b]\), which quantity is fixed and which quantity varies?
- A graph makes a region of area \(5\) below the axis before the endpoint. What signed contribution does that region make?
- A region of area \(3\) below the axis is followed by a region of area \(7\) above it. What is the accumulation value at the end of both regions, relative to the starting value \(0\)?
- What condition on \(f\) ensures the defining integral exists for every endpoint in \([a,b]\)?
