Reading Accumulation from a Graph
An accumulation function records the net effect of an integrand over an interval. When the integrand is given as a graph, its value can often be found without first finding an algebraic formula for the integrand. Instead, read the regions between the graph and the horizontal axis, calculate their areas, and combine those areas with the correct signs.
The key word is net. A region above the horizontal axis contributes positively; a region below it contributes negatively. If the accumulation starts from a nonzero initial value, add the net signed area to that starting value.
In a graph-reading problem, the variable \(t\) is the input to the integrand, and \(x\) is the endpoint where the accumulated value is being evaluated. The integral records the signed area between the graph of \(f\) and the horizontal axis, not the total geometric area regardless of whether regions lie above or below the axis.
Signed Areas and the Direction of the Bounds
Suppose first that \(a\le x\). Over the interval from \(a\) to \(x\), each part of the graph above the horizontal axis contributes its geometric area, and each part below the axis contributes the negative of its geometric area. Thus the integral is the area above the axis minus the area below it, with all regions restricted to the interval from \(a\) to \(x\).
If \(x<a\), the bounds run in the opposite direction. Reverse the bounds to describe the ordinary left-to-right areas, and then apply a minus sign to the entire result. The area above the axis on \([x,a]\) is not, by itself, the value of \(\int_a^x f(t)\,dt\).
These formulas keep two different signs clear. The graph’s position determines whether a region contributes positively or negatively. The direction of the bounds determines whether the integral is taken from left to right or in reverse.
Why Signed Area Gives the Integral
The connection follows from the definition of a definite integral as a limit of sums. Divide an interval from \(a\) to \(x\), where \(a\le x\), into small subintervals. Over one subinterval of width \(\Delta t\), the graph’s contribution is approximated by a rectangle of signed height \(f(t_i)\) and width \(\Delta t\). Its signed area is \(f(t_i)\Delta t\).
Rectangles above the axis have positive height and positive signed area. Rectangles below the axis have negative height and negative signed area. Adding the rectangle contributions gives a Riemann sum; as the subdivisions become finer, that sum approaches the definite integral. This is why adding areas with signs gives the integral, while adding all geometric areas as positive numbers generally does not.
For reversed bounds, reversing the order of the endpoints changes the sign of the integral. The geometric regions themselves do not change; the total signed area for the left-to-right interval is multiplied by \(-1\).
Calculating Common Graph Regions
Before combining contributions, identify the shape of each region and use a formula that matches it. For a triangle, multiply its base length by its height and divide by \(2\). For a rectangle, multiply its width by its height. These heights are distances from the horizontal axis, so use a positive height when calculating geometric area; apply the sign separately according to the graph’s position.
A curved region can also have an exact area if its graph forms a familiar geometric shape, such as part of a circle. If the shape is not recognizable and no formula or graph information gives its exact area, do not treat a visual estimate as an exact value. Use only the information supplied by the problem.
Worked Examples
Worked Example: Combining Triangles and a Rectangle
Let \(A(x)=2+\int_0^x f(t)\,dt\). From the graph, suppose \(f\) forms an above-axis triangle on \([0,2]\) with base \(2\) and height \(3\), an above-axis rectangle on \([2,4]\) with width \(2\) and height \(3\), and a below-axis triangle on \([4,5]\) with base \(1\) and height \(2\). Find \(A(5)\).
First calculate the geometric area of each region. The first triangle has area
It is above the axis, so its signed contribution is \(+3\). The rectangle has area
It is also above the axis, so its signed contribution is \(+6\). The last triangle has geometric area
It lies below the axis, so its signed contribution is \(-1\). Therefore, the integral from \(0\) to \(5\) is
Add the initial value \(2\):
Check: The signed area is \(8\), and the initial value is \(2\), so their sum is \(10\). The below-axis region reduces the accumulation by \(1\); treating it as positive would incorrectly give \(12\).
Worked Example: Evaluating an Accumulation with Reversed Bounds
Let \(B(x)=7+\int_5^x f(t)\,dt\). On the interval \([1,5]\), suppose the graph has an above-axis rectangle from \(1\) to \(3\), of width \(2\) and height \(2\), followed by a below-axis triangle from \(3\) to \(5\), of base \(2\) and height \(2\). Find \(B(1)\).
The graph information describes the interval from \(1\) to \(5\), but the integral in \(B(1)\) runs from \(5\) to \(1\). First compute the signed area in the left-to-right direction. The rectangle contributes
The triangle’s geometric area is
It is below the axis, so its signed contribution is \(-2\). Consequently,
Reverse the bounds to obtain the integral that appears in \(B(1)\):
Now add the initial value:
Check: Moving from \(5\) back to \(1\) negates the left-to-right signed total of \(2\), giving \(-2\). Adding that change to \(7\) gives \(5\). The area above the axis does not make the reversed integral positive.
Worked Example: A Semicircle and a Rectangle
Let \(D(x)=1+\int_0^x f(t)\,dt\). Suppose the graph of \(f\) from \(t=0\) to \(t=4\) is the upper half of a circle of radius \(2\), with its diameter on the horizontal axis. From \(t=4\) to \(t=6\), suppose the graph is a horizontal segment at height \(-\frac{3}{2}\). Find \(D(6)\).
The curved region is an upper semicircle, so its geometric area is half the area of a circle of radius \(2\):
It lies above the axis, so its signed contribution is \(+2\pi\). The next region is a rectangle of width \(6-4=2\) and height \(\frac{3}{2}\) in magnitude. Its geometric area is
Because that rectangle lies below the axis, its signed contribution is \(-3\). Therefore, the total integral is
Add the initial value \(1\):
Check: The curved region increases the accumulation by \(2\pi\), and the rectangle decreases it by \(3\). Starting at \(1\) gives \(1+2\pi-3=2\pi-2\), as calculated.
A Common Sign Mistake
A frequent error is to add every geometric area as a positive number. For example, in the first worked example, this incorrect calculation would give
This treats the below-axis triangle as an increase of \(1\), even though its signed contribution is \(-1\). The correction is to separate geometric area from signed contribution: the triangle’s geometric area is \(1\), but the integral contribution is \(-1\). Thus the correct calculation is
A second sign error can occur with reversed bounds. If the integrand is above the axis on \([x,a]\), it is still incorrect to claim that \(\int_a^x f(t)\,dt\) is positive. Reversing the bounds reverses the sign, regardless of where the graph lies.
Units and Reasonableness Checks
If the integrand represents a rate, area combines the rate’s units with the horizontal variable’s units. For example, a rate in liters per minute multiplied by a time width in minutes gives liters. An accumulation value should have the same units as its initial value, so adding an area measured in liters to an initial quantity measured in liters is consistent.
After calculating, check the direction of each contribution. Above-axis regions increase the accumulation in a left-to-right integral; below-axis regions decrease it. If the bounds are reversed, negate the whole left-to-right signed total. These checks do not replace the calculation, but they often reveal a sign or endpoint error.
Check Your Understanding
Use signed area and the direction of the bounds carefully in each question.
- A graph is above the axis and forms a triangle with base \(4\) and height \(3\) on \([0,4]\). What is its contribution to \(\int_0^4 f(t)\,dt\)?
- A rectangle below the axis has width \(5\) and height \(2\) in magnitude. What is its signed contribution when the bounds run from left to right across it?
- On \([1,4]\), the signed area is \(-6\). What is \(\int_4^1 f(t)\,dt\)?
- If \(A(x)=3+\int_0^x f(t)\,dt\) and the signed area from \(0\) to \(x\) is \(-2\), what is \(A(x)\)?
- Why is the total geometric area not always equal to the definite integral?
