Reading a Slope from a Rate Graph
An accumulation function can describe how much a quantity has changed since a chosen starting point. Its graph may be difficult to construct from scratch, but a key feature is often available immediately: the slope of the accumulation function at a point is the value of its integrand at that point.
This gives a direct way to answer questions about an accumulation graph using only the graph or table of the integrand. You do not need to calculate the accumulated amount first. The essential task is to read the integrand’s value at the specified input accurately, including its sign.
What the Slope Tells You
The equation \(A'(c)=f(c)\) is a point-by-point statement. To find the slope of \(A\) at \(c\), read the height of the graph of \(f\) at \(x=c\), or look up \(f(c)\) in a table. The sign and size of that value carry important information:
- If \(f(c)>0\), then \(A\) has a positive slope at \(c\).
- If \(f(c)<0\), then \(A\) has a negative slope at \(c\).
- If \(f(c)=0\), then \(A\) has a horizontal tangent at \(c\).
- A larger value of \(|f(c)|\) means a steeper slope for \(A\) at that point.
A horizontal tangent means the instantaneous slope is zero; by itself, it does not tell you whether the accumulation function has a local maximum, a local minimum, or neither. To decide that, you would need additional information about slopes near the point. Here, the goal is to find the slope at the specified point, not to classify the point beyond what that slope establishes.
Why the Integrand Gives the Slope
The connection follows from the definition of a derivative. Let \(A(x)=\int_a^x f(t)\,dt\), with \(f\) continuous near \(c\). For a nonzero change \(h\), the difference quotient for \(A\) at \(c\) is
The quotient is the average value of \(f\) over the short interval from \(c\) to \(c+h\). To see why it approaches \(f(c)\), subtract \(f(c)\) from both sides of the average:
Its absolute value is at most the greatest difference between \(f(t)\) and \(f(c)\) on the interval. More precisely,
As \(h\) approaches zero, every \(t\) between \(c\) and \(c+h\) approaches \(c\). Continuity of \(f\) at \(c\) means that the maximum difference on the right approaches zero. Therefore, the difference quotient approaches \(f(c)\), so \(A'(c)=f(c)\). This argument also works when \(h<0\), using the signed-integral convention.
Worked Examples: Reading Slopes from Graphs and Tables
Worked Example: A Piecewise-Linear Integrand Graph
Suppose the graph of a continuous integrand \(f\) consists of straight segments joining the points \((-2,-1)\), \((0,3)\), and \((2,1)\). Let \(A(x)=\int_{-2}^{x}f(t)\,dt\). Find the slopes of \(A\) at \(x=-1\), \(x=0\), and \(x=1\).
First, use \(A'(x)=f(x)\). On the segment from \((-2,-1)\) to \((0,3)\), the change in \(f\) is \(3-(-1)=4\), while the change in \(x\) is \(0-(-2)=2\). Thus the segment’s slope is \(4/2=2\). Moving one unit right from \(x=-2\) to \(x=-1\), the graph height rises by \(2\), so \(f(-1)=-1+2=1\).
At \(x=0\), the graph passes through \((0,3)\), so \(f(0)=3\). On the segment from \((0,3)\) to \((2,1)\), the change in \(f\) is \(1-3=-2\), and the change in \(x\) is \(2-0=2\). Its slope is \(-2/2=-1\). Moving one unit right from \(x=0\) to \(x=1\), the graph height changes by \(-1\), so \(f(1)=3-1=2\).
Therefore, the accumulation-function slopes are
Check each result against the graph: the integrand’s heights at \(-1\), \(0\), and \(1\) are respectively \(1\), \(3\), and \(2\). Each height matches the corresponding slope. The base point \(-2\) does not change this slope rule.
Worked Example: Using a Table of Integrand Values
Suppose \(f\) is continuous and the table gives values of \(f\). If \(A(x)=\int_0^x f(t)\,dt\), find the slope of \(A\) at \(x=2\) and \(x=4\).
| \(x\) | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| \(f(x)\) | 5 | -3 | 0 | 2 |
Read the requested entries directly from the row for \(f(x)\). The table gives \(f(2)=-3\) and \(f(4)=2\). Since \(A'(x)=f(x)\),
At \(x=2\), \(A\) has a negative slope; at \(x=4\), it has a positive slope. Check by comparing each slope with the table entry in the same column: \(-3\) matches \(f(2)\), and \(2\) matches \(f(4)\). No interpolation is needed because the table supplies the values at the exact requested inputs.
Worked Example: Slope, Sign, and Units
A tank is being filled or drained. Let \(f(t)\) be the water’s net inflow rate, in liters per minute, and let \(A(t)=\int_0^t f(s)\,ds\) be the change in the amount of water, in liters, since \(t=0\). A graph of \(f\) shows that \(f(3)=-4\) liters per minute. Find the slope of \(A\) at \(t=3\) and interpret it.
Because \(A'(t)=f(t)\), substitute the integrand’s value at the requested time:
The negative slope means that at \(t=3\) minutes, the amount of water is decreasing at an instantaneous rate of \(4\) liters per minute. The units are consistent: the slope of an amount measured in liters with respect to time measured in minutes has units of liters per minute.
Check the result against the graph: at \(t=3\), the rate graph is \(4\) units below the horizontal axis, so its value is \(-4\), not \(4\). That signed value is the slope of \(A\).
Common Mistake: Confusing a Slope with an Accumulated Value
The statement \(A'(c)=f(c)\) does not say that \(A(c)=f(c)\). The integrand value gives the slope of the accumulation function at \(c\); the value \(A(c)\) represents the accumulated change from the base point to \(c\). Those are different quantities.
For example, suppose a table gives \(f(2)=-3\). A common incorrect calculation is: “Therefore \(A(2)=-3\).” This replaces the derivative with the function value. The correct conclusion is
This tells us the slope of \(A\) at \(2\), not the value \(A(2)\). The accumulated value would require information about \(f\) over the interval from the base point to \(2\); one value from a table is not enough to determine it. Always check what the question asks for: if it asks for a slope, report \(A'(c)\); if it asks for an accumulated value, the slope rule alone is not sufficient.
A Reliable Reading Process
When a question asks for accumulation slopes from an integrand graph or table, use this sequence:
Locate the specified value \(c\) on the horizontal axis or in the table’s input row.
Find the height \(f(c)\) on the graph, or the entry in the table’s \(f(x)\) row under \(c\). Keep the sign.
State \(A'(c)=f(c)\). That value is the requested slope.
A positive value means positive slope, a negative value means negative slope, and zero means a horizontal tangent. Include units when the quantities have units.
This process is useful because graphs and tables often make slopes accessible even when the full formula for an accumulation function is unavailable. It also keeps the interpretation grounded in the quantity that directly controls the slope: the integrand’s value at the point, rather than the total area accumulated before reaching that point.
Check Your Understanding
Assume throughout that \(f\) is continuous and \(A(x)=\int_a^x f(t)\,dt\). Use the relationship between the integrand and the accumulation-function slope.
- If a graph shows \(f(5)=7\), what is \(A'(5)\), and what is the sign of the slope?
- A table gives \(f(-2)=0\). What does this tell you about the tangent to \(A\) at \(x=-2\)?
- If \(f(4)=-6\) units per second, state \(A'(4)\) with units and interpret its sign.
- A student says that \(f(3)=2\) implies \(A(3)=2\). Identify the error and write the conclusion that does follow.
- At a point where the integrand graph is above the horizontal axis, must the accumulation function have a positive slope? Explain using \(A'(x)=f(x)\).
