Why Exponential and Logarithmic Boundaries Need Care
Exponential and logarithmic functions are closely connected, but that does not mean their graphs always intersect or enclose a finite region. The exponential function \(y=e^x\) is defined for every real \(x\), whereas the natural logarithm \(y=\ln x\) is defined only for \(x>0\). Their relative positions and the boundaries of the region must be checked before an area integral is written.
When the curves do enclose a region, the main choices are familiar from area between curves: use vertical slices to integrate with respect to \(x\), or horizontal slices to integrate with respect to \(y\). With exponential and logarithmic graphs, a useful choice can often avoid solving for an inverse function that would make the setup harder.
Check the Intersections and the Domain
To find where two graphs meet, set their \(y\)-values equal and solve for the coordinate. For example, if the graphs are \(y=e^x\) and \(y=1+K\ln(x+1)\), solve \(e^x=1+K\ln(x+1)\), while also respecting the logarithm's domain \(x>-1\). Candidate points must be substituted into both expressions to verify that they are intersections.
Do not assume that exponential and logarithmic graphs intersect just because one is an inverse of the other. The inverse of \(y=e^x\) is \(y=\ln x\), which is its reflection across \(y=x\); that relationship does not say that the two graphs meet. In fact, \(e^x>\ln x\) for every \(x>0\). Those two graphs alone do not enclose a finite region.
A finite region may instead be bounded by a pair of intersections, by the curves together with vertical or horizontal lines, or by other specified boundaries. Once the boundaries are known, determine their order throughout the region. The correct area integrand must be nonnegative.
Worked Example: A Region Bounded by Two Intersections
Worked Example: A Shifted Logarithm and an Exponential
Find the area enclosed by \(y=e^x\) and \(y=1+K\ln(x+1)\), where \(K=\dfrac{e-1}{\ln 2}\).
First check the proposed intersections. At \(x=0\), the exponential has value \(e^0=1\), and the logarithmic graph has value \(1+K\ln 1=1\). At \(x=1\), the exponential has value \(e\), and the logarithmic graph has value \(1+K\ln 2=1+(e-1)=e\). Thus the graphs meet at \(x=0\) and \(x=1\).
To establish which graph is above between those points, let \(g(x)=1+K\ln(x+1)-e^x\). Then \(g(0)=g(1)=0\), and \(\displaystyle g''(x)=-\frac{K}{(x+1)^2}-e^x<0\) on \([0,1]\). Thus \(g'\) is strictly decreasing. For any \(0<x<1\), the Mean Value Theorem applied on \([0,x]\) and \([x,1]\) gives points \(u\in(0,x)\) and \(v\in(x,1)\) such that \(\displaystyle \frac{g(x)}{x}=g'(u)>g'(v)=\frac{-g(x)}{1-x}\). Multiplying by \(x(1-x)>0\) gives \(g(x)>0\). Therefore the logarithmic graph is above the exponential graph between the intersections.
Use vertical slices from \(x=0\) to \(x=1\):
An antiderivative is \(\displaystyle F(x)=x+K\big((x+1)\ln(x+1)-(x+1)\big)-e^x\), since differentiating gives \(1+K\ln(x+1)-e^x\). Evaluate the endpoints:
The approximation is rounded to four decimal places. It is positive, as required for a geometric area, and the antiderivative check confirms the integral's integrand.
Choosing the Slice Direction
With vertical slices, the area is found by integrating upper curve minus lower curve with respect to \(x\). With horizontal slices, it is right curve minus left curve with respect to \(y\). The boundaries and the limits must be expressed in the variable of integration.
For a graph \(y=e^x\), solving for \(x\) gives \(x=\ln y\). For \(y=1+K\ln(x+1)\), solving for \(x\) gives \(x=e^{(y-1)/K}-1\). If a horizontal-slice setup uses either of these curves, the inverse expressions and their domains must be correct. Sometimes vertical slices keep the original equations simpler; sometimes horizontal slices do.
Worked Example: Horizontal Slices Avoid Extra Work
Worked Example: Reversing the Slices for the Same Region
Use horizontal slices to find the area of the region from the previous example. The boundary heights range from \(y=1\) to \(y=e\). Solving the logarithmic equation for \(x\) gives \(x=e^{(y-1)/K}-1\); solving the exponential equation gives \(x=\ln y\).
For interior heights, the logarithmic boundary is to the left and the exponential boundary is to the right. This follows because the logarithmic graph was above the exponential graph between their intersections, and both graphs increase there. Thus the horizontal width is \(\ln y-\big(e^{(y-1)/K}-1\big)\).
An antiderivative is \(G(y)=y\ln y-y-Ke^{(y-1)/K}+y\). Its derivative is \(\ln y-e^{(y-1)/K}+1\). Since \((e-1)/K=\ln2\), we have \(e^{(e-1)/K}=2\). Therefore,
This matches the vertical-slice result. The matching values provide a useful check that the horizontal boundaries and limits were chosen correctly.
Not Every Region Comes from an Intersection
An exponential and a logarithmic graph can form part of a bounded region even when they do not meet. In that case, additional boundaries, such as vertical lines, may close the region. The endpoints then come from those specified boundaries, not from solving an intersection equation.
Worked Example: Exponential and Logarithmic Curves with Vertical Boundaries
Find the area between \(y=e^x\) and \(y=\ln x\), from \(x=1\) to \(x=e\). On this interval, \(e^x>1\) and \(\ln x\leq1\), so \(e^x\) is strictly above \(\ln x\). The vertical lines \(x=1\) and \(x=e\) close the region.
An antiderivative is \(H(x)=e^x-x\ln x+x\), because \(H'(x)=e^x-(\ln x+1)+1=e^x-\ln x\). Evaluating gives
The derivative check verifies the integrand, and the strict ordering on the interval ensures that this integral gives positive geometric area.
A Common Sign Mistake
In the first example, a tempting but incorrect setup is to integrate \(e^x-\bigl(1+K\ln(x+1)\bigr)\), treating the exponential as the top curve. That reverses the boundaries: on \(0<x<1\), we proved that the logarithmic graph is above the exponential. The reversed integral evaluates to \(K-e\approx-0.2393\), a signed area rather than the geometric area.
The correction is to use upper minus lower: \(\bigl(1+K\ln(x+1)\bigr)-e^x\). A negative result is a signal to check the boundary order, not to report a negative geometric area.
A Reliable Setup
Record the exponential and logarithmic expressions and any additional lines that close the region.
Solve equality for intersections, or use the stated vertical or horizontal boundaries. Check logarithm domains and verify candidate intersections in the original equations.
Compare the complexity of top-minus-bottom in \(x\) with right-minus-left in \(y\). Invert a graph only when needed for the chosen direction.
Confirm which boundary is above or to the right on the whole interval. If the order changes, split the integral at the change; otherwise integrate the nonnegative difference.
Differentiate the proposed antiderivative, substitute the limits carefully, and check that the final area is nonnegative.
These checks matter because an algebraically valid integral can still describe the wrong region if its limits or boundary order are incorrect. For exponential and logarithmic graphs in particular, confirming the domain and testing the intersections prevents an assumed relationship from replacing the actual geometry.
Check Your Understanding
Use the definitions and examples above to answer the following questions.
- Why do \(y=e^x\) and \(y=\ln x\), by themselves, not enclose a finite region?
- For vertical slices, which boundary value is subtracted from which?
- In the first worked example, how does the second derivative help establish which curve is above between the intersections?
- For the horizontal slices in the second example, what are the left and right \(x\)-boundaries as functions of \(y\)?
- If an area integral evaluates to a negative number, what should you check before interpreting the result?
