Exponential and Logarithmic Curves
The exponential function \(y=e^x\) and the natural logarithm \(y=\ln x\) are inverse functions. Their graphs are reflections of one another across the line \(y=x\). That relationship is useful when rewriting a boundary in terms of \(y\), but it does not mean that the two graphs intersect or that a region involving them is symmetric.
To find area, identify the actual region first. Check the domain, determine which curve is above or to the right on each part of the region, and use the corresponding slice length. Vertical slices measure upper minus lower; horizontal slices measure right minus left.
Do the Two Graphs Intersect?
Before looking for an enclosed area between two curves, check whether they meet. The logarithm is defined only for \(x>0\). On that domain, \(e^x\) is always greater than \(\ln x\), so the graphs \(y=e^x\) and \(y=\ln x\) do not intersect.
One way to see this is to use the inequalities \(e^x\ge x+1\) and \(\ln x<x\) for \(x>0\). The first follows because \(e^x-x-1\) has its minimum value \(0\) at \(x=0\); the second follows because \(x-1-\ln x\) has its minimum value \(0\) at \(x=1\). Therefore,
Thus, the two curves alone do not enclose a bounded region. To define a finite area, some additional boundary is needed, such as vertical lines \(x=a\) and \(x=b\). Once the region is specified, test curve order over its interval rather than assuming that inverse functions must cross.
Worked Example: Establishing the Curve Order
Compare \(y=e^x\) and \(y=\ln x\) on \(1\le x\le e\). The domain condition for \(\ln x\) is satisfied because every \(x\) in this interval is positive. From the inequalities above, \(e^x>\ln x\) throughout the interval. In particular, at the test point \(x=1\),
The test point confirms the order, and the inequality establishes that the order does not change anywhere in the interval. The exponential curve is above the logarithmic curve, so the vertical slice length is \(e^x-\ln x\).
Area with Vertical Slices
Consider the region \(R\) bounded by \(y=e^x\), \(y=\ln x\), and the vertical lines \(x=1\) and \(x=e\). A vertical slice stays between the same two curves as \(x\) runs from \(1\) to \(e\). Since \(e^x\) is above \(\ln x\), the area is the integral of upper minus lower.
Worked Example: Finding the Area Using \(dx\)
The area is
An antiderivative of \(\ln x\), for \(x>0\), is \(x\ln x-x\), since differentiating gives \(\ln x+1-1=\ln x\). Therefore,
The result is positive, as an area must be. The subtraction in the integral also has the correct order at every \(x\): upper curve \(e^x\) minus lower curve \(\ln x\).
Rewriting the Same Region for Horizontal Slices
A horizontal slice requires describing the region's left and right edges as functions of \(y\). The original restrictions are \(1\le x\le e\), \(\ln x\le y\), and \(y\le e^x\). For \(y>0\), monotonicity of the exponential and logarithm gives the equivalent inequalities
Combining these with \(1\le x\le e\), a horizontal slice has left endpoint \(\max(1,\ln y)\) and right endpoint \(\min(e,e^y)\). The region reaches from \(y=0\), where its only point is \((1,0)\), up to \(y=e^e\), the value of the upper curve at \(x=e\). The left and right boundaries change at \(y=1\) and \(y=e\), so the area integral must be split there.
Here \(\ln y\le0\) when \(y>0\), so the left edge is \(x=1\). Also \(1\le e^y\le e\), so the right edge is \(x=e^y\).
Here \(\ln y\le1\), so the left edge remains \(x=1\). Since \(e^y\ge e\), the right edge is \(x=e\).
Now \(\ln y\ge1\), so the left edge is \(x=\ln y\). The right edge remains \(x=e\).
Worked Example: Finding the Same Area Using \(dy\)
Using the three horizontal ranges, the slice lengths are \(e^y-1\), \(e-1\), and \(e-\ln y\), respectively. Thus,
Evaluate each part. The first is
The second is
For the final integral, an antiderivative of \(e-\ln y\) is \((e+1)y-y\ln y\), because its derivative is \(e+1-(\ln y+1)=e-\ln y\). Therefore,
Adding the three contributions gives
This matches the vertical-slice result. Both integrals describe the same region \(R\), so their equality is a check on the bounds and calculations. The horizontal description is not a claim that \(R\) is symmetric about \(y=x\). Its reflection across \(y=x\) would be a different region; the two regions are reflections of one another and consequently have equal areas.
Why the Slice Direction Matters
A vertical slice of \(R\) always runs from the logarithmic curve to the exponential curve. Its boundaries are simple, giving one integral. A horizontal slice encounters three different boundary patterns, so it requires three integrals. The horizontal method is valid, but here it takes more bookkeeping.
This is a useful general principle: choose the direction that makes slice lengths and bounds simplest. A change from \(dx\) to \(dy\) does not automatically preserve the same numerical limits or the same boundary expressions. The limits must describe the full range of the chosen slicing variable, and each slice must stay inside the intended region.
A Common Mistake: Reversing the Difference
Because \(e^x\) and \(\ln x\) are both familiar functions, it can be tempting to subtract them in either order. But area integrands must be nonnegative. Reversing the curves in the vertical setup produces
which is negative. That is not an area; it signals that the lower curve was subtracted from the upper curve in the wrong order. On \(1\le x\le e\), \(e^x>\ln x\), so the correction is
Another error is to assume that horizontal limits are simply \(y=1\) and \(y=e\) because the vertical boundaries are \(x=1\) and \(x=e\). The actual vertical range of \(R\) is \(0\le y\le e^e\). Horizontal limits come from the lowest and highest points of the entire region, not just from the vertical boundary values.
Check Your Understanding
Use the region and reasoning in this tutorial to check whether you can identify the curves, bounds, and slice lengths.
- Why do \(y=e^x\) and \(y=\ln x\) fail to intersect for \(x>0\)?
- For the region bounded by these curves and \(x=1,\ x=e\), which curve is above the other?
- Write the area integral for that region using vertical slices.
- For \(1\le y\le e\), what are the left and right horizontal boundaries of the region?
- Why must the horizontal-slice setup be split at \(y=1\) and \(y=e\)?
- What does a negative value from an area integral usually indicate about the integrand's order?
