From Vertical Distance to Area
When two curves bound a region, the area depends on the vertical distance between them at each \(x\)-value. If one curve stays above the other across the entire interval, that distance has a consistent expression: the upper function minus the lower function.
The main task is therefore to identify the order of the curves correctly and use that order throughout the integral. The earlier material on ordering curves with test points helps establish which function is above. Here, we use that information to calculate the area.
The Area Formula
Let \(f\) and \(g\) be continuous on the closed interval \([a,b]\), where \(a<b\). Suppose that \(f(x)\ge g(x)\) for every \(x\) in the interval. The curve \(y=f(x)\) is then an upper boundary, and \(y=g(x)\) is a lower boundary. Their vertical separation at \(x\) is \(f(x)-g(x)\).
The condition \(f(x)\ge g(x)\) matters: it makes the integrand nonnegative, so the integral accumulates geometric area rather than cancelling positive and negative contributions. If the functions have the same value at some points but the order otherwise remains the same, the formula still applies. At such points, the vertical gap is zero.
If the functions describe physical quantities, their units should also be considered. For example, if \(x\) is measured in metres and \(f(x)\) and \(g(x)\) in metres, the area has units of square metres. More generally, the units of an area integral are the units of the vertical difference multiplied by the units of \(x\).
Why Upper Minus Lower Gives the Area
A vertical slice of the region at a particular \(x\)-value extends from \(y=g(x)\) to \(y=f(x)\). Its height is therefore
Over a short interval of width \(\Delta x\), the slice is approximately a thin rectangle: its area is approximately its height times its width. Adding these approximate slice areas gives a Riemann sum. As the widths of the slices approach zero, continuity ensures that the sum approaches the definite integral.
Because every slice height is nonnegative, these contributions add to the region's geometric area. This explains both why the integrand is upper minus lower and why the order condition is essential.
A Reliable Setup and Evaluation
For a problem that states or establishes that one curve stays above the other, use the following sequence. The bounds \(a\) and \(b\) specify the horizontal interval; the integrand records the vertical distance within that interval.
Use the given horizontal limits \(a\) and \(b\) as the bounds of integration.
Use the stated order or a valid comparison on the interval to decide which function has the greater \(y\)-value.
Subtract the lower function from the upper function, then integrate from \(a\) to \(b\).
Use an antiderivative, substitute both bounds, and confirm that the result is nonnegative.
Worked Examples
Worked Example: A Line Above a Parabola
Find the area between \(y=x+2\) and \(y=x^2\) on \([0,1]\), given that \(x+2\) is the upper function there.
The upper function is \(f(x)=x+2\), the lower function is \(g(x)=x^2\), and the bounds are \(0\) and \(1\). Thus,
An antiderivative of \(x+2-x^2\) is \(\frac{x^2}{2}+2x-\frac{x^3}{3}\), since differentiating it gives \(x+2-x^2\). Evaluate at the bounds:
The answer is positive, as expected for area. Differentiating the antiderivative confirms the integrand, \(x+2-x^2\), so the evaluation uses the correct function difference.
Worked Example: A Line Above a Parabola on a Symmetric Interval
Find the area between \(y=4-x\) and \(y=x^2\) on \([-1,1]\). First confirm the stated order: for \(-1\le x\le1\), \(4-x\ge3\), while \(x^2\le1\). Therefore \(4-x\) is above \(x^2\) throughout the interval.
Set up the area using upper minus lower:
An antiderivative is \(4x-\frac{x^2}{2}-\frac{x^3}{3}\), because its derivative is \(4-x-x^2\). Substituting the upper bound gives
Substituting the lower bound gives
Subtract the lower-bound value from the upper-bound value:
The result is positive. As a check, differentiating \(4x-\frac{x^2}{2}-\frac{x^3}{3}\) gives \(4-x-x^2\), the vertical gap used in the integral.
Worked Example: A Trigonometric Curve Above a Horizontal Line
Find the area between \(y=2+\sin x\) and \(y=1\) on \([0,\pi]\). Since \(\sin x\ge0\) on this interval, \(2+\sin x\ge2>1\). The trigonometric curve is therefore always above the horizontal line.
The upper-minus-lower difference is \((2+\sin x)-1=1+\sin x\), so
An antiderivative is \(x-\cos x\), because \(\frac{d}{dx}(x-\cos x)=1+\sin x\). Evaluate:
This is positive. Differentiation returns \(1+\sin x\), confirming the antiderivative and the integrand.
A Common Mistake: Reversing the Difference
The order of subtraction is not interchangeable. For the first example, a common incorrect setup is to subtract the upper curve from the lower curve:
An antiderivative is \(\frac{x^3}{3}-\frac{x^2}{2}-2x\). Evaluating gives
The negative answer is not the geometric area. It is the signed integral of lower minus upper. Correct the setup by subtracting the lower function from the upper function:
A useful check is to examine the integrand before evaluating. If the order is correct throughout the interval, upper minus lower should be nonnegative there, and the resulting area cannot be negative.
What to Keep in Mind
The formula is concise, but it relies on two decisions: using the correct interval and using the correct vertical order. The bounds describe where the region begins and ends horizontally; the function difference describes its slice height. Neither should be chosen from the antiderivative or from the appearance of an answer.
Once the order is established, the calculation follows a consistent pattern:
This method turns a region into an accumulation of vertical distances. It is especially useful because the curves need not be lines or simple geometric shapes: as long as the functions are continuous and their order is fixed on the interval, the same setup applies.
Check Your Understanding
For each question, identify the upper and lower functions before setting up the integral.
- On \([0,2]\), \(y=5\) is above \(y=x+1\). Write the area integral. Do not evaluate it.
- On \([0,1]\), \(y=3-x^2\) is above \(y=1\). Write and evaluate the area integral.
- For an area integral, what does a negative result suggest about the order of subtraction?
- Why does continuity help justify adding the vertical slice areas with a definite integral?
- If the functions meet at an endpoint but one remains above the other throughout the interval, can the upper-minus-lower formula still be used? Explain briefly.
