Why Check an Area Integral?
A definite integral can be evaluated correctly and still give the wrong answer to an area problem. The error may be in the bounds, in the order of the boundary curves, or in a failure to split the interval where the curves exchange order. Checking the sign of the proposed integrand is a quick way to detect some of these problems before doing the full calculation.
For a region between two graphs, a vertical slice has height equal to the upper curve minus the lower curve. If the graphs are \(y=f(x)\) and \(y=g(x)\), then the slice height is \(f(x)-g(x)\) when \(f\) is above \(g\). The area on an interval \([a,b]\) is found by integrating those nonnegative slice heights, provided the boundaries and bounds correctly describe the region.
Three Checks Before Evaluating
A proposed setup should be checked in a deliberate order. First determine the horizontal extent of the region. Next identify the vertical boundaries on each interval. Then test the sign of the resulting difference. Finding intersection points helps with all three checks: they can mark the region's endpoints and any locations where the curves exchange order.
Find the relevant left and right endpoints of the region. Make sure the interval includes the entire region, not just part of it, and that the bounds run from left to right.
On each interval, compare the candidate upper and lower curves. Use a test value strictly inside the interval, rather than relying on a sketch alone.
Subtract lower from upper. The difference should be nonnegative throughout the interval. If its sign changes, split the interval where the order changes and write a separate area integral on each piece.
Testing a point is especially informative when the curves are continuous. If their difference is nonzero throughout an interval, its sign cannot change there: a change from positive to negative would require the difference to equal zero somewhere in between. Thus, when all intersections in an interval have been accounted for, one interior test point determines the order throughout that interval.
Worked Example: Bounds That Miss Part of the Region
Worked Example: Bounds That Miss Part of the Region
Consider the region bounded by \(y=4-x^2\) and the \(x\)-axis. A proposed setup uses bounds from \(0\) to \(2\). Check whether those bounds capture the entire region, and find the correct area.
First find where the graph meets the \(x\)-axis by setting \(4-x^2=0\):
The region extends from \(x=-2\) to \(x=2\), so bounds \(0\) to \(2\) include only its right-hand portion. To check which curve is above the axis, use the interior test point \(x=0\): \(4-0^2=4>0\). The curve is above the \(x\)-axis between the two intersections. Therefore, the full area setup is:
An antiderivative is \(4x-\frac{x^3}{3}\), since its derivative is \(4-x^2\). Evaluating at the bounds gives:
Check: the integrand is nonnegative throughout \([-2,2]\), since \(4-x^2\geq 0\) there, and the answer is positive. The proposed bounds \(0\) to \(2\) would give \(16/3\), only half the full area. Indeed, \(\int_0^2(4-x^2)\,dx=16/3\); the omitted interval \([-2,0]\) contributes another \(16/3\).
What a Negative Integrand Tells You
Suppose a proposed vertical-slice setup is \(\int_a^b(g(x)-f(x))\,dx\). At a test point \(c\) inside the interval, if \(g(c)-f(c)<0\), then \(g(c)<f(c)\): the curve labelled as the lower boundary is actually above the one labelled as the upper boundary at that point. The expression has the wrong order there.
A negative value at one point is not just a minor issue to overlook. If the integrand is continuous, it is negative on some interval around that point, so the integral of the proposed difference cannot represent nonnegative slice heights throughout. Correct the order or split the interval if the curves exchange order.
Worked Example: Reversing the Boundary Order
For the curves \(y=x^2\) and \(y=2x\), a proposed area integral is \(\int_0^2(x^2-2x)\,dx\). Check the order and correct the setup for the region between the curves on \([0,2]\).
The curves intersect where \(x^2=2x\), so \(x(x-2)=0\) and the endpoints are \(0\) and \(2\). Test \(x=1\): the values are \(1^2=1\) and \(2(1)=2\). Thus \(2x\) is above \(x^2\) at this point. The proposed integrand has the wrong order, as its value at \(1\) is \(1-2=-1\).
Use upper minus lower to obtain the corrected setup:
An antiderivative is \(x^2-\frac{x^3}{3}\), whose derivative is \(2x-x^2\). Therefore:
Check the sign across the interval: \(2x-x^2=x(2-x)\geq 0\) for \(0\leq x\leq 2\), as required for slice heights. The original integral is \(-4/3\), the negative of the corrected area. Differentiating the antiderivative confirms the integrand, and the positive result agrees with the sign check.
When One Test Point Is Not Enough
A single test point determines the boundary order only on an interval where the curves do not intersect. If the curves meet inside the proposed bounds, check whether their order changes there. When it does, the area must be written as a sum of integrals with the boundaries in the correct order on each subinterval. Adding the integrals preserves the area; using one signed difference across both pieces may cause positive and negative contributions to cancel.
Worked Example: An Order Change Hidden by Cancellation
For \(y=x^3\) and \(y=x\) on \([-1,1]\), a proposed setup is \(\int_{-1}^{1}(x-x^3)\,dx\). Check whether the curves stay in the same order, then find the area.
Intersections satisfy \(x^3=x\), or \(x(x-1)(x+1)=0\), so they occur at \(-1,0,1\). The point \(x=0\) divides the interval. On \((-1,0)\), test \(x=-\frac12\): \(x^3=-\frac18\) and \(x=-\frac12\), so \(x^3>x\). On \((0,1)\), test \(x=\frac12\): \(x=\frac12\) and \(x^3=\frac18\), so \(x>x^3\). The curves exchange order at \(0\).
Thus the area requires two integrals, each with upper minus lower:
An antiderivative of \(x^3-x\) is \(\frac{x^4}{4}-\frac{x^2}{2}\); its derivative is \(x^3-x\). The first contribution is:
An antiderivative of \(x-x^3\) is \(\frac{x^2}{2}-\frac{x^4}{4}\); its derivative is \(x-x^3\). The second contribution is:
Therefore \(A=\frac14+\frac14=\frac12\). Each piece has a nonnegative integrand in its own interval. By contrast, the proposed single integral equals zero: \(x-x^3\) is negative on \((-1,0)\) and positive on \((0,1)\), so its signed contributions cancel. Zero is not the area of this nonzero region.
Why the Sign Check Works
The area of a region built from vertical slices is the accumulation of slice heights times slice widths. A slice height is a length, so it cannot be negative. If \(f(x)\geq g(x)\) throughout \([a,b]\), then every height \(f(x)-g(x)\) is nonnegative, and its definite integral gives the area between the curves on that interval.
This also explains why bounds matter. Integrating correctly ordered slice heights over too small an interval leaves out part of the region; integrating outside the region may include extra contributions. Bounds and signs check different parts of a setup, so both must be verified.
A Practical Audit
Before evaluating a proposed area integral, ask:
- Do the bounds match the full left-to-right extent of the intended region?
- Have all interior intersections that could change boundary order been identified?
- Does a test point in each interval confirm which curve is above?
- Is the integrand upper minus lower, and is it nonnegative throughout each interval?
- If the integral is negative or zero unexpectedly, could reversed order or cancellation be the cause?
A negative result does not mean that geometric area itself is negative. It means the signed integral, as written, does not correctly represent the desired area. Recheck the bounds and boundary order before changing the calculation.
Check Your Understanding
For each question, focus first on the setup and sign rather than on rapid integration.
- The curves \(y=3\) and \(y=x^2\) intersect at \(x=-\sqrt{3}\) and \(x=\sqrt{3}\). Which curve is upper at \(x=0\), and what is the correct area integrand on this interval?
- A proposed integral for an area is negative. Name two setup problems that could explain the sign.
- Why should the interval be split at an interior intersection when the curves exchange order there?
- For \(y=x\) and \(y=x^2\) on \([0,1]\), which curve is upper at \(x=\frac12\), and what order should the integrand use?
- If an interval's bounds omit part of the region but the integrand is nonnegative everywhere on those bounds, is the setup correct? Explain.
