Why a Crossing Changes the Integral
When two curves cross between the endpoints of an interval, neither curve is necessarily above the other throughout the interval. A single integral of one function minus the other can then count some regions as negative, even though geometric area is nonnegative. The remedy is to find where the curves meet, check their order between consecutive intersections, and split the integral wherever the upper curve changes.
This is a setup skill as much as an integration skill. The bounds and the order of subtraction must match the geometry on each piece. A correct antiderivative cannot repair an integral whose signs or intervals were chosen incorrectly.
Intersections and Curve Order
An intersection occurs at an \(x\)-value where the functions have the same output: \(f(x)=g(x)\). To find possible division points, solve this equation and keep the solutions that lie within the interval of interest. Include the interval endpoints as bounds as well.
Knowing the intersection points alone is not enough. On each interval between consecutive division points, determine which function is larger. One reliable method is to choose a test value strictly inside that interval and compare the two function values. Equivalently, examine the sign of the difference \(f(x)-g(x)\): a positive value means \(f\) is above \(g\), and a negative value means \(g\) is above \(f\).
For continuous functions, if the difference changes from positive to negative, it must equal zero somewhere between those test values. Thus, once all equality points in the interval have been found, the sign cannot switch inside one of the subintervals without another intersection. This is why checking one test value per subinterval is sufficient.
Solve \(f(x)=g(x)\), and retain the solutions in the integration interval.
Place the interval endpoints and interior intersections in increasing order.
Use a test value to identify the upper and lower functions on that piece.
Integrate upper minus lower on each piece, then add the nonnegative areas.
Worked Example: Two Curves Switch Order Once
Worked Example: Two Curves Switch Order Once
Find the area between \(y=2x\) and \(y=x^2\) on \(0\le x\le 3\).
Step 1: Find the intersections. Set the functions equal:
Both values are in the interval. The division points are \(0,2,3\).
Step 2: Check the curve order. On \((0,2)\), test \(x=1\): \(2(1)=2\) and \(1^2=1\), so \(2x\) is above \(x^2\). On \((2,3)\), test \(x=2.5\): \(2x=5\) and \(x^2=6.25\), so \(x^2\) is above \(2x\).
Step 3: Set up and evaluate the area.
The area is \(\frac{8}{3}\) square units. As a check, both integrands are nonnegative on their stated intervals: the first because \(2x\ge x^2\) on \([0,2]\), and the second because \(x^2\ge 2x\) on \([2,3]\). The sum therefore represents geometric area rather than cancellation.
Why Splitting Works
The area formula follows from adding the areas of separate regions. On any interval where \(f(x)\ge g(x)\), the vertical distance between the graphs at \(x\) is \(f(x)-g(x)\). Adding these distances over the interval by integration gives that region’s area. If \(g(x)\ge f(x)\), the vertical distance is instead \(g(x)-f(x)\). The absolute value in \(|f(x)-g(x)|\) expresses both cases at once.
Here is the justification for replacing the absolute-value integral by a sum. Suppose the equality points inside \([a,b]\), in order, are \(c_1,\ldots,c_n\), and set \(c_0=a\), \(c_{n+1}=b\). On each open interval \((c_i,c_{i+1})\), the difference \(f-g\) has a constant sign if all its zeros have been included. If \(f-g\ge 0\) there, then \(|f-g|=f-g\); if \(f-g\le 0\), then \(|f-g|=g-f\). Integrating these equalities over each piece and using additivity of definite integrals gives the area as a sum of upper-minus-lower integrals.
This reasoning also explains an important detail: an intersection is a candidate division point, but it does not always mark a change in which function is higher. If the order remains the same on both sides, the integral does not need to be split there. Splitting there is still valid, provided the same upper-minus-lower expression is used on both pieces.
Worked Example: Several Intersections and a Sign Check
Worked Example: Several Intersections and a Sign Check
Find the area between \(y=x^3\) and \(y=x\) on \([-1,2]\).
Step 1: Find and order the intersections.
The intersections are \(x=-1,0,1\). Together with the endpoints, they give the pieces \([-1,0]\), \([0,1]\), and \([1,2]\).
Step 2: Determine which curve is higher. At \(x=-\frac12\), \(x^3=-\frac18\) and \(x=-\frac12\), so \(x^3\) is higher on \((-1,0)\). At \(x=\frac12\), \(x=\frac12\) and \(x^3=\frac18\), so \(x\) is higher on \((0,1)\). At \(x=\frac32\), \(x^3=\frac{27}{8}\) and \(x=\frac32\), so \(x^3\) is higher on \((1,2)\).
Step 3: Integrate upper minus lower on each piece. An antiderivative of \(x^3-x\) is \(\frac{x^4}{4}-\frac{x^2}{2}\).
The area is \(\frac{11}{4}\) square units. Each term in the sum is nonnegative, which checks that the curve order was used consistently on all three pieces.
A Crossing Is Not the Same as a Switch
Graphs can meet without exchanging which one is higher. For example, \(y=x^2\) and \(y=0\) meet at \(x=0\), but \(x^2\ge 0\) on both sides of zero. The point of contact is an intersection, but the upper curve remains \(x^2\) throughout. A test on either side confirms that there is no change in order.
For completeness, the area between these graphs on \([-1,1]\) is
One could split at zero and add the two integrals; that would give the same result. The practical rule is to split when the upper-minus-lower expression changes, not automatically at every point where the graphs meet.
Worked Example: Avoiding Cancellation Across a Crossing
Worked Example: Avoiding Cancellation Across a Crossing
Find the area between \(y=x^2-1\) and \(y=0\) on \([-2,2]\).
Step 1: Find the intersections. Solve \(x^2-1=0\), so \((x-1)(x+1)=0\), giving \(x=-1\) and \(x=1\).
Step 2: Identify the upper curve on each piece. For a test value \(x=-1.5\), \(x^2-1=1.25>0\), so the parabola is above the \(x\)-axis on \([-2,-1]\). At \(x=0\), \(x^2-1=-1<0\), so the \(x\)-axis is above the parabola on \([-1,1]\). At \(x=1.5\), \(x^2-1=1.25>0\), so the parabola is again above the axis on \([1,2]\).
Step 3: Add the three region areas.
The area is \(4\) square units. A useful check is that the signed integral of \(x^2-1\) over the full interval is only \(\frac43\): the positive outer regions and negative middle region partly cancel. The geometric area is larger because all three region areas must be added positively.
Common Mistake: Integrating One Difference Across the Whole Interval
A frequent error is to find intersections, but then integrate \(f-g\) unchanged from the left endpoint to the right endpoint. For the last example, that incorrect calculation would be
This is a valid signed integral, but it is not the geometric area between the curves. The middle interval contributes negatively because \(x^2-1<0\) there. The correction is to reverse the subtraction on that piece:
Before evaluating an area integral, check that each integrand is nonnegative over its stated interval. If one piece is negative, either the curves were subtracted in the wrong order or the interval needs further division.
Writing the Setup Clearly
In a free-response solution, make the reasoning visible. Show the equation used to find intersections, state the resulting division points, and identify the upper curve on each interval. Then write the complete sum before evaluating it. This makes it possible to distinguish an intersection error, an order error, and an integration error.
For a problem with division points \(a=c_0<c_1<\cdots<c_n=b\), a clear setup has the form
The upper and lower functions can differ from one interval to the next. The bounds must be in increasing order, and the result should be nonnegative. If the evaluated result is negative, revisit the curve order and the subtraction before assuming the antiderivative is at fault.
Check Your Understanding
For each question, identify how intersections and curve order determine the area setup.
- The curves are \(y=x^2\) and \(y=2x\) on \([0,3]\). At what \(x\)-values do they intersect, and which curve is above on each resulting interval?
- Why does \(\int_a^b(f(x)-g(x))\,dx\) fail to give geometric area when \(f-g\) is negative on part of \([a,b]\)?
- The graphs \(y=x^2\) and \(y=0\) meet at \(x=0\). Does their order switch there? Explain how you would check.
- Write, but do not evaluate, the geometric-area integral for \(y=x^3\) and \(y=x\) on \([-1,2]\).
- What should you check if one of the integrands in a proposed sum of area integrals is negative on its interval?
