Intersections Organize the Area
Trigonometric graphs repeat, cross, and often create several regions within one period. The key to finding their total area is to locate the intersections first. Between consecutive intersections, one graph stays above the other, so the area on that interval can be calculated without ambiguity.
For periodic graphs, a chosen period gives a natural interval to study. It does not remove the need to check crossings inside that interval: the upper graph can change more than once, and different regions need not have equal areas.
The absolute value matters because definite integrals record signed accumulation: an interval where \(f\) lies below \(g\) contributes a negative integral of \(f-g\), even though geometric area is positive. In practice, find the intersections, determine the order of the graphs between them, and integrate the positive difference on each piece.
A Reliable Method for One Period
For two trigonometric functions on a chosen interval \([a,b]\), begin by solving \(f(x)=g(x)\) within that interval. Use identities where they simplify the equation, and include interval endpoints if they are intersections. Then order all the intersection values and endpoints. On each resulting open interval, test one interior point to determine which graph is above the other. Finally, add the areas from the pieces.
Solve \(f(x)-g(x)=0\), retaining only the solutions in the chosen interval.
List the endpoints and crossings in increasing order. Each adjacent pair gives one piece to check.
Test a point inside each piece. Integrate upper graph minus lower graph there.
Sum the nonnegative piece areas, and verify the antiderivative and the signs used.
A crossing generally changes which graph is higher; a tangency may not. A root of \(f-g\) is therefore a point to inspect, not proof by itself that the sign changes. If two graphs have a common period \(T\), their arrangement repeats after \(T\), but individual regions within that period can still differ.
Worked Example: Sine and Cosine on One Period
Worked Example: Sine and Cosine on One Period
Find the area between \(y=\sin x\) and \(y=\cos x\) on \([0,2\pi]\), a common period of the two graphs.
First solve the intersection equation. The identity \(\sin x-\cos x=\sqrt{2}\sin(x-\pi/4)\) gives $$ \sin x-\cos x=0 \quad\Longleftrightarrow\quad \sin(x-\pi/4)=0. $$ Thus \(x=\pi/4+k\pi\) for integers \(k\). The intersections inside the interval are \(x=\pi/4\) and \(x=5\pi/4\).
Test the intervals \((0,\pi/4)\), \((\pi/4,5\pi/4)\), and \((5\pi/4,2\pi)\). At \(x=0\), \(\sin x-\cos x=-1\), so cosine is higher on the first interval. At \(x=\pi/2\), \(\sin x-\cos x=1\), so sine is higher on the second. At \(x=3\pi/2\), \(\sin x-\cos x=-1\), so cosine is higher on the third.
Therefore, $$ A=\int_0^{\pi/4}(\cos x-\sin x)\,dx +\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx +\int_{5\pi/4}^{2\pi}(\cos x-\sin x)\,dx. $$ Antiderivatives are \(\sin x+\cos x\) for \(\cos x-\sin x\), and \(-\cos x-\sin x\) for \(\sin x-\cos x\). Differentiating gives \(\cos x-\sin x\) and \(\sin x-\cos x\), respectively.
The first piece is $$ (\sin(\pi/4)+\cos(\pi/4))-(\sin 0+\cos 0) =\sqrt{2}-1. $$ The second piece is $$ (-\cos(5\pi/4)-\sin(5\pi/4)) -(-\cos(\pi/4)-\sin(\pi/4)) =\sqrt{2}-(-\sqrt{2})=2\sqrt{2}. $$ The third piece is $$ (\sin(2\pi)+\cos(2\pi)) -(\sin(5\pi/4)+\cos(5\pi/4)) =1-(-\sqrt{2})=1+\sqrt{2}. $$ Adding the positive piece areas gives $$ A=(\sqrt{2}-1)+2\sqrt{2}+(1+\sqrt{2})=4\sqrt{2}. $$ The result is positive, as area must be. The sign tests also show that each integral used upper graph minus lower graph.
Repeated Regions Need Not Have Equal Areas
Two graphs can produce several lobes in a single period. The intersections still give the correct partition, but the areas of the resulting pieces must be evaluated rather than assumed equal. The next example has three interior crossings in one common period.
Worked Example: A Faster Oscillation Crossing a Slower One
Find the area between \(y=\sin(2x)\) and \(y=\sin x\) on \([0,2\pi]\).
Set the functions equal and use the sine-difference identity: $$ \sin(2x)-\sin x =2\cos\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right). $$ The factor \(\sin(x/2)\) vanishes at \(x=0\) and \(x=2\pi\). The other factor vanishes when \(3x/2=\pi/2+k\pi\), giving \(x=\pi/3+2k\pi/3\). The crossings in the interval are therefore \(0,\pi/3,\pi,5\pi/3,2\pi\).
Testing \(x=\pi/6\) gives \(\sin(2x)-\sin x>0\). Testing \(x=\pi/2\) gives a negative difference, and testing \(x=4\pi/3\) gives a positive difference. Testing \(x=11\pi/6\) gives a negative difference. Thus the difference alternates signs across the four pieces.
An antiderivative of \(\sin(2x)-\sin x\) is $$ F(x)=-\frac{1}{2}\cos(2x)+\cos x. $$ Indeed, \(F'(x)=\sin(2x)-\sin x\). Its values at the partition points are $$ F(0)=\frac{1}{2},\qquad F(\pi/3)=\frac{3}{4},\qquad F(\pi)=-\frac{3}{2},\qquad F(5\pi/3)=\frac{3}{4},\qquad F(2\pi)=\frac{1}{2}. $$
Apply the sign on each piece. Where the difference is positive, use \(F(\text{right})-F(\text{left})\); where it is negative, reverse that subtraction. The four areas are $$ \begin{aligned} A_1&=F(\pi/3)-F(0)=\frac{3}{4}-\frac{1}{2}=\frac{1}{4},\\ A_2&=F(\pi/3)-F(\pi)=\frac{3}{4}-\left(-\frac{3}{2}\right)=\frac{9}{4},\\ A_3&=F(5\pi/3)-F(\pi)=\frac{3}{4}-\left(-\frac{3}{2}\right)=\frac{9}{4},\\ A_4&=F(5\pi/3)-F(2\pi)=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}. \end{aligned} $$ Hence $$ A=A_1+A_2+A_3+A_4 =\frac{1}{4}+\frac{9}{4}+\frac{9}{4}+\frac{1}{4}=5. $$ The two small end regions have area \(1/4\) each, while the middle regions have area \(9/4\) each. Their total is \(5\), consistent with the positive areas on all four pieces.
Why the Partition Method Works
Here is the reason the intersection-based procedure is valid. Let \(h(x)=f(x)-g(x)\), and suppose the consecutive partition points are \(c_0<c_1<\cdots<c_n\), with no intersections inside any interval \((c_{i-1},c_i)\). If \(h\) took both positive and negative values inside one such interval, continuity and the intermediate value theorem would force it to equal zero somewhere between those values. That would be an additional intersection inside the interval, contrary to the partition.
To see the formula, if \(f-g\) is positive on a piece, then \(f\) is above \(g\), and the area is \(\int(f-g)\). If \(f-g\) is negative, then \(g\) is above \(f\), and the area is \(-\int(f-g)\). In either case, the area is the absolute value of the signed integral. This also explains why a crossing should be included in the partition even when two adjacent pieces happen to have equal areas.
Symmetry and a Phase Shift
Trigonometric identities can expose a symmetry that makes several area pieces easier to recognize. For example, subtracting two sine functions of the same frequency often gives a shifted cosine. Its zeros locate the crossings, and its repeated shape can make the lobes symmetric. The calculation still needs to respect the chosen interval and its endpoints.
Worked Example: Two Sines with a Phase Shift
Find the area between \(y=\sin x\) and \(y=\sin(x+\pi/3)\) on \([0,2\pi]\).
The difference is $$ \sin x-\sin(x+\pi/3) =-2\sin(\pi/6)\cos(x+\pi/6) =-\cos(x+\pi/6). $$ Thus intersections occur where \(\cos(x+\pi/6)=0\), namely at \(x=\pi/3\) and \(x=4\pi/3\) within the interval. At \(x=0\), the difference is negative; at \(x=\pi\), it is positive; at \(x=3\pi/2\), it is negative. The area is therefore $$ A=\int_0^{\pi/3}\cos(x+\pi/6)\,dx +\int_{\pi/3}^{4\pi/3}-\cos(x+\pi/6)\,dx +\int_{4\pi/3}^{2\pi}\cos(x+\pi/6)\,dx. $$
An antiderivative of \(\cos(x+\pi/6)\) is \(\sin(x+\pi/6)\), and differentiating it returns the integrand. The first piece is $$ \sin(\pi/2)-\sin(\pi/6)=1-\frac{1}{2}=\frac{1}{2}. $$ The middle piece is $$ -\sin(3\pi/2)+\sin(\pi/2)=1+1=2. $$ The final piece is $$ \sin(13\pi/6)-\sin(3\pi/2) =\frac{1}{2}-(-1)=\frac{3}{2}. $$ The total area is $$ A=\frac{1}{2}+2+\frac{3}{2}=4. $$ The two end regions differ in size on this chosen interval, but the exact piece calculations sum to \(4\). This is a useful reminder that a phase shift can change how the interval cuts through a repeating graph, even though the graphs themselves remain periodic.
A Common Mistake: Integrating the Difference Once
A frequent error is to integrate \(f-g\) over the entire period without checking its sign. For the sine-and-cosine example, this gives $$ \int_0^{2\pi}(\sin x-\cos x)\,dx =[-\cos x-\sin x]_0^{2\pi} =(-1)-(-1)=0. $$ The zero is the signed total: positive and negative contributions cancel. It is not the geometric area, which is \(4\sqrt{2}\).
The correction is to split at the crossings \(\pi/4\) and \(5\pi/4\), use the upper function minus the lower function on each piece, and add the three positive results. Equivalently, keep the absolute value in \(\int_0^{2\pi}|\sin x-\cos x|\,dx\). Dropping the absolute value before handling sign changes is what causes the mistake.
Check Your Understanding
For each question, identify the intersections and sign intervals before calculating or describing the area.
- On \([0,2\pi]\), what equation would you solve first to find the crossings of \(y=\cos x\) and \(y=-\cos x\)?
- Why is \(\int_a^b(f(x)-g(x))\,dx\) not always the geometric area between the graphs?
- For \(y=\sin x\) and \(y=\sin(2x)\) on \([0,2\pi]\), what are the interior intersection values?
- What does a sign test between consecutive intersections tell you?
- If two periodic graphs have a common period, what repeats after that period, and what still must be checked within it?
