Why Trigonometric Curves Need Careful Setup
Trigonometric graphs oscillate, so two curves can meet more than once on a specified interval. Their order can switch at a crossing, and a single integral of “upper curve minus lower curve” will give the geometric area only if that order remains correct throughout the interval.
The main tasks are therefore to find every intersection in the interval, arrange those points from left to right, and test the order between consecutive intersections. Exact angles in radians are especially useful: they keep the bounds and evaluated integrals exact.
Finding and Organizing the Intersections
For curves \(y=f(x)\) and \(y=g(x)\), their intersections satisfy \(f(x)=g(x)\). With trigonometric functions, use identities and solve the resulting equation for all solutions in the specified interval. Periodicity matters: a general solution may produce several values in the interval, and none should be omitted.
Some equations simplify by factoring or using a standard identity. For example, \(\sin(2x)=2\sin x\cos x\), while \(\cos(2x)=2\cos^2x-1\). After solving, substitute each candidate into the original equation and check that its angle lies within the interval. Include intersections at the endpoints when listing all intersections on a closed interval; they are bounds, not interior crossings.
Once the intersection values are ordered, choose a test point in each open interval between successive values. Comparing \(f\) and \(g\) there determines which is above on that interval.
Why One Test Point Determines the Order
Let \(h(x)=f(x)-g(x)\), where both curves are continuous. Suppose \(h\) has no zero in an open interval between consecutive intersections. If \(h\) were positive at one point and negative at another, the Intermediate Value Theorem would require \(h\) to equal zero somewhere between those points. That would be another intersection, contrary to the way the interval was chosen. Thus \(h\) has one sign throughout the interval, and a single test point determines the order there.
On an interval where \(h>0\), the vertical distance between the curves is \(f(x)-g(x)\); where \(h<0\), it is \(g(x)-f(x)\). In either case that distance is \(|h(x)|\). Splitting at all intersections therefore gives the area as the sum of integrals of the appropriate positive differences.
In practice, the absolute value is handled by splitting the interval at the intersections and writing the difference in the correct order on each piece. This setup does not assume that the curves cross at every intersection: if they touch and keep the same order, the test-point signs on either side will agree.
Worked Example: Sine and Cosine on a Full Period
Worked Example: Sine and Cosine on a Full Period
Find the intersections of \(y=\sin x\) and \(y=\cos x\) on \([0,2\pi]\), determine their order, and find the area between the curves on that interval.
1. Find the intersections. Set \(\sin x=\cos x\). The solutions are \(x=\frac{\pi}{4}+k\pi\), where \(k\) is an integer. The values in \([0,2\pi]\) are \(\frac{\pi}{4}\) and \(\frac{5\pi}{4}\). Substitution checks them: both sine and cosine equal \(\frac{\sqrt{2}}{2}\) at \(\frac{\pi}{4}\), and both equal \(-\frac{\sqrt{2}}{2}\) at \(\frac{5\pi}{4}\).
2. Determine the order. At \(x=0\), \(\sin 0=0<1=\cos 0\), so cosine is above sine on \((0,\frac{\pi}{4})\). At \(x=\frac{\pi}{2}\), \(\sin x=1>0=\cos x\), so sine is above on \((\frac{\pi}{4},\frac{5\pi}{4})\). At \(x=\frac{3\pi}{2}\), \(\sin x=-1<0=\cos x\), so cosine is above on \((\frac{5\pi}{4},2\pi)\).
3. Integrate each positive difference.
For the first and third integrals, an antiderivative of \(\cos x-\sin x\) is \(\sin x+\cos x\). For the middle integral, an antiderivative of \(\sin x-\cos x\) is \(-\cos x-\sin x\). Differentiating these expressions returns their respective integrands. The evaluated areas are positive and sum to \(4\sqrt{2}\) square units.
Worked Example: A Sine Curve and a Horizontal Line
Worked Example: A Sine Curve and a Horizontal Line
Find the area between \(y=\sin x\) and \(y=\frac12\) on \([0,2\pi]\).
1. Find and check the intersections. The equation \(\sin x=\frac12\) has solutions \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\) in this interval. At both angles, sine is \(\frac12\).
2. Determine the order. At \(x=0\), \(0<\frac12\), so the horizontal line is above the sine curve before \(\frac{\pi}{6}\). At \(x=\frac{\pi}{2}\), \(1>\frac12\), so the sine curve is above between the intersections. At \(x=\pi\), \(0<\frac12\), so the line is above after \(\frac{5\pi}{6}\).
3. Set up the area. The positive region between the intersections has area
The signed integral of \(\sin x-\frac12\) over the full interval is
Thus the signed contribution outside the positive middle region is \(-\pi-P\), so its geometric area is \(\pi+P\). The total area is \(P+(\pi+P)=2P+\pi\), which gives
As a check, the antiderivative \(-\cos x-\frac{x}{2}\) differentiates to \(\sin x-\frac12\). The final area is positive, as required for geometric area.
Worked Example: A Double-Angle Curve and Cosine
Worked Example: A Double-Angle Curve and Cosine
Find the area between \(y=\cos(2x)\) and \(y=\cos x\) on \([0,2\pi]\).
1. Find all intersections. Using \(\cos(2x)=2\cos^2x-1\), set \(2\cos^2x-1=\cos x\). Factoring gives
Therefore \(\cos x=-\frac12\) or \(\cos x=1\). On \([0,2\pi]\), these give \(x=\frac{2\pi}{3}\), \(x=\frac{4\pi}{3}\), and the endpoints \(x=0,2\pi\). Substitution into the original equation verifies each value.
2. Determine the order. At \(x=\frac{\pi}{3}\), \(\cos(2x)=-\frac12<\frac12=\cos x\). At \(x=\pi\), \(\cos(2x)=1>-1=\cos x\). At \(x=\frac{5\pi}{3}\), \(\cos(2x)=-\frac12<\frac12=\cos x\). Thus cosine is above \(\cos(2x)\) on the first and last pieces, and \(\cos(2x)\) is above cosine in the middle.
3. Integrate the positive differences. An antiderivative of \(\cos(2x)-\cos x\) is \(F(x)=\frac12\sin(2x)-\sin x\). Hence
For verification, \(F'(x)=\cos(2x)-\cos x\). Also, \(F(0)=F(2\pi)=0\), \(F(\frac{2\pi}{3})=-\frac{3\sqrt{3}}{4}\), and \(F(\frac{4\pi}{3})=\frac{3\sqrt{3}}{4}\). These values give the three positive contributions shown above.
Common Mistake: Keeping the Same Subtraction Order
A frequent error is to integrate \(f-g\) across the whole interval even though the curves switch order. In the first example, using \(\int_0^{2\pi}(\sin x-\cos x)\,dx\) gives \(0\). That is the net signed difference, not the geometric area: positive and negative parts cancel.
The correction is to locate every intersection, test the order on each resulting interval, and reverse the subtraction wherever the lower curve changes. Here the correct three pieces give \(4\sqrt{2}\), a positive area. A final sign check helps catch the error: a geometric area cannot be negative, and cancellation in a signed integral does not mean the region has zero area.
A Reliable Setup
Solve \(f(x)=g(x)\), using identities where helpful, and retain every solution in the specified interval.
Include relevant interval endpoints and distinguish them from interior crossings.
Compare the curve values at one interior point to identify the upper and lower graphs.
Split at intersections and add the upper-minus-lower integrals.
Verify intersections, antiderivatives, and that the final geometric area is nonnegative.
Check Your Understanding
For each question, identify the interval bounds carefully and check the order of the curves before writing an area integral.
- Find all intersections of \(y=\sin x\) and \(y=\cos x\) on \([0,\pi]\).
- On which intervals in \([0,2\pi]\) is \(y=\sin x\) above \(y=\frac12\)?
- Why is one test point sufficient to determine curve order between consecutive intersections, provided the curves are continuous?
- What does a signed integral of \(f-g\) measure when the curves change order, and why can it differ from geometric area?
- For \(y=\cos(2x)\) and \(y=\cos x\) on \([0,2\pi]\), which curve is above at \(x=\pi\)?
