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Concept 07 / 10Integral Calculus

Properties of Definite Integrals: Symmetry, King's Property and Periodicity

Additivity over intervals, the property integral f(x) = integral f(a + b - x) and its use for integrals that do not have elementary antiderivatives, even and odd functions on symmetric intervals, the half-interval property, periodic functions, and comparison inequalities.

Figure 1
Many definite integrals in JEE cannot be done by finding an antiderivative, yet they have simple values because of symmetry. The properties below are the tools. P1 (dummy variable): ∫abf(x)dx=∫abf(t)dt\int_{a}^{b}f(x)dx = \int_{a}^{b}f(t)dt. P2 (reversal): ∫abf=−∫baf\int_{a}^{b}f = -\int_{b}^{a}f. P3 (additivity): ∫abf=∫acf+∫cbf\int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f for any cc. This is how piecewise functions, modulus and greatest-integer integrands are handled: split at the break points. P4 (King's property): ∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b - x)dx. Substituting x→a+b−xx \to a + b - x reflects the interval onto itself about its midpoint, as in the figure, so the area is unchanged. The special case ∫0af(x)dx=∫0af(a−x)dx\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a - x)dx is used constantly. P5 (symmetric interval): ∫−aaf(x)dx=∫0a[f(x)+f(−x)]dx\int_{-a}^{a}f(x)dx = \int_{0}^{a}[f(x) + f(-x)]dx. So it equals 2∫0af2\int_{0}^{a}f if ff is even, and 00 if ff is odd. P6 (half interval): ∫02af(x)dx=∫0a[f(x)+f(2a−x)]dx\int_{0}^{2a}f(x)dx = \int_{0}^{a}[f(x) + f(2a - x)]dx. So it equals 2∫0af2\int_{0}^{a}f if f(2a−x)=f(x)f(2a - x) = f(x), and 00 if f(2a−x)=−f(x)f(2a - x) = -f(x). P7 (periodicity): if ff has period TT, then ∫aa+Tf=∫0Tf\int_{a}^{a + T}f = \int_{0}^{T}f for every aa, and ∫0nTf=n∫0Tf\int_{0}^{nT}f = n\int_{0}^{T}f for integer nn. How King's property is used. Let I=∫abf(x)dxI = \int_{a}^{b}f(x)dx. Apply P4 to get a second expression for II, then add the two. If the sum of the integrands simplifies (typically to a constant or something easily integrable), 2I2I is found. The classic example is I=∫0π/2sin⁡nxsin⁡nx+cos⁡nxdxI = \int_{0}^{\pi/2}\frac{\sin^{n}x}{\sin^{n}x + \cos^{n}x}dx. Replacing xx by π2−x\frac{\pi}{2} - x swaps sin⁡\sin and cos⁡\cos, so I=∫0π/2cos⁡nxcos⁡nx+sin⁡nxdxI = \int_{0}^{\pi/2}\frac{\cos^{n}x}{\cos^{n}x + \sin^{n}x}dx. Adding, 2I=∫0π/21 dx=π22I = \int_{0}^{\pi/2}1\,dx = \frac{\pi}{2}, so I=π4I = \frac{\pi}{4} for every nn. In general, ∫abf(x)f(x)+f(a+b−x)dx=b−a2\int_{a}^{b}\frac{f(x)}{f(x) + f(a + b - x)}dx = \frac{b - a}{2}. A second classic: integrals with a factor xx. For I=∫0πxf(sin⁡x)dxI = \int_{0}^{\pi}xf(\sin x)dx, King's property gives I=∫0π(π−x)f(sin⁡x)dxI = \int_{0}^{\pi}(\pi - x)f(\sin x)dx. Adding, 2I=π∫0πf(sin⁡x)dx2I = \pi\int_{0}^{\pi}f(\sin x)dx, so I=π2∫0πf(sin⁡x)dxI = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)dx. The awkward factor xx disappears. A useful standard value obtained this way: ∫0π/2ln⁡sin⁡x dx=∫0π/2ln⁡cos⁡x dx=−π2ln⁡2\int_{0}^{\pi/2}\ln\sin x\,dx = \int_{0}^{\pi/2}\ln\cos x\,dx = -\frac{\pi}{2}\ln 2. Inequalities and estimates. If f(x)≤g(x)f(x) \le g(x) on [a,b][a, b], then ∫abf≤∫abg\int_{a}^{b}f \le \int_{a}^{b}g. If m≤f(x)≤Mm \le f(x) \le M on [a,b][a, b], then m(b−a)≤∫abf≤M(b−a)m(b - a) \le \int_{a}^{b}f \le M(b - a). And ∣∫abf∣≤∫ab∣f∣\left|\int_{a}^{b}f\right| \le \int_{a}^{b}|f|. These give bounds when exact evaluation is impossible, for example π4≤∫0π/2dx4−sin⁡2x≤π23\frac{\pi}{4} \le \int_{0}^{\pi/2}\frac{dx}{\sqrt{4 - \sin^{2}x}} \le \frac{\pi}{2\sqrt{3}} from 3≤4−sin⁡2x≤2\sqrt{3} \le \sqrt{4 - \sin^{2}x} \le 2. Key Takeaways: - ∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b - x)dx; add the two forms and simplify. - ∫abf(x)f(x)+f(a+b−x)dx=b−a2\int_{a}^{b}\frac{f(x)}{f(x) + f(a + b - x)}dx = \frac{b - a}{2}. - ∫−aaf=2∫0af\int_{-a}^{a}f = 2\int_{0}^{a}f (even) or 00 (odd). - ∫02af=2∫0af\int_{0}^{2a}f = 2\int_{0}^{a}f if f(2a−x)=f(x)f(2a - x) = f(x); 00 if f(2a−x)=−f(x)f(2a - x) = -f(x). - ∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dx\int_{0}^{\pi}xf(\sin x)dx = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)dx. - Periodic: ∫0nTf=n∫0Tf\int_{0}^{nT}f = n\int_{0}^{T}f; ∫aa+Tf\int_{a}^{a+T}f is independent of aa. - ∫0π/2ln⁡sin⁡x dx=−π2ln⁡2\int_{0}^{\pi/2}\ln\sin x\,dx = -\frac{\pi}{2}\ln 2. - Pitfall: using the odd-function property when the interval is not symmetric about 00. Brain-Teaser 1: Evaluate ∫0πxsin⁡x1+cos⁡2xdx\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}dx. Solution: This is ∫0πxf(sin⁡x)dx\int_{0}^{\pi}xf(\sin x)dx-type, since cos⁡2x=1−sin⁡2x\cos^{2}x = 1 - \sin^{2}x. So I=π2∫0πsin⁡x1+cos⁡2xdxI = \frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x}{1 + \cos^{2}x}dx. Put u=cos⁡xu = \cos x: π2∫−11du1+u2=π2⋅π2=π24\frac{\pi}{2}\int_{-1}^{1}\frac{du}{1 + u^{2}} = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^{2}}{4}. Without King's property, the factor xx would force integration by parts into an integral with no elementary antiderivative. Brain-Teaser 2: Evaluate ∫−π/2π/2cos⁡x1+exdx\int_{-\pi/2}^{\pi/2}\frac{\cos x}{1 + e^{x}}dx. Solution: By P5, I=∫0π/2[cos⁡x1+ex+cos⁡x1+e−x]dxI = \int_{0}^{\pi/2}\left[\frac{\cos x}{1 + e^{x}} + \frac{\cos x}{1 + e^{-x}}\right]dx. Since 11+ex+11+e−x=11+ex+exex+1=1\frac{1}{1 + e^{x}} + \frac{1}{1 + e^{-x}} = \frac{1}{1 + e^{x}} + \frac{e^{x}}{e^{x} + 1} = 1, this is ∫0π/2cos⁡x dx=1\int_{0}^{\pi/2}\cos x\,dx = 1. The integrand is neither even nor odd, but pairing xx with −x-x still makes it collapse. Brain-Teaser 3: Evaluate ∫0100π∣sin⁡x∣ dx\int_{0}^{100\pi}|\sin x|\,dx and ∫010{x} dx\int_{0}^{10}\{x\}\,dx. Solution: ∣sin⁡x∣|\sin x| has period π\pi, and ∫0π∣sin⁡x∣dx=2\int_{0}^{\pi}|\sin x|dx = 2, so the first integral is 100×2=200100 \times 2 = 200. {x}\{x\} has period 11 and ∫01x dx=12\int_{0}^{1}x\,dx = \frac{1}{2}, so the second is 10×12=510 \times \frac{1}{2} = 5. Periodicity turns a long interval into a single period multiplied by the number of periods.

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