Basic Properties of Definite Integrals

Definite integrals satisfy several powerful properties that often simplify evaluation without explicitly finding an anti-derivative.

Property 0: Dummy Variable Property

∫abf(x) dx=∫abf(t) dt\int_a^b f(x)\,dx = \int_a^b f(t)\,dt The value of a definite integral depends on the function and the limits, not on the name of the variable of integration.

Property 1: Reversal of Limits

∫abf(x) dx=−∫baf(x) dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx If the upper and lower limits are interchanged, the sign of the integral changes.

A direct consequence is: ∫aaf(x) dx=0\int_a^a f(x)\,dx = 0

Property 2: Splitting Property / Additivity

If cc lies between aa and bb, then ∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx This is especially useful for piecewise functions, modulus functions, and functions whose sign changes at some point x=cx=c.

The King's Rule and its Variations

One of the most useful properties of definite integrals is the following substitution symmetry:

Property 3: King's Rule

∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx This follows from the substitution t=a+b−xt=a+b-x.

A very important special case is when the lower limit is 00.

Property 4: Special Case

∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx This is heavily used in trigonometric integrals, especially when a=π/2a=\pi/2, because then complementary-angle identities can be used: sin⁡(π2−x)=cos⁡x,cos⁡(π2−x)=sin⁡x.\sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x.

Properties for Symmetric Limits and Even/Odd Functions

When limits are symmetric, additional simplifications become available.

Property 5

∫02af(x) dx=∫0af(x) dx+∫0af(2a−x) dx\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx This is obtained by splitting at x=ax=a and then applying the substitution x=2a−tx=2a-t in the second part.

Property 6: Symmetry about x=ax=a

From Property 5, ∫02af(x) dx={2∫0af(x) dx,if f(2a−x)=f(x),0,if f(2a−x)=−f(x)\int_0^{2a} f(x)\,dx = \begin{cases} 2\int_0^a f(x)\,dx, & \text{if } f(2a-x)=f(x), \\ 0, & \text{if } f(2a-x)=-f(x) \end{cases} So if the function is symmetric about the midpoint x=ax=a, the integral doubles; if it is antisymmetric about x=ax=a, the total integral is zero.

Property 7: Even and Odd Functions

For symmetric limits [−a,a][-a,a], ∫−aaf(x) dx={2∫0af(x) dx,if f(−x)=f(x) (even),0,if f(−x)=−f(x) (odd)\int_{-a}^a f(x)\,dx = \begin{cases} 2\int_0^a f(x)\,dx, & \text{if } f(-x)=f(x) \text{ (even)}, \\ 0, & \text{if } f(-x)=-f(x) \text{ (odd)} \end{cases}

Examples:

  • Even functions: x2x^2, cos⁡x\cos x, ∣x∣|x|
  • Odd functions: xx, x3x^3, sin⁡x\sin x

For odd functions, the positive and negative contributions cancel over symmetric intervals.

Example 1: Splitting Property with Modulus Function

Evaluate: ∫−12∣x3−x∣ dx\int_{-1}^{2} |x^3-x|\,dx

Solution: Step 1: Find the points where the expression inside the modulus is zero. x3−x=x(x2−1)=x(x−1)(x+1)x^3-x = x(x^2-1)=x(x-1)(x+1) So the zeros are x=−1,0,1x=-1,0,1.

Step 2: Determine the sign of x3−xx^3-x on the relevant intervals.

  • On (−1,0)(-1,0), take x=−12x=-\tfrac12: (−12)3−(−12)=−18+12=38>0\left(-\frac12\right)^3-\left(-\frac12\right)=-\frac18+\frac12=\frac38>0 So x3−x>0x^3-x>0 here.
  • On (0,1)(0,1), take x=12x=\tfrac12: (12)3−12=18−12=−38<0\left(\frac12\right)^3-\frac12=\frac18-\frac12=-\frac38<0 So x3−x<0x^3-x<0 here.
  • On (1,2)(1,2), take x=32x=\tfrac32: (32)3−32>0\left(\frac32\right)^3-\frac32>0 So x3−x>0x^3-x>0 here.

Step 3: Split the integral. ∫−12∣x3−x∣ dx=∫−10(x3−x) dx+∫01−(x3−x) dx+∫12(x3−x) dx\int_{-1}^{2}|x^3-x|\,dx = \int_{-1}^{0}(x^3-x)\,dx + \int_0^1 -(x^3-x)\,dx + \int_1^2 (x^3-x)\,dx

Step 4: Use the anti-derivative ∫(x3−x) dx=x44−x22.\int (x^3-x)\,dx = \frac{x^4}{4}-\frac{x^2}{2}. Hence, I=[x44−x22]−10−[x44−x22]01+[x44−x22]12.I = \left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0} - \left[\frac{x^4}{4}-\frac{x^2}{2}\right]_0^{1} + \left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^{2}.

Step 5: Compute each part.

  • First part: (0)−(14−12)=14\left(0\right)-\left(\frac14-\frac12\right)=\frac14
  • Second part: −(14−12−0)=14-\left(\frac14-\frac12-0\right)=\frac14
  • Third part: (164−42)−(14−12)=(4−2)−(−14)=2+14=94\left(\frac{16}{4}-\frac{4}{2}\right)-\left(\frac14-\frac12\right)= (4-2)-\left(-\frac14\right)=2+\frac14=\frac94

Step 6: Add them. I=14+14+94=114I=\frac14+\frac14+\frac94=\frac{11}{4}

Answer: 114\frac{11}{4}

Example 2: King's Rule on [0,π/2][0,\pi/2]

Evaluate: I=∫0π/2sin⁡xsin⁡x+cos⁡x dxI=\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx

Solution: Step 1: Let I=∫0π/2sin⁡xsin⁡x+cos⁡x dx.I=\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.

Step 2: Apply Property 4: ∫0π/2f(x) dx=∫0π/2f(π2−x) dx.\int_0^{\pi/2} f(x)\,dx = \int_0^{\pi/2} f\left(\frac{\pi}{2}-x\right)\,dx. Then I=∫0π/2sin⁡(π/2−x)sin⁡(π/2−x)+cos⁡(π/2−x) dx.I=\int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2-x)}}{\sqrt{\sin(\pi/2-x)}+\sqrt{\cos(\pi/2-x)}}\,dx. Using complementary-angle identities, sin⁡(π2−x)=cos⁡x,cos⁡(π2−x)=sin⁡x.\sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x. So, I=∫0π/2cos⁡xcos⁡x+sin⁡x dx.I=\int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.

Step 3: Add the two expressions for II. 2I=∫0π/2sin⁡x+cos⁡xsin⁡x+cos⁡x dx=∫0π/21 dx.2I = \int_0^{\pi/2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx = \int_0^{\pi/2} 1\,dx.

Step 4: Evaluate. 2I=[x]0π/2=π2.2I = \left[x\right]_0^{\pi/2}=\frac{\pi}{2}. Hence, I=π4.I=\frac{\pi}{4}.

Answer: π4\frac{\pi}{4}

Example 3: General King's Rule

Evaluate: I=∫π/6π/311+tan⁡x dxI=\int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\tan x}}\,dx

Solution: Step 1: Rewrite the integrand in a symmetric form. 11+tan⁡x=11+sin⁡xcos⁡x=cos⁡xcos⁡x+sin⁡x.\frac{1}{1+\sqrt{\tan x}} = \frac{1}{1+\frac{\sqrt{\sin x}}{\sqrt{\cos x}}} = \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}. Thus, I=∫π/6π/3cos⁡xcos⁡x+sin⁡x dx.I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.

Step 2: Here, a=π6,b=π3,a+b=π2.a=\frac{\pi}{6}, \qquad b=\frac{\pi}{3}, \qquad a+b=\frac{\pi}{2}. By King's Rule, I=∫π/6π/3cos⁡(π/2−x)cos⁡(π/2−x)+sin⁡(π/2−x) dx.I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\pi/2-x)}}{\sqrt{\cos(\pi/2-x)}+\sqrt{\sin(\pi/2-x)}}\,dx. Using complementary-angle identities, I=∫π/6π/3sin⁡xsin⁡x+cos⁡x dx.I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.

Step 3: Add the two forms. 2I=∫π/6π/31 dx=[x]π/6π/3=π3−π6=π6.2I = \int_{\pi/6}^{\pi/3} 1\,dx = \left[x\right]_{\pi/6}^{\pi/3}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}. Therefore, I=π12.I=\frac{\pi}{12}.

Answer: π12\frac{\pi}{12}

Example 4: Even and Odd Function Property

Evaluate: ∫−11sin⁡5xcos⁡4x dx\int_{-1}^{1} \sin^5 x\cos^4 x\,dx

Solution: Step 1: Let f(x)=sin⁡5xcos⁡4x.f(x)=\sin^5 x\cos^4 x. Check parity: f(−x)=sin⁡5(−x)cos⁡4(−x)=(−sin⁡x)5(cos⁡x)4=−sin⁡5xcos⁡4x=−f(x).f(-x)=\sin^5(-x)\cos^4(-x)=(-\sin x)^5(\cos x)^4=-\sin^5 x\cos^4 x=-f(x). So f(x)f(x) is an odd function.

Step 2: Since the limits are symmetric, by Property 7, ∫−11f(x) dx=0.\int_{-1}^{1} f(x)\,dx = 0. Hence, ∫−11sin⁡5xcos⁡4x dx=0.\int_{-1}^{1} \sin^5 x\cos^4 x\,dx=0.

Answer: 00

Example 5: Using King's Rule to Eliminate xx

Evaluate: I=∫0πxsin⁡x1+cos⁡2x dxI=\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx

Solution: Step 1: Let I=∫0πxsin⁡x1+cos⁡2x dx.I=\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx.

Step 2: Apply Property 4 with a=πa=\pi: I=∫0π(π−x)sin⁡(π−x)1+cos⁡2(π−x) dx.I=\int_0^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx. Now, sin⁡(π−x)=sin⁡x,cos⁡(π−x)=−cos⁡x,cos⁡2(π−x)=cos⁡2x.\sin(\pi-x)=\sin x, \qquad \cos(\pi-x)=-\cos x, \qquad \cos^2(\pi-x)=\cos^2 x. Therefore, I=∫0π(π−x)sin⁡x1+cos⁡2x dx.I=\int_0^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx.

Step 3: Add both expressions for II. 2I=∫0ππsin⁡x1+cos⁡2x dx=π∫0πsin⁡x1+cos⁡2x dx.2I=\int_0^{\pi} \frac{\pi\sin x}{1+\cos^2 x}\,dx = \pi \int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx.

Step 4: Use substitution t=cos⁡xt=\cos x, so dt=−sin⁡x dxdt=-\sin x\,dx. When x=0x=0, t=1t=1. When x=πx=\pi, t=−1t=-1. Thus, 2I=π∫1−1−dt1+t2=π∫−11dt1+t2.2I=\pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^{1} \frac{dt}{1+t^2}.

Step 5: Integrate. 2I=π[tan⁡−1t]−11=π(π4−(−π4))=π⋅π2=π22.2I=\pi [\tan^{-1} t]_{-1}^{1} = \pi \left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}. Hence, I=π24.I=\frac{\pi^2}{4}.

Answer: π24\frac{\pi^2}{4}

Example 6: Algebraic Use of Property 4

Evaluate: ∫02x2−x dx\int_0^2 x\sqrt{2-x}\,dx

Solution: Step 1: Let I=∫02x2−x dx.I=\int_0^2 x\sqrt{2-x}\,dx. Apply Property 4 with a=2a=2: I=∫02(2−x)2−(2−x) dx=∫02(2−x)x dx.I=\int_0^2 (2-x)\sqrt{2-(2-x)}\,dx = \int_0^2 (2-x)\sqrt{x}\,dx.

Step 2: Expand. I=∫02(2x1/2−x3/2) dx.I=\int_0^2 (2x^{1/2}-x^{3/2})\,dx.

Step 3: Integrate term by term. I=[2⋅x3/23/2−x5/25/2]02=[43x3/2−25x5/2]02.I=\left[ 2\cdot\frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2} \right]_0^2 = \left[ \frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2} \right]_0^2.

Step 4: Substitute x=2x=2. Since 23/2=22,25/2=42,2^{3/2}=2\sqrt2, \qquad 2^{5/2}=4\sqrt2, we get I=43(22)−25(42)=823−825.I = \frac{4}{3}(2\sqrt2) - \frac{2}{5}(4\sqrt2)=\frac{8\sqrt2}{3}-\frac{8\sqrt2}{5}.

Step 5: Simplify. I=82(13−15)=82⋅215=16215.I=8\sqrt2\left(\frac13-\frac15\right)=8\sqrt2\cdot\frac{2}{15}=\frac{16\sqrt2}{15}.

Answer: 16215\frac{16\sqrt2}{15}

Example 7: Property 6 with Symmetry about Midpoint

Evaluate: ∫02πcos⁡5x dx\int_0^{2\pi} \cos^5 x\,dx

Solution: Step 1: Let f(x)=cos⁡5xf(x)=\cos^5 x and write the integral as I=∫02πcos⁡5x dx.I=\int_0^{2\pi} \cos^5 x\,dx. This is an integral over [0,2a][0,2a] with 2a=2π2a=2\pi, so a=πa=\pi.

Step 2: Check symmetry: f(2π−x)=cos⁡5(2π−x)=cos⁡5x=f(x).f(2\pi-x)=\cos^5(2\pi-x)=\cos^5 x=f(x). So by Property 6, I=2∫0πcos⁡5x dx.I=2\int_0^{\pi} \cos^5 x\,dx.

Step 3: Now consider J=∫0πcos⁡5x dx.J=\int_0^{\pi} \cos^5 x\,dx. Here 2a=π2a=\pi, so a=π/2a=\pi/2. Check: f(π−x)=cos⁡5(π−x)=(−cos⁡x)5=−cos⁡5x=−f(x).f(\pi-x)=\cos^5(\pi-x)=(-\cos x)^5=-\cos^5 x=-f(x). Hence by Property 6, J=0.J=0. Therefore, I=2J=0.I=2J=0.

Answer: 00