Properties of Definite Integral

Change of variable of integration is immaterial so long as limits of integration remain the same .

Properties :

1. $ \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b}f(t)dt $

2. $ \displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx $

3. $ \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx $

, where the point c may lie between a and b or it may be exterior to (a, b)

Note:This property is useful when f (x) is not continuous in [a, b] because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the subintervals.

Illustration : Evaluate : $ \displaystyle \int_{0}^{\pi}min[|tanx| , |\frac{4}{\pi}x -2|]dx $

Solution:

From the graph it is clear that the function changes its definition at x = π/4 , π/2 and 3π/4 . Hence, the given function can be written as :

$  = \int_{0}^{\pi/4} tanx dx  + \int_{\pi/4}^{\pi/2} -(\frac{4}{\pi}x -2 ) dx +  \int_{\pi/2}^{3\pi/4} (\frac{4}{\pi}x -2 ) dx  +  \int_{3\pi/4}^{\pi} (- tanx ) dx $

$  = [ln secx]_{0}^{\pi/4}  +  [2x-\frac{2x^2}{\pi}] _{\pi/4}^{\pi/2} +  [\frac{2x^2}{\pi}-2x] _{\pi/2}^{3\pi/4} –   [ln secx]_{3\pi/4}^{\pi} $

$ = ln\sqrt{2} + (\pi – \frac{\pi}{2}-\frac{\pi}{2} + \frac{\pi}{8})  + (\frac{9 \pi}{8} -\frac{3 \pi}{2} -\frac{\pi}{2} + \pi) + \ln\sqrt{2}$

$= ln2 + \frac{\pi}{4}$

 

Illustration : For x > 0 ,  Let $ \displaystyle f(x) = \int_{1}^{x} \frac{ln t}{1 + t} dt $

Find the function f(x) + f(1/x) and show that f(e) + f(1/e) = 1/2 . Here lnt = loget

Solution:  $ \displaystyle f(x) = \int_{1}^{x} \frac{ln t}{1+t} dt $

$ \displaystyle f(\frac{1}{x}) = \int_{1}^{1/x} \frac{ln t}{1+t} dt $

Let $ \displaystyle t = \frac{1}{y} $

=> $ \displaystyle dt = -\frac{1}{y^2} dy $

$ \displaystyle f(\frac{1}{x}) = \int_{1}^{x} \frac{ln (1/y)}{1+ (1/y)} ( -\frac{1}{y^2}) dy $

$ \displaystyle f(x) = \int_{1}^{x} \frac{ln y}{y(1+y)} dy $

$ \displaystyle f(x) = \int_{1}^{x} \frac{ln t}{t(1+t)} dt $

$ \displaystyle f(x) + f(1/x)= \int_{1}^{x} \frac{ln t}{1+t} (1+\frac{1}{t})dt $

$ \displaystyle = \int_{1}^{x} \frac{ln t}{t}dt $

$ \displaystyle = \frac{1}{2}(ln x)^2 $

$ \displaystyle f(e) + f(1/e) = \frac{1}{2}(ln e)^2 = \frac{1}{2} $

Note: The expression for f(x) changes at the end points of each of the sub-interval. You might at times be puzzled about the end points as limits of integration. You may not be sure for x=0 as the limit of the 1st integral or the next one. In fact, it is immaterial, as the area of the line is always zero. Therefore even if you write

$ \displaystyle = \int_{-1}^{0} (1-2x)dx $

here as 0 is not included in its domain it is deemed to be understood that you are approaching x = 0 from the left in the first integral and from the right in the second integral. Similarly for x = 1

Exercise :Evaluate :

(a) $ \displaystyle \int_{1}^{3} |x^2-2x |dx $

(b) $ \displaystyle \int_{-4}^{3} |x^2-4 |dx $

(c) $ \displaystyle \int_{-\pi/3}^{\pi/3} \frac{\sqrt{1+sin2x}}{|cosx|} dx $

(d) $ \displaystyle \int_{0}^{\pi/3} |\frac{sin^{-1}sinx}{cos^{-1}cosx} |dx $

Properties:

$ \displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx $

In particular  $ \displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx $

Illustration : Evaluate:

(i) $ \displaystyle \int_{0}^{\pi/2}(sinx-cosx)log(sinx+cosx)dx $

(ii) $ \displaystyle \int_{0}^{2}\frac{dx}{(17+8x-4x^2)(e^{6(1-x)})+1} $

Solution: (i) Here $\displaystyle I = \int_{0}^{\pi/2}(sinx-cosx)log(sinx+cosx)dx $

Using Property ;

$ \displaystyle I = \int_{0}^{\pi/2}(sin(\pi/2 -x) -cos(\pi/2 – x))log(sin(\pi/2 -x ) + cos(\pi/2 – x))dx $

$ \displaystyle I = \int_{0}^{\pi/2}(cosx – sinx)log(cosx + sinx)dx $

$ \displaystyle I = – \int_{0}^{\pi/2}(sinx-cosx)log(sinx+cosx)dx $

⇒  I = − I

⇒  2I = 0

⇒  I = 0

Sol:(ii) Let $ \displaystyle I = \int_{0}^{2}\frac{dx}{(17+8x-4x^2)(e^{6(1-x)})+1} $ …(i)

Using , $ \displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx $

Also $ \displaystyle I = \int_{0}^{2}\frac{dx}{(17+8x-4x^2)(e^{-6(1-x)})+1} $ …(ii)

Adding, we get $ \displaystyle 2I = \int_{0}^{2}\frac{1}{(17+8x-4x^2)}(\frac{1}{e^{6(1-x)}+1}+\frac{1}{e^{-6(1-x)}+1})dx $

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Properties:

$ \displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a/2} [f(x)+f(a-x)] dx $

In Particular ,  If f(x) = f(a-x) then ,

$ \displaystyle \int_{0}^{a}f(x)dx = 2\int_{0}^{a/2} f(x) dx $

If f(x) = -f(a-x) then ,

$ \displaystyle \int_{0}^{a}f(x)dx = 0 $

Illustration: $\large \int_{0}^{\pi} \frac{x dx}{1 + cos^2 x}$

Sol: Let $\large  I = \int_{0}^{\pi} \frac{x dx}{1 + cos^2 x}$

$\displaystyle I = \int_{0}^{\pi} \frac{\pi – x dx}{1 + cos^2 (\pi – x)}$ ; (by using property)

$\displaystyle I = \int_{0}^{\pi} \frac{\pi dx}{1 + cos^2 x} – I $

$\displaystyle 2 I = \int_{0}^{\pi} \frac{\pi dx}{1 + cos^2 x}  $

$\displaystyle 2I =  2\pi \int_{0}^{\pi/2} \frac{dx}{1 + cos^2 x} $ ; (by using property)

$\displaystyle 2I =  2\pi \int_{0}^{\pi/2} \frac{sec^2 xdx}{2 + tan^2 x} $

Let tan x = t so that for x → 0, t → 0 and for x → π/2, t → ∞.

Hence we can write , $\large I = \pi \int_{0}^{\infty} \frac{dt}{2 + t^2} $

$\displaystyle = \frac{\pi}{\sqrt{2}} [tan^{-1}\frac{t}{\sqrt{2}}]_{0}^{\infty}$

$\displaystyle = \frac{\pi}{\sqrt{2}} . \frac{\pi}{2}$

$ \displaystyle = \frac{\pi^2}{2 \sqrt{2}}$

Properties :

$\displaystyle \int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(a-x) ] dx  $

In particular ,

$\displaystyle \int_{-a}^{a} f(x) dx =  2 \int_{0}^{a} f(x)  dx  , if \; f(x) \;is \;even $

$\displaystyle = 0 \; , if \; f(x) \; is \; odd $

Illustration: Evaluate : $\displaystyle \int_{-7/2}^{-5/2}\frac{2x^3 + 18x^2 + 58x + 66}{|(x+2) (x+3) (x+4)|} dx $

Solution: Put x = t -3  ⇒ dx = dt

$\displaystyle I = \int_{-1/2}^{1/2}\frac{2t^3 + 4 t}{|t(t^2 – 1)|} dt$

Which is clearly an odd function. Hence I = 0.

Illustration :If the function f : [–1 , 1] → R is continuous and even, then show that

$\displaystyle \int_{0}^{\pi/2} f(cos2x) cosx dx = \sqrt{2} \int_{0}^{\pi/4} f(sin2x) cosx dx $

Solution:  $\displaystyle  I = \int_{0}^{\pi/2} f(cos2x) cosx dx $

$ \displaystyle I = \int_{0}^{\pi/2} f(cos2(\frac{\pi}{2}-x)) cos(\frac{\pi}{2}-x) dx $

$\displaystyle  I = \int_{0}^{\pi/2} f(cos2x) sinx dx $ ; (f is an even function)

Hence , $\displaystyle  2I = \int_{0}^{\pi/2} f(cos2x) (sinx + cosx ) dx $

$ \displaystyle = \sqrt{2}\int_{0}^{\pi/2} f(cos2x) cos(x – \frac{\pi}{4}) dx $

Putting x = u + π/4

$\displaystyle = \sqrt{2}\int_{-\pi/4}^{\pi/4} f(cos2(u + \frac{\pi}{4})) cosu du $

$ \displaystyle = \sqrt{2}\int_{-\pi/4}^{\pi/4} f(-sin2u) cosu du $

$\displaystyle = \sqrt{2}\int_{-\pi/4}^{\pi/4} f(sin2u) cosu du $ , as f is even

$ \displaystyle 2 I = 2 \sqrt{2}\int_{0}^{\pi/4} f(sin2u) cosu du $

$\displaystyle  I = \sqrt{2}\int_{0}^{\pi/4} f(sin2u) cosu du $

Properties:

If f (x) is a periodic function with period T, then

(a) $\displaystyle \int_{a}^{a + nT} f(x)dx = n \int_{0}^{T} f(x)dx $ ; Where  n ∈ I

(i) In particular, if a = 0 ,

$\displaystyle \int_{0}^{nT} f(x)dx = n \int_{0}^{T} f(x)dx $ ; Where  n ∈ I

(ii) If n = 1,

$\displaystyle \int_{a}^{a+T} f(x)dx = n \int_{0}^{T} f(x)dx $

(b) $\displaystyle \int_{mT}^{nT} f(x)dx = (m-n) \int_{0}^{T} f(x)dx $ ; Where  n , m ∈ I

(c) $\displaystyle \int_{a + nT}^{b + nT} f(x)dx = n \int_{a}^{b} f(x)dx $ ; Where  n ∈ I

Solved Example : $\displaystyle \int_{0}^{4 \pi} |cos x | dx $

Solution: Note that |cos x| is a periodic function with period  π . Hence the given integral

$\displaystyle 4 \int_{0}^{\pi} |cos x | dx $ ; (using property )

$\displaystyle 4 [\int_{0}^{\pi/2} |cos x | dx + \int_{\pi/2}^{\pi} |cos x | dx ] $

$\displaystyle 4 [\int_{0}^{\pi/2} cos x dx – \int_{\pi/2}^{\pi} cos x dx ] $

$\displaystyle  = 4[sinx]_{0}^{\pi/2} – 4[sinx]_{\pi/2}^{\pi} $

= 4[1+1] = 8

Also Read:

→ Definite Integral : Fundamental Theorem Of Calculus
→ Properties of Definite Integral
→ Differentiation under the Integral sign : Leibnitz’s Rule
→ Inequalities involving in definite integrals

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