Definite IntegralThis section becomes easier to move on to the concept once they have fully grasped the idea of indefinite integrals and had opportunities to practice solving them. In some ways, indefinite integrals can be seen as a way to introduce the theory withdefinite integralsoffering more scope for lecturers and students to apply this knowledge to real world problems.

Many studentsgrasp integrationbut make silly errors when completing the two part calculation. Simplepeer assessmentcould help students to think about setting each step of their work minimizing these small but significant slips. This work could be made more engaging by inviting students to mix and match integration problems withgraphical representations.

This section needs a lot of practice to get better understanding. Broadly, this section makes the root of everyscientific organization and studies.

DEFINITE INTEGRATION2.1GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRALIf f(x) > 0 for all x[a, b]; thenbaf(x) is numerically equal to the area bounded by the curve y = f(x), then x-axis and the straight lines x = a and x = b i.e.baf(x)In generalbaf(x) dxrepresents to algebraic sum of the figures bounded by the curvey = f(x), the x-axis and the straight line x = a and x = b. The areas above x-axis are taken place plus sign and the areas below x-axis are taken with minus sign i.e,i.e.baf(x) dx area OLAarea AQM area MRB+area BSCDNote:baf(x) dx, represents algebraic sum of areas means, that if area of function y=f(x) is asked between a to b. =>Area bounded=ba|f(x)|dx and not been represented bybaf(x) dx

e.g., If some one asks the area of y=x3between-1 to 1. Then y=x3could be plotted as; Area=01x3dx +10x3dx = 1/2 or, using above definition Area=11|x3| dx = 210x3dx = 2 [x4/ 4]10= 1/2 But if, we integrate x3between-1 to 1.=>10x3dx = 0which does not represent area. Thus, students are adviced to make difference between area and definite Integral.

FUNDAMENTAL THEOREM OF CALCULUS (NEWTON-LEIBNITZ FORMULA)This theorem state that If f(x) is a continuous function on [a, b] and F(x) is any anti derivative of f(x) on [a, b] i.e.F'(x) = f (x) (a, b), thenO'baf(x)dx = F(b) F(a)

The function F(x) is the integral of f(x) and a and b are the lower and the upper limits of integration.Illustration 1: Evaluate22dx / 4 + x2directly as well as by the substitution x=1 /t. Examine as to why the answer do not tally?Solution: l =22dx / 4 + x2 = [1/2 tan1(x/2)]22= 1/2 [tan1(1) tan1(1)] = 1/2 [/4 (/4)] =/4 => l =/4 On the other hand; if x=1/t then, l =22dx / 4 + x2=1/21/2dt / t2(4 + 1 / t2) =1/21/2dt/ 4t2+ 1 = [1/2 tan1(2t)]1/21/2 = 1/2 tan1 (1/2 tan1(1) =/8 /8 = /4I=/4when x= 1/t In above two results l=-/4 is wrong. Since the integrand 1/4+x2> 0 and therefore the definite integral of this function cannot be negative. Since x=1/t is discontinuous at t=0, the substitution is not valid (I =/4). Note: It is important the substitution must be continuous in the interval of integration.Illustration 2:Let =0dx /x4+ 7x2+ 1 and =0x2dx /x4+ 7x2+ 1then show that= Solution:=0dx /x4+ 7x2+ 1put x = 1/t => dx = 1/t2then =0/t2dt / 1/t4+ 7/t2+ 1 =0dtt2/t4+ 7t2+ 1 =0t2dt /t4+ 7t2+ 1 =

PROPERTIES OF DEFINITE INTEGRATION1. Change of variable of integration is immaterial so long as limits of integration remain the same i.e.baf(x)dx =baf(t) dt2.baf(x)dx = baf(x)dx3.baf(x)dx =caf(x)dx +baf(x)dxGenerally we break the limit first at the points where f(x) is discontinuous and second at the points where definition of f(x) changes.Illustration 3: Evaluate5/120[tanx]dx, where [.] is the greatest integer function.Solution:Let I =5/120[tanx] dx

Value of tan x at x = 5/12 Value of tan x at x = 0 is 0 Integers between 0 and 2+ 3are 1, 2, 3 tan x = 1, tan x = 2, tan x = 3 =>x = tan-11, x = tan-12, x = tan-13I =+tan10[tanx] dx +tan12tan11[tanx]dx +tan13tan12[tanx] dx +15/12tan13[tanx]dx =tan100 dx +tan1 2tan11 1 dx +tan1 3tan122 dx +15/12tan133dx =0 + (tan12 tan11) + 2 (tan13 tan12) + 3 (5 /12 tan13) = 5 / 4 /4 tan13 tan12

== [tan1(3 + 2 / 1 6) +] = tan-1(-1) = /44.a0f(x)dx =a0f(ax) dx.Illustration 4:If f, g, h be continuous function on [0, a] such that f(a - x) = f(x), g(a - x) = - g(x) and 3h(x) - 4h(a - x) = 5, then prove thata0f(x) g(x) h(x)dx = 0.Solution: I = = a0f(x) g(x) h(ax) dx

7I = 3I + 4I =a0f(x) g(x) {3h(x)4h(ax)} dx

= 5a0f(x) g(x) dx= 0, since f (a x) g (a x) = f (x) g (x) => I = 0Illustration 5:0x sin 2 x sin ( / 2 cos x) dx

Solution: Let I =0x sin 2x sin (/2 cosx) dx .(1) =0( x)sin 2 ( x) sin (/2 cos ( x)) dx =0( x) (sin 2x) (sin 2x) sin ( /2 cosx) dx

=0( x) sin 2x sin (/2 cos x) dx .(2) Adding (1) & (2), we get 2I =0sin2 x sin (/2 cos x) dx

=> I =/202 sin x cos x sin (/2 cos x) dx

Put/2cos x = z =>/2 sin x dx = dz

=0/2z sin z sz = 8/.5.a0 f(x) dx =a/20 [f(x) + f(ax)] dx

Special cases:If f (x) = f (a x), thena0 f(x)dx = 21/20 f(x) dx.If f (x) = f (a x), thena0 f(x) dx = 0.6.aa f(x) dx =a0 [f(x0 + f(x)]dx

Special case:aa f(x) dx = 2a0 f(x)dx, if f(x) is even,aa f(x) dx = 2, if f(x) is oddIllustration 6:Evaluate44x2/ (x2+ 16) (1+ex5) dx

Solution: Let I =44f(x) / (1 + ex5) dx (f(x) = x2/ x2+ 16)

2I =44f(x) / (1 + ex5)dx +44f(x) / (1 + ex5)dx =dx =44f(x) dx

I =40x2/ x2+ 16dx = 4 tan-11Illustration 7:Find the value ofis0dx / 1 + 5cosx+22log (5 x / 5 + x) dx is

Solution: Let I = I1+ I2 Consider I1=0dx / 1 + 5cosx (1) Now I1=0dx / 1+ 5cos(x)=0dx / 1+ 5cosx=05cosxdx / 5cosx+ 1(2) Adding (1) and (2) , we get 2I1=0dx / 1+ 5cosx+05cosxdx / 5cosx+ 1 =01.dx = I1=/2 Consider I2=22log (5x / 5+x) dx

Let g(x) = log (5x / 5+x) Now g(-x) = log(5 (x) / 5+x) = log 5x/5+x = g(x) g(x) is an odd function22g(x)dx = 0=>I2= 0 I = I1+ I2=/2+ 0 =/27.10f((ba) x + a) dx

Illustration 8: Evaluate2/31/3e9(x2/3)2dx

Solution:l1=54e(x+5)2dx

= (54)10e((5+4) x4+5)2dx

10e(x1)2dx (i) Again let I2=2/31/3e9(x2/3)2dx

I2= (2/3 1/3)10e9[(2/3 1/2) x + 1/3 2/3]2dx

= 1/310e(x1)2dx

where I = I1+ 3I2 l1+ 3(l1/3) = l1 l1 I = 02/31/3e9(x2/3)2dx= 08. If f (x) is a periodic function with period T, thena+nTaf(x)dx = nT0f(x) dx where n 1,

In particular, (i) if a = 0,nT0f(x) dx = nT0f(x) dxwhere nI (ii)Ifn = 1,a +Taf(x)dx =T0f(x)dx

Illustration 9:Evaluate100|sinx|dx.Solution: LetI =100|sin x|dx Weknowthat |sinx| is aperiodicfunctionwithperiod p Hence I = 100|sin x| dx [applying prop. 8]Illustration 10:If f(x) is a function satisfying f(x+a)+f(x)=0 for all xR and constant a such thatc+bbf(x)dx is independent of b, then find the least positive value of c.Solution: We have f(x + a) + f(x) for all xR (i) => f(x + a + a) + f(x + a) = 0 [Replacing x by x + a] .(ii) => f(x + 2a) + f(x+ a) = 0 .(iii) Subtracting (i) from (ii), we get f(x+ 2a) - f(x) = 0 for all xR. => f(x + 2a) = f(x) for all xR So, f(x) is periodic with period 2a It is giventhatc+bbf(x)dx is independent of b. The minimum value of c is equal to the period of f(x) i.e., 2a.

2.2DIFFERENTIATION UNDER THE INTEGRAL SIGNLeibnitzs RuleIf g is continuous on [a, b] and f1(x) and f2(x) are differentiable functions whose values lie in [a, b], then d/dx f2(x)f1(x)g(t)dt = g(f2(x))f2'(x) g(f1(x))f1'(x)Illustration 11:If f(x) = cos x -x0(xt) thenshow that f(x) + f (x) = cos x

Solution:f' (x) = sin x (xf (x) +x0f(t)dt) + xf(x) = sin x x0f(t)dt f" (x) = cos x f(x)Illustration 12If a function f(x) is definedxR such that. Prove thataxf(t)/t dx. Prove thata0g(x) dx =a0f(x)dxSolution: g(x) =axf(t)/t dt Diffrentiate w.r.t. x g(x) = F(x)/x F(x) = -x g(x)a0f(x)dx = a0xg'(x) dx =a0f(x)dx= ag(a) +a0g(x) dx =a0g(x)dx, [g(a) = 0]Illustration 13 Determine a positive integer n5, such that10ex(x1)ndx = 16 6e

Solution:Let ln=10ex (x1)ndx

Integrating by parts, In=In= [ex(x1)n]1010ex.n(x1)dx = (1)n+1= nln1 (i) Also, l1=10ex.n(x1)n1dx = [(x1)ex]1010exex.1dx = (1)[ex]10= 1(e1) = 2 e From (i), I2= ( -1)3- 2I1= -1 - 2(2 - e) = -5 + 2e and I3= (-1)4- 3I2= 1 - 3(-5 + 2e) = 16 - 6e Which is given . n = 3

DEFINITE INTEGRAL AS LIMIT OF A SUMAn alternative way of describing baf(x)dx is that the definite integralbaf(x) dxis a limiting case of the summation of an infinite series, provided f(x) is continuous on [a, b] i.e.,baf(x)dx = limn->h n1r=0f(a+rh) where h = ba/n. The converse is also true i.e., if we have an infinite series of the above form, it can be expressed as a definite integral.Method to express the infinite series as definite integral:(i) Express the given series in the form 1/n f (r/n)

(ii) Then the limit is its sum when n > , i.e.limn->h 1/nf(r/n)(iii) Replace r/n by x and 1/nby dx andlimn->by the sign of .(iv) The lower and the upper limit of integration are the limiting values of r/nfor the first and the last term of r respectively.Some particular cases of the above are(a)limn->nr=11/n f(r/n) orlimn->n1r=01/n f(r/n) =10f(x)dx

(b)limn->pnr=11/n f(r/n) =f(x)dx where=limn->r/n = 0 (as r = 1) and =limn->r/n = p (as r = pn)

Illustration 14 Show that (A)limn->(1/n+1 + 1/n+2 + 1/n+3 + ... + ... 1/n+n)= ln2. (B)limn->1p+ 2p+ 3p+ ... + / np+1=1/p+1 (p > 0)Solution: (A) LetI =limn->(1/n+1 + 1/n+2 + 1/n+3 + ... + 1/n+n) =limn->(1 / n+1 + 1/n+2 + 1/n+3 + ... + 1/n+n)=limn->1/nnr=11 / 1+r/n Now=limn->1/n= 0 (asr = 1) and=limn->r/n= 1(asr = n) =>l 101/1+x dx= [In (1+x)]10=> I = ln2. (B) 1p+ 2p+ 3p+ ... + np/ np+1=nr=11p/n.np=nr=11+n(r/n)p Take f(x) = xp; Let h = 1/nso that as n > ; h > 0 limn->nr=11/n f(0 + r/n)

=10f(x)dx =10xpdx

= 1/p+1

Section c AREA AS DEFINITE INTEGRALLet f (x) be a continuous function in (a, b). Then the area bounded by the curve y = f (x),x axis and lines x = a and x = b is given by the formulaA = |baf(x)dx|,provided f (x)>0 (or f(x)1 < sin2x/4 + cos x/4 < 2[sin2x/4 + cos x/4] = 1 Now we have to find out the area enclosed by the circle |z| = 2, parabola (y -3/4) = (x+1/2)2, line y = 1and x-axis.Required area is shaded area in the figure. Hence required area=3 1 + (3 1) +01(x2+ x + 1).dx + 2234x2dx

=(2/3 +3 1/6) sq. unitsIllustration 17:Let f(x) = Max. {sinx, cos x,1/2} then determine the area of the region bounded by the curves y = f(x), x-axis, y-axis and x = 2.Solution: f(x)= Max {sin x, cos x,1/2} interval value of f(x) for 0g(x) > 0x > 1 and g(x)>0 x (0, 1]=>ln x>f(x) x (0, 1]19:, where [.] denotes the greatest integer function, and xR+, is equal to (A) 1/In2 ([x] + 2[x]1) (B)1/In2 ([x] + 2[x]) (C)1/In2 ([x] 2[x]) (D)1/In2 ([x] + 2[x]+1)Solution:Let n0

20: Let I1=then (A) I1> I2> I3> I4 (B) I2> I3> I4> I1 (C) I3> I4> I1> I2 (D) I2> I1> I3> I4

21: Area bounded by y = g(x), x-axis and the lines x = -2, x = 3, where and f(x) = x2-, is equal to(A) 113/24 sq.units (B) 111/24 sq.units (C) 117/24 sq.units (D) 121/24 sq,units

22: Areaof the region which consists of all the points satisfying the conditions |xy| + |x+y|2, is equal to (A) 4(7 ln8)sq. units (B) 4(9 ln8)sq. units (C) 2(7 ln8)sq. units (D) 2(9 ln8)sq. unitsSolution:The expression|xy| + |x+y|2. represents the region lying inside the hyperbola xy =2 Required area = 4(73 In2) sq.units = 4(7 In8) sq.units

23: Area bounded by the parabola (y - 2)2= x 1, the tangent to it at the point P (2, 3) and the x-axis is equal to (A) 9 sq. units (B) 6 sq. units (C) 3 sq. units (D) None of theseSolution: (y - 2)2= (x - 1)=>2(y - 2).= 1=> dy/dx = 1/2(y2) Thus equation of tangent at P(2, 3) is, (y 3) = 1/2 (x2) i.e., x = 2y 4 Required area= ((y2)3/3 y2+ 5y)30= 9 sq. units24: Two lines draw through the point P(4, 0) divide the area bounded by the curvesy = 2 x/4 and x axis, between the linea x = 2, and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to (A)2 2/ (B)2/ (C)2/ (D)42/Solution:Area bounded by y =2 .sin x/4and x-axis between the lines x = 2 and x = 4, Let the drawn lines are L1: y m1(x - 4) = 0 andL2: y m2(x - 4) = 0, meeting the line x = 2 at the points A and B respectively Clearly A = (2, - 2m1); B= (2, -2m2)25: If A =is equal to (A) 1/+2 A (B) 1/2 + 1/+2 A

(C)1/2 1/+2 A (D) 1/2 + 1/+2 + A

26. The value of the integralis (A) 1 (B)/12 (C)/6 (D) none of theseSolution: Using the propertyf (a + b x) dx, the given integral

Hence (B) is the correct answer.27. IfI =dx then (A) 0 (B) 2 (C)/2 (D) 2 /2

Hence (D) is the correct answer.

28. If I =, then (A) 0 < I < 1 (B) I >/2 (C) I 2Solution:Since x [0,/2]=>1