INTEGRATION
Integration or the process of finding the antiderivative is one of the most important operations in calculus. It is the opposite process of differentiation.
ANTIDERIVATIVE OR INDEFINITE INTEGRAL
If f x
` a
is the derivative of the function F x , that is F. x = f x for all x in the
` a ` a ` a ` a ` a
domain of f, then F x is called the antiderivative or indefinite integral of f x . We write Z f x dx = F x + C
` a ` a
The antiderivative of a given function is not unique. For example,
F x = x 2 @ 2, f x = 2x
` a ` a
G x = x 2 + 5, and
` a
H x = x2
` a
are all antiderivatives of
as
b c d b c d b c ` a ` a ` a ` a dff ff ff ff ff ff ff ff ff F. x = G. x = H. x = f x or ff x 2 @ 2 = ff x 2 + 5 = ff x 2 = 2x for all x in the dx dx dx
domain of f .
We note that, all the antiderivatives of the function f differ only by a constant (the derivative of the constant value is always zero). Geometrically, this means that the graphs of F x , G x and H x derivatives will be same for any x = x 0 . All indefinite integrals of f x = 2x are then expressed by the general form of the antiderivative which is F x = x 2 + C where C is the constant of integration or an arbitrary constant.
` a ` a ` a ` a ` a
are identical except for their vertical position. All their
The notation Z is used for integration. The symbol Z f x dx is used to denote the
` a
indefinite integral of the function f(x). The function f(x) is called the integrand.
So, we can write Z 2x dx = x 2 + C .
The indefinite integral of a function is sometimes called the general antiderivative of the function.
INTEGRATION BY INSPECTION
We shall solve the first few questions by method of inspection, that is, comparing the integrand with some known derivatives. f x = sin x
` a
Example : Find the indefinite integral of the function Solution :
We know that the derivative of the function so we write
Z sin x dx = @ cos x + C
F x = @ cos x
` a
is F. x = sin x
` a
Example : Find the indefinite integral of the function Solution :
We know that dx
b dff c ff 3 ff ff
f x = x2
` a
x = 3x , or
2 3
dx
f 3g dff ff ff xff ff ff ff f
3
1f f f = A 3x 2 = x 2 3
As the derivative of F x = xff ff f Z x 2 dx = ff+ C 3
3
` a
` a xff ff ff f is F. x = x 2 3
STANDARD INTEGRATION FORMULAS: COMPARISON WITH DIFFERENTIATIONS
In the above cases of integration by inspection, the antiderivatives were found by comparing the integrand with some known derivatives. But it is not always possible to do
so, as in most cases the integrand does not match with our known derivatives. So, we need some integration formulas. As integration is the reverse process of differentiation, we can make the first few integration formulas directly from the corresponding derivative formulas. These are given below :
For Polynomial Functions:
Differentiation Formula dff ff` ff ff xn = nA xn@1
a
Corresponding Integration Formula xffff ffff ffff Z x n dx = fffff C + n+1
Z dx = x + C
n+1
dx dx
dff ff` a ff ff
x = 1
For Trigonometric Functions:
Differentiation Formula
a dff ff` ff ff sin x = cos x dx
Corresponding Integration Formula
Z cos x dx = sin x + C Z sin x dx = @ cos x + C Z sec 2 x dx = tan x + C Z csc 2 x dx = @ cot x + C Z sec x tan x dx = sec x + C Z csc x cot x dx = @ csc x + C
dx
dff ff` ff ff dff ff` ff ff dff ff` ff ff dff ff` ff ff dff ff` ff ff
cos x = @ sin x
a
dx dx
tan x = sec 2 x
a a
cot x = @csc 2 x sec x = sec x tan x
a a
dx dx
csc x = @ csc x cot x
For Exponential & Logarithmic Functions: Differentiation Formula dff ff` x a ff ff e = ex a = a x ln a Corresponding Integration Formula
Z e x dx = e x + C
dx dx
a 1f f f ln x = f x dx
dff ff` x a ff ff dff ff` ff ff
afff fff fff Z a x dx = fff+ C a>0, a ≠ 1 ln a 1f f f Z fdx = ln |x| + C x
x
INTEGRATION OF COMBINATION OF FUNCTIONS
The following rules of integration for addition, subtraction and scaler multiples of functions can be used. Note that, unlike differentiation, product and quotients of functions are not covered in these (which are to be solved using substitution or integration by parts or by some other methods).
Z c f x dx = c Z f x dx
` a ` a
Z f x F g x dx = Z f x dx F Z g x dx
B ` a ` aC ` a ` a
Example : Evaluate Z x 3 dx Solution :
3+1 xffff 1f ffff fff f f Z x 3 dx = ffff+ C = fx 4 + C
3+1
4
1f fff fff fff f w dx w w w w w Example : Evaluate Z pwf x Solution :
1f f f f f w w w w w w w xfffff 1f fff fff ff f ff2fff ff fff f Z fwf = Z x @ 2 dx = ffffff C = 2 p x + C w dx w w w w w + 1f px f f f @ +1 2
f f @ f+ 1
1
SUBSTITUTION METHOD: CHANGE OF VARIABLES
Till now we have discussed integrals of those functions which are readily obtained from derivatives of some known functions. They could be solved using the standard formulas. But some integrals can not be evaluated directly as no standard formula matches the form of the given function.
Z tan3 x sec 2 x dx .
wwww wwww wwww www www www www www Z x q x 2 + 5 dx ,
` a2 lnffff fffff ffff fx ff ff Z
Some examples are
x
dx ,
One way of solving them is to substitute some new variables in the integral and thus transforming it to standard form with the new variable. After choosing the suitable variable, we have to rewrite the integral in terms of the new variable, so that one or more of the standard formulas can be used. After the integration process is over, the result is to be written in terms of the original variable.
Example : Evaluate Z 3x x 2 + 1 dx
b
c4
Solution :
Let t = x 2 + 1 dtf ff ff ff f [ = 2x dx 1f f [ x dx = fdt 2
4 3f f Z 3x x 2 + 1 dx = Z ft 4 dt
b
c
2
5 3f tff 3f f f f f ff ff = A f C = fft 5 + C + 2 5 10 c5 3fb ff ff = ff x 2 + 1 + C 10
Example : Evaluate Z sin 3x dx Solution :
Let 3x = t dx = 1f f f dt 3 3
1f f Z sin 3x dx = Z fsin t dt 1f 1f f f f f = @ cos t + C = @ cos 3x + C 3 3
Example : Evaluate Z tan2 2x @ 3 dx
` a
Solution :
Let 2x @ 3 = t 1f f f dx = dt 2
` a 1f f Z tan2 2x @ 3 dx = Z ftan2 t dt
2
c 1f b 1f f f f f = Z sec 2 t @ 1 dt = J Z sec 2 t dt @Z dt K 2 2 @ A 1f f f tan t @ t + C 2 B a ` aC 1f ` f f = tan 2x @ 3 @ 2x @ 3 + C 2
H
I
=
OTHER INTEGRATION FORMULAS
Apart from the standard formulas given above, the following formulas are also used in integration (these are obtained by using the substitution method) :
Z tan x dx = @ ln |cos x| + C Z cot x dx = ln |sin x| + C Z sec x dx = ln |sec x + tan x| + C Z csc x dx = @ ln|csc x + cot x| + C
dx xf ffffff 1f fffff fffff f f f f Z ffffff = farctan f + C 2 2
a a x +a L M dx 1ff L xffff @ aM ffffff ffffff fffff ff ff fffff ffff M Z ffffff = ff lnL fffff + C L M 2 2 2a L x + a M x @a
L M dx 1ff L fffff + xM ffffff ffffff ffffff fffff ff L affff ff ff ffff ffff M Z 2 M+ = lnL 2a La @ xM a @ x2 q x2 +
C
L M dx ffffffff ffffffff fffffff ffffff Z fwwwwf = lnLx + q x 2 + a 2 M + C wwww wwww wwww wwww wwww wwww wwww L M
a2
L L
wwww M wwww wwww wwww wwww wwww wwww wwww
L M dx fffffff fffff ffffffff fffffff Z ffwwwff= lnLx + q x 2 @ a 2 M + C wwwww wwww wwww wwww wwww wwww wwww w L M q x2 @ a2
wwwww wwww M wwww wwww wwww wwww wwww wwww
dx xf fffffff fffff f f ffffffff fffffff Z ffwwwff= arcsin f+ C wwwww wwww wwww wwww wwww wwww wwww w
qa 2 @ x 2
a
L M 1f 1f f f f f Z q x 2 + a 2 dx = fx q x 2 + a 2 + fa 2 lnLx + q x 2 + a 2 M + C L M
wwwww wwwww wwwww wwww wwww wwww wwww wwww
wwwww wwwww wwwww wwww wwww wwww wwww wwww
2
2
wwwww wwwww wwwww wwww wwww wwww wwww wwww q x 2 @ a 2 dx Z wwwww wwwww wwwww wwww wwww wwww wwww wwww
=
1f 1f xf f f f f f f Z q a 2 @ x 2 dx = fx q a 2 @ x 2 + fa 2 arcsin f + C 2 2 a dx xf fffffffff ffffff f f f f ffffffffff 1f fffffffff Z fffwwwwf = farcsec f+ C wwwww wwww wwww wwww wwww wwww wwww x q x2 @ a2 a a
wwwww wwwww wwwww wwww wwww wwww wwww wwww L 1f q 2 1f 2 L f f f f f f 2 x x @ a @ a lnLx L
L
2
2
wwwww wwwww wwwww wwww wwww wwww wwww wwww
wwwww wwww M wwww wwww wwww wwww wwww wwww q x2 @ a2 M + M + M
wwww M wwww wwww wwww wwww wwww wwww wwww
C
INTEGRATION BY PARTS
Another useful integration technique for indefinite integrals which do not fit in the basic formulas is integration by parts. We may consider this method when the integrand is a product of two functions.
When u and v are differentiable functions of x which is the variable of integration, then d uv = u dv + v du,
` a
or
u dv = d uv @ v du
` a
Integrating both sides we get the following formula for Integration by parts,
Z u dv = uv @ Z v du
While using this method of integration, the given integral is to be separated into two parts, one part being u and the other part, combined with dx, being dv. For this reason this method is called integration by parts. (a) We have to first choose dv which must be readily integrable to find v. The u function will be the remaining part of the integrand that will be differentiated to find du. (b) The purpose is to find an integral Z v du which is easier to integrate than the original integral Z u dv .
Example : Evaluate Z x sec 2 x dx Solution :
Let u = x so du = dx and dv = sec 2 x dx using hence then v =Z dv =Z sec 2 x dx = tan x
Z u dv = uv @ Z v du Z x sec 2 x dx = x tan x @Z tan x dx
= x tan x @ @ ln |cos x| + C = x tan x + ln |cos x| + C
b c
Example : Evaluate Z ln x 2 + 4 dx
b c
Solution : Let u = ln x 2 + 4
b c
so
2x fffff fffff ffff du = fffffdx 2 x +4
and dv = dx using
b
then v =Z dv =Z dx = x
Z u dv = uv @ Z v du
c b c
2
2x ff fffff fffff ff Z ln x 2 + 4 dx = x ln x 2 + 4 @ Z fffffdx 2 x +4
F 8 G fffff fffff ffff = x ln x 2 + 4 @ Z 2 @ fffff dx 2 x +4 b c b c F 8 G fffff fffff ffff = x ln x 2 + 4 @ Z 2 @ fffff dx 2 x +4 b c xf f f f = x ln x 2 + 4 @ 2x + 4 arctan + C 2
INTEGRATION USING TRIGONOMETRIC IDENTITIES
When the integrand involves trigonometric functions which can not be solved by the basic integration formulas, we need to use the different trigonometric identities (together with substitution if needed) to get them into a form in which the basic integration formula scan be applied. The trigonometric identities often used for integration are
1 1f 1 ffff fff fff fff fff f fff fff fff sec t = ffff csc t = fff cot t = ffff cos t sin t tan t 2 sin t + cos 2 t = 1 tan2 t + 1 = sec 2 t cot 2 t + 1 = csc 2 t 1fffffffff @ cos 2α 2 ffffffff ffffffff sin α = fffffffff 2 1ffffffff + cos 2α ffffffff ffffffff cos 2 α = fffffffff 2 1f f sinαcosα = fsin 2α 2 D c b cE 1f b f f sinαcosβ = sin α + β + sin α @ β 2 D c b cE 1f b f cosαsinβ = f sin α + β @ sin α @ β 2 D c b cE 1f b f f cosαcosβ = cos α + β + cos α @ β 2 D c b cE 1f b f sinαsinβ = f cos α @ β @ cos α + β 2
sinff ffff fff ft tan t = ffff cos t
3 Example : Evaluate Z sin x cos 3 x dx
Solution :
Z sin3 x cos 3 x dx = Z sin3 x cos 2 x A cos xdx
= Z sin x 1 @sin x A cos x dx let sin x = t, cos x dx = dt
3
b
2
c
B
C
= Z t 3 @ t 5 dt
b c
1f 4 1f 6 f f f f f t @ t +C 6 4 1f 4 1f 6 f f f f f = sin x @ sin x + C 4 6 =
Example : Evaluate Z sec 6 x dx Solution :
Z sec 6 x dx = Z sec 4 x Asec 2 x dx
= Z sec 2 x Asec 2 x dx
b b b
c2
= Z tan2 x + 1 Asec 2 x dx = Z t 2 + 1 A dt = Z t 4 + 2t 2 + 1 dt
b c c2
c2
B
let tan x = t [ sec 2 x dx = dt
C
=
1f 5 2f 1f 5 2f 3 f f f f f f f f t + t +t = tan x + tan3 x + tan x + C 3 5 3 5
TRIGONOMETRIC SUBSTITUTION
The following two substitutions are useful in many cases : 1 A For Z sin x cos n x dx : If m is odd, we substitute u = cos x A If n is odd we substitute
m
u = sin x A 2 A For Z tanm x sec n x dx : If m is odd, we substitute u = sec x A If n is even we substitute u = tan x A
Some more integrations may be simplified by the following substitution :
wwww wwww wwww wwww wwww wwww wwww wwww q a 2 @ x 2 , substitute x = a sin t 1 A For the integrand in the form wwww wwww wwww wwww wwww wwww wwww www q a 2 + x 2 , substitute x = a tan t 2 A For the integrand in the form wwww wwww wwww wwww wwww wwww wwww wwww q x 2 @ a 2 , substitute x = a sec t 3 A For the integrand in the form
For the integrands in the forms q a 2 @ b x 2 , q a 2 + b x 2 and q b x 2 @ a 2 , af af af f f f f f f f f f we have to substitute x = sin t, x = tan t, x = sec t respectively A b b b
2 2 2
wwwwww wwwwww wwwwww wwwww wwwww wwwww wwwww wwwww
wwwwww wwwwww wwwwww wwwww wwwww wwwww wwwww wwwww
wwwwww wwwwww wwwwww wwwww wwwww wwwww wwwww wwwww
The above substitutions have been used in making the follwoing formulas: wwwww wwwww wwww wwww wwww wwww wwww wwww wwwww wwwww wwww wwww wwww wwww wwww wwww wwww M wwww wwww wwww wwww wwww wwww www L M 1f 1f L f f Z q x 2 + a 2 dx = fx q x 2 + a 2 + fa 2 lnLx + q x 2 + a 2 M + C L M 2 2
M 1f 1f L f f Z q x 2 @ a 2 dx = fx q x 2 @ a 2 @ fa 2 lnLx + q x 2 @ a 2 M + C M L wwwww wwwww wwwww wwww wwww wwww wwww wwww wwwww wwwww wwwww wwww wwww wwww wwww wwww
2 2
2
wwwww wwwww wwwww wwww wwww wwww wwww wwww wwwww wwwww wwwww wwww wwww wwww wwww wwww 1f 1f xf f f f f Z q a 2 @ x 2 dx = fx q a 2 @ x 2 + fa 2 arcsin f + C
L
wwww M wwww wwww wwww wwww wwww wwww wwww
2
a
3 Example : Evaluate Z sin x cos 4 x dx
Solution :
Z sin3 x cos 4 x dx
=Z sin x cos 4 x sin x dx
2
=Z 1 @cos 2 x cos 4 x sin x dx
b b c
taking cos x = t, sinx dx = @ dt
=Z 1 @ t 2 t 4 @ dt
c `
1f 7 1f 5 f f f f f t @ t +C 7 5 1f 1f 7 f f f f f = cos x @ cos 5 x + C 5 7 =
a
Example : Evaluate Z tan3 x sec 4 x dx
Solution :
Z tan3 x sec 4 x dx
=Z tan2 x sec 3 x A sec x tan x A dx =Z t 2 + 1 t 3 dt
b c
taking sec x = t, sec x tan x dx = dt
1f 1f f f f f f = t5 + t4 + C 5 4 1f 1f f f f f f = sec 5 x + sec 4 x + C 4 5
fffdxffff fffffffff fffffffff ffffff ff wwww wwww wwww wwww www www www Example : Evaluate Z fffwwwwf 2q x 4 @ x2
xf f f Solution : Taking a right triangle with one acute angle t such that sin t = f 2 x 2
t
wwww wwww wwww wwww wwww www www www q4 @ x 2
fffdxffff fffffffff fffffffff ffffff ff ffffffffff fffdx fffff fffffffff ffffff ff Z fffwwwwf Z fffwwwwf wwww = wwww wwww wwww www www www wwww wwww wwww wwww wwww wwww www
x 2 q4 @ x 2
x 2 q2 @ x 2
2
B
taking x = 2 sin t, dx = 2 cos t dt
C
2 cos fdt fffffffffff fffffffffff ffffft ffff f = Z fff2ffffffff 4 sin t A 2 cos t 1f f f Z = f cosec 2 t dt 4 1f f f = @ fcot t + C 4wwww wwww wwww wwww wwww www www www
q4 @ x 2 fffffff fffffff fffffff = @ fffffff + C 4x
INTEGRATION USING PARTIAL FRACTIONS
Pffff x ffff ffff ffff ` a where A rational function is defined as the ratio of two polynomials in the form Q x
` a ` a
P(x) and Q(x) are polynomials in x and Q x ≠ 0. If the degree of P(x) is less than the
degree of Q(x), then the rational function is called proper, otherwise, it is called
improper. Like the case of improper fractions reducible to proper fractions, improper
rational functions can be reduced to proper rational functions by long division process.
Pffxff x x3 ffff ffff fffff fffff ffff fffff fffff ffff 1 f ` a . and fffffis a proper rational fraction. For example , fffff= x + fffff 2 2 Q x x @1 x @1
Pffff xf Pffff ` a Pffxff xf ffff fff ffff fff ffff ffff 1 f ` ` = ` a Thus, if fffafis improper, then fffaf T x + fffffwhere T(x) is a polynomial in x Q x Q x Q x
` a
` a
` a
` a
The rational fractions, which we shall consider here for integration, will be those whose denominator can be factored into linear and quadratic factors. We shall write the integrand as a sum of simpler rational functions by a method called
partial fraction decomposition.
Pffff xf Pffff x ffff fff ffff ffff ` ` a Assume that we want to evaluate Z ffffdx , where fffafis a proper rational function. Q x Q x
` a
` a
We can write the integrand as a sum of simpler rational function as per some of the examples in the following table :
Form of the rational function px + q ffffffffffff fffffffffff fffffffffff fffffffffff ` a` a,a ≠b x @a x @b pxf+fff fffffff ffffff ffffff f fq
`
x @a
px + qx + r fffffffffffffffff fffffffffffffffff fffffffffffffffff fffffffffffffffff ` a` a` a,a ≠b ≠c
2
a2
Form of the partial fractions B A fffff fffff ffff ffff ffff ffff ffff ffff + x @a x @b A B fffff fffffff ffff ffffff ffff ffffff ffff ffffff +` x @ a x @ aa2
B C A fffff fffff ffff ffff ffff ffff ffff ffff ffff ffff ffff ffff x @a
x @a x @b x @c
2
+
x @b
+
x @c x @b
px + qx + r ffffffffffff ffffffffffff ffffffffffff fffffffffff ` a` a2 , a ≠ b x @a x @b
2
px + qx + r ffffffffffffffff ffffffffffffffff ffffffffffffffff ffffffffffffffff c ` ab 2 x @ a x + bx + c
fffffffff fffff1fff fx fffffff f +f Example : Evaluate Z fffffffffdx 2 x @ 5x + 6
2
x @a
A Bx + C fffff fffffffff F where x 2 + bx + c ffff fffffffff ffff fffffffff ffff ffffffff
B C A fffff fffff fffffff ffff ffff ffffff ffff ffff ffffff ffff ffff ffffff + +` a2 x @a x @b
+
x 2 + bx + c
cannot be factored further
G
Solution : By long division we get x2 + 1 5x @ f 5x @ 5 fffffffff fffffffff fffffffff fffffffff fffffffff fffffffff fffff5fff fffffffffff fffffffffff fffffffffff a` a = 1 + fffffffff= 1 + `ffffffffffff 2 2 x @2 x @3 x @ 5x + 6 x @ 5x + 6 5x @ 5 A B fffffffffff fffffffffff fffffffffff ffff ffff ffff ffff ffff ffff a` a Let `ffffffffffff= fffff fffff + x @2 x @3 x @2 x @3 ` a ` a 5x @ 5 = A x @ 3 + B x @ 2 For x = 2, @ A = 5, so A = @ 5 For x = 3, B = 10 x2 + f 5 fffffffff fffff1fff fffffffff ffff ffff ffff ffff ffff 10ff Thus fffffffff= 1 @ fffff fffff + ff 2 x @2 x @3 x @ 5x + 6 x +1 fffffffff fffffffff fffffffff Z fffffffffdx 2
2
x @ 5x + 6
f1ff ffff ffff f f1ff ffff ffff f dx dx = Z dx @ 5 Z fffff + 10 Z fffff x @2 x @3 = x @ 5 ln |x @ 2| + 10 ln |x @ 3| + C
dx fffff fffff ffff Example : Evaluate Z fffff 4 x @1 Solution :
x @1
1 A B ffffffffffffffffff ffff ffff Cxfffff fffffffffffffffff ffff ffff ffffff fffffffffffffffff ffff ffff fffff f+ D c= Let `ffffffffffffffffff ffff+ ffff+ ffffff a` ab 2 2 x @1 x + 1 x + 1 x @1 x + 1 x + 1 For x = 1,
` ab
1 1 1 fffff fffffffffffff ffffffffffffffffff fffff fffffffffffff ffffffffffffffffff fffff fffffffffffff fffffffffffffffff ffff cb c c = b ffffffffffff= ` fffffffffffffffff a` ab 4 x2@ 1 x2 + 1 x @ 1 x + 1 x2 + 1
1 = A x + 1 x 2 + 1 + B x @ 1 x 2 + 1 + Cx + D x @ 1 x + 1 1f f f 1 = 4A, A = f 4
c ` ab c ` a` a`
a
For x = @ 1, For x = 0, For x = 2,
1f f f 1 = @ 4B, B = @ f 4 f g 1f 1f 1f f f f f f f f 1 = A @ B @ D = @ @ @ D, D = @ f 4 4 2
` a 15f 5f 1f ff f f f f 1 = 15A + 5B + 3 2C + D = ff f+ 3 2C @ f = 6C + 1, C = 0 @ f 4 4 2 f g
1f ffff 1f ffff 1 1 1 fffff fffff ffff f ffff f ffff f ffff f ffff Z fffffdx = f ffffdx @ f ffffdx Z Z 4 x @1 4 x @1 4 x+1
e 1f f f f 0 A x @ dx 2 fffffffffff fffffffffff ffffffffff + Z fffffffffff 2
d
x +1
1f 1f 1f fffff dx f f f f f ffff f Z 2 fff = fln|x @ 1| @ fln|x + 1| @ f fffff 4 4 2 x +1 1f 1f 1f f f f f f = fln|x @ 1| @ fln|x + 1| @ farctan x + C 4 L 4 2 M M @ff f 1f f 1f Lxffff 1f f f ffff ff L M = flnL fffM@ arctan x + C 4 Lx + 1M 2
INTEGRATION OF HYPERBOLIC FUNCTIONS
The following formulas are used for hyperbolic functions :
Z sinh x dx = cosh x + C Z cosh x dx = sinh x + C Z tanh x dx = ln |cosh x | + C Z coth x dx = ln |sinh x | + C
Z sech 2 x dx = tanh x + C Z csch 2 x dx = @ coth x + C Z sech x tanh x dx = @ sech x + C Z csch x coth x dx = @ csch x + C
dxff xf ffffffff fffffff fffffff f f f Z fffwwff= sinh@ 1 f + C wwww wwww wwww wwww wwww wwww wwww w
qx 2 + a 2
a
dx xf ffffffff ffffffff fffffff ffffff f f Z fwwwwf= cosh@ 1 f + C, x >a>0 wwww wwww wwww wwww wwww wwww wwww
qx 2 @ a 2
a
dx xf ffffff 1f fffff fffff f f f f Z ffffff= ftanh@ 1 f + C, 2 2 a @x a a dx 1f xf ffffff fffff fffff f f f f Z ffffff= @ fcoth@ 1 f + C, 2 2 x @a a a
x 2 < a2 x 2 > a2
APPLICATIONS OF INDEFINITE INTEGRALS
A. FINDING EQUATIONS OF FAMILY OF CURVES :
If we know the equation y = f x of a curve, the slope of it at any point given by
` a
f. x . Conversely, when the slope of the curve at any point is given by the
` a
` a
b
x, f x
` ac
will be
dyf ` a ff ff f m = ff= f. x , dx
` a
equation
of
the
family
of
curves
will
be
given
by
y =Z f. x dx = f x + C .
To find out a particular curve from the family of curves, we need an initial condition, generally given as the coordinates of one point on the required curve, which we can use to find the arbitrary constant C.
Example : Find the equation of the curve whose slope at any point (x,y) is given by
4xf ff ff f m = @ ff and one point on the curve is 9y
f
w w w w w w 3fw pw fw w fw w p2 , 2 . What kind of a curve is this? 2
g
Solution : 4ff xf ff f m = @ ff 9y dyf 4ff xf ff ff ff f ff f = @ ff 9y dx 9y dy = @ 4x dx
Z 9y dy = @Z 4x dx
2 2 yff xff ff f ff f 9 ff= @ 4 ff+ C 2 2 4x 2 + 9 y 2 = 2C this is the general equation of the family of curve A
putting the point
4
f g 9f f f ` a
f
w w w w w w 3fw pw fw w fw w p2 , 2 in this curve we get 2
g
2
+ 9 2 = 2C
2C = 36 4x 2 + 9 y 2 = 36
# The equation of the curve is This is an Ellipse A
B. DISTANCE, VELOCITY AND ACCELERATION :
In many applications the distance, velocity and acceleration are functions of time. In applications of derivatives, we have seen that instantaneous velocity is the derivative of the distance function and instantaneous acceleration is the derivative of the velocity
function. As integration is the inverse operation of differentiation, the indefinite integral
of the acceleration function represents the velocity function and the indefinite integral of the velocity function represents the distance function.
If the distance function, velocity function and the acceletaion function of time are ` a ` a ` a s t , v t and a t respectively ` a ` a ` a ` a then s. t = v t and v . t = a t conversely, Z v t dt = s t + C
` a ` a
and Z a t dt = v t + C
` a ` a
In all the distance-velocity-acceleration problem one side is taken positive, and the opposite is negative. While throwing up a ball, if we take the upward direction positive, then s t ≥ 0 for any point in the motion, v t ≥ 0 in the upward motion and v t ≤ 0 in the downward motion, downwards.
` a ` a ` a
a t < 0 as the acceleration due to gravity is always
` a
Example : A ball is thrown upward from the ground at a velocity 192 ft/sec2 . When will the velocity be zero? How high will it reach?
Solution :
We know acceleration due to gravity is @ 32 ft s@ 2 a t = @ 32 v t =Z a t dt =Z @ 32 dt = @ 32 t + C putting t = 0 in this, we get ` a ` a v 0 = @ 32 0 + C
` a B ` a ` a ` a
F negative sign as it is always G
donwards
192 = C as initial velocity was 192, v 0 = 192
so v t = @ 32 t + 192 ` a when the velocity is zero, v t = 0 @ 32 t + 192 = 0 t =6 The velocity will be zero after 6 seconds A s t =Z v t dt =Z @ 32t + 192
` a ` a ` a
` a
C
= @ 16t 2 + 192t + C1 at t = 0, s 0 = 0, C1 = 0
` a
s t = @ 16t 2 + 192t As it reaches the highest position at t = 6,
` a
s 6 = @ 16 6 + 192 6 = 576 It will reach a height of 576 ft A
` a ` a2 ` a
DEFINITE INTEGRALS
THE RIEMANN SUMS AND DEFINITE INTEGRAL
Let a function f be continuous on the closed interval [a,b] that is from x=a to x=b. We divide this interval in “n” subintervals by inserting (n-1) points x1 ,x 2 , …,x n @ 1 between a and b and relabel a as x 0 and b as x n . We denote the length of the first subinterval @ A @ A x 0 ,x1 by Δ1 x = x1 @ x 0 , of second subinterval x1 ,x 2 by Δ2 x = x 2 @ x1 , …… of last @ A subinterval x n @ 1 ,x n by Δn x = x n @ x n @ 1 .
B B Now we pick a point x1 from x 0 ,x1 and form the product f x1 Δ1 x ; pick a point x B 2
from x1 ,x 2 and form the product f x B Δ2 x and go on tin this manner until we form all 2 the products
@ A b c
B f x1 Δ1 x, f x B Δ2 x, ' , f x B Δn x n 2
@
A
b
c
The sum of these products
n k=1
b
c
b
c
b
c
B S n = X f x B Δk x = f x1 Δ1 x + f x B Δ2 x + ' + f x B Δn x n k 2
b
c
b
c
b
c
b
c
is called a Riemann Sum.
A Riemann Sum may therefore be thought as the sum of n products. If f x >0 on a,b ,
then the Riemann Sum is a positive real number, and if Riemann Sum is a negative real number.
f x <0 on a,b , then the
` a
B
` a
C
B
C
FUNDAMENTAL THEOREM & DEFINITE INTEGRAL
Evaluating the definite integrals using the limits of Riemann sums is generally very time consuming. Here the fundamental theorem of calculus helps us to evaluate definite integrals without using Riemann sums.
If f x is continuous on the interval a,b and if F. x = f x over a,b , then
b
` a
B
C
` a
` a
B
C
Z f x dx = F x | = F b @ F a
a
` a
` ab
a
` a
` a
3
Example : Evaluate Z x 2 dx
2
Solution :
3 xff ff f As ffis the antiderivative of x 2 : 3
8f 19f xff 3ff 2ff ff f3 f f f f ff f f Z x 2 dx = ff| = ff ff 9 @ f= ff @ f= 3
2
3
3
3
3
3
3
3
3
2 2π
Example : Evaluate Z cos x dx
πf f f f f 3
Solution : As sin x is the antiderivative of cos x :
2π πf f f Z cos x dx = sin x | = sin 2π @ sin ff πf f f f f 3 πf f f f f 3 2π
3
=0@
3
w w w w w w pff f3 f fff ff ff
2
=@
w w w w w w pff f3 f fff ff ff
2
Example : Evaluate Z 3 p x dx
1
w w w w w w w
Solution :
As 2 x 2 is the antiderivative of 3 p x :
w w w w w w w Z 3 p x dx
1 3 3f f f f f
w w w w w w w
= 2 x | = 2 3 @1
1
3f3 f f f f 2
d
3f f f f f 2
e 3f f f f f 2
= 6 p3 @ 2
w w w w w w
PROPERTIES OF DEFINITE INTEGRALS
Certain properties of definite integrals are needed to solve related problems. The common properties are :
a
1 A Z f x dx = 0
a b
` a
a
2 A Z f x dx = @ Z f x dx
a b
` a
b b
` a
3 A Z c f x dx = c Z f x dx
a
` a
a
` a
If f x
4 A Sum Rule : Z f x + g x dx = Z f x dx + Z g x dx
a a
` a
and g x
b
` a
are continuous on the interval of integration a ≤ x ≤ b, then
b b
B ` a
` aC
` a
a b
` a
b
b
5 A Difference Rule : Z f x @ g x dx = Z f x dx @ Z g x dx
a a
B ` a
` aC
` a
a
` a
If a, b, c are any three points in a closed interval, then
b c b
6 A Z f x dx = Z f x dx + Z f x dx
a
` a
a
` a
c
` a
7 A The Mean Value Theorem of Definite Integrals : If f x is continuous on the closed interval a,b , then atleast one number c exists in the open interval a,b such that
b
` a
b
c
B
C
Z f x dx = f c b @ a
a
` a B
` a`
a ` a
The value of f c is called the average or mean value of the function f x on the interval a,b and f c =
` a fffff ` a f1ffZ ffff ffff f
a b
C
` a
b@a
f x dx
APPLICATIONS OF DEFINITE INTEGRAL
There are many applications of definite integrals in solving problems and these are discussed in a separate chapter.