Introduction
Integral is one of the important concepts of calculus. Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions as shown in the figure.
It is also used to find volumes, central points and many useful things.
The functions that could possibly have given function as a derivative are called anti derivatives or primitive of the function. Again, the formula that gives all these anti derivatives is called the indefinite integral of the function and such process of finding anti derivatives is called integration.
The development of integral calculus is used to solve the problems of the following types:
(a) The problem of finding a function whenever its derivative is given,
(b) The problem of finding the area bounded by the graph of a function under certain conditions.
These two problems lead to the two forms of the integrals, e.g., indefinite and definite integrals, which together constitute the Integral Calculus. The definite integral is also used to solve many interesting problems from various disciplines like economics, finance and probability.
Integration as an Inverse Process of Differentiation
Integration is the inverse process of differentiation. Let f(x) be a function. Then the collection of all primitives is called the indefinite integral of f(x) and denoted by ʃ f(x) dx.
Thus, d[ф(x) + C]/dx = f(x) => ʃ f(x) dx = ф(x) + C
Where ф(x) is primitive of f(x) and C is an arbitrary constant known as constant of integration.
The following is a list to find integrals of other functions.
Some properties of indefinite integral
(i) The process of differentiation and integration are inverses of each other in the sense of the following results :
d[ʃ f(x) dx]/dx = f(x)
and ʃ f’(x) dx = f(x) + C, where C is any arbitrary constant.
(ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent.
i.e. d[ʃ f(x) dx]/dx = d[ʃ g(x) dx]/dx
=> d[ʃ f(x) dx - ʃ g(x) dx]/dx = 0
=> ʃ f(x) dx - ʃ g(x) dx = C, where C is any real number
=> ʃ f(x) dx = ʃ g(x) dx + C
So, the families of curve [ʃ f(x) dx + C1, C1 є R] and [ʃ g(x) dx + C2, C2 є R] are identical.
Hence, ʃ f(x) dx and ʃ g(x) dx are equivalent.
(iii) ʃ [f(x) + g(x)] dx = ʃ f(x) dx + ʃ g(x) dx
(iv) For any real number k, ʃ [k * f(x)] dx = k * ʃ f(x) dx
(v) ʃ [k1 * f1(x) + k2 * f2(x) + . . . . .+ kn * fn(x)] dx = k1 * ʃ f1(x) dx + k2 * ʃ f2(x) dx + . . . . . .+ kn * ʃ fn(x) dx
Problem: Find the integral of the following functions:
(a) ʃ (ax2 + bx + c) dx (b) ʃ (2x2 + ex) dx
Solution:
(a) ʃ (ax2 + bx + c) dx = ʃ ax2 dx + ʃ bx dx + ʃ c dx
= aʃ x2 dx + b ʃ x dx + c ʃ dx
= ax3/3 + bx2/2 + cx + C
So, ʃ (ax2 + bx + c) dx = ax3/3 + bx2/2 + cx + C
(b) ʃ (2x2 + ex) dx = ʃ 2x2 dx + ʃ ex dx
= 2ʃ x2 dx + ʃ ex dx
= 2x3/3 + ex + C
So, ʃ (2x2 + ex) dx = 2x3/3 + ex + C
Methods of Integration
There are certain methods for solving the integration:
- Integration by Substitution
- Integration using Partial Fractions
- Integration by Parts
- Integration by Substitution
The given integral ʃ f(x) dx can be transformed into another form by changing the independent variable x to t by substituting x = g (t).
Let I = ʃ f(x) dx
Put x = g(t) so that dx/dt = g′(t).
We write dx = g′(t) dt
Thus I = ʃ f(x) dx = ʃ f{g(t)} * g’(t) dt
Problem: Integrate the following functions:
(a) 2x/(1 + x2) (b) (log x)2/x
Solution:
(a) Let 1 + x2 = t
=> 2x dx = dt
Now, ʃ 2x/(1 + x2) dx = ʃ dt/t
= log|t| + C
= log|1 + x2| + C
= log(1 + x2) + C
(b) Let log x = t
=> dx/x = dt
Now, ʃ [(log x)2/x] dx = ʃ t2 dt
= t3/3 + C
= (log|x|)3/3 + C
There are some important trigonometries integral which are listed below:
(i) ʃ tan x dx = log|sec x| + C
(ii) ʃ cot x dx = log|sin x| + C
(iii) ʃ sec x dx = log|sec x + tan x| + C
(iv) ʃ cosec x dx = log|cosec x – cot x| + C
Integrals of Some Particular Functions
To find the integral ʃ dx/(ax2 + bx + c)
We write ax2 + bx + c = a[x2 + bx/a + c/a] = a[(x + b/2a)2 + (c/a – b2/4a2)]
Put x + b/2a = t so that dx = dt and write c/a – b2/4a2 = ±k2
Now, ʃ dx/(ax2 + bx + c) = (1/a) * ʃ dt/(t2 ± k2)
If c/a – b2/4a2 is +ve then we put + k2 and if If c/a – b2/4a2 is +ve then we put - k2
Problem: Find the integral
1/(9x2 + 6x + 5)
Solution:
1/(9x2 + 6x + 5) = 1/{(3x + 1)2 + 22}
Let 3x + 1 = t
=> 3 dx = dt
=> dx = dt/3
So, ʃ dx/(9x2 + 6x + 5) = (1/3) * ʃ dt/(t2 + 22)
= (1/3) * [(1/2) * tan-1(t/2)] + C
= (1/6) * tan-1{(3x + 1)/2)} + C
To find the integral of type ʃ dx/√(ax2 + bx + c):
To integrate ʃ dx/√(ax2 + bx + c), we do the same process as to find the integral of type ʃ dx/(ax2 + bx + c) and then apply the standard formula to obtain the solution.
Problem: Find the integral
1/√(x2 + 2x + 2)
Solution:
1/√(x2 + 2x + 2) = 1/√{(x + 1)2 + 12}
Let x + 1 = t
=> dx = dt
So, ʃ dx/√(x2 + 2x + 2) = ʃ dt/√(t2 + 12)
= log[t + √(t2 + 1)] + C
= log[(x + 1) + √{(x + 1)2 + 1)}] + C
= log[(x + 1) + √(x2 + 2x + 2)] + C
To find the integral of type ʃ [(px + q)/(ax2 + bx + c)] dx where p, q, r and s are constant:
To integrate ʃ [(px + q)/(ax2 + bx + c)] dx, we have to find real numbers A and B such that
px + q = A * d(ax2 + bx + c)/dx + B
=> px + q = A(2ax + b) + B
To determine the value of A and B, we equate from both sides the coefficients of x and the constant terms. After finding A and B, the integral is reduced to one of the known forms.
Problem: Find the integral
(4x + 1)/(2x2 + x - 3)
Solution:
Let 4x + 1 = A * d(2x2 + x - 3)/dx + B
=> 4x + 1 = A(4x + 1) + B
=> 4x + 1 = 4Ax + A + B
Equating the coefficients of x and constant term on both sides, we get
4A = 4
=> A = 1
A + B = 1
=> 1 + B = 1
=> B = 0
Let 2x2 + x – 3 = t
=> (4x + 1)dx = dt
Now, ʃ [(4x + 1)/(2x2 + x - 3)]dx = ʃ dt/t
= log t + C
= log(2x2 + x - 3) + C
To find the integral of type ʃ [(px + q)/√(ax2 + bx + c)] dx where p, q, r and s are constant:
To integrate ʃ [(px + q)/√(ax2 + bx + c)] dx, we do the same process as to find the integral of type
ʃ [(px + q)/(ax2 + bx + c)] dx and transform the integral into known standard forms.
Problem: Find the integral
(5x + 3)/√(x2 + 4x + 10)
Solution:
Let (5x + 3) = A * d(x2 + 4x + 10)/dx + B
=> (5x + 3) = A(2x + 4) + B
Equating the coefficients of x and constant term on both sides, we get
2A = 5
=> A = 5/2
And 4A + B = 3
=> B = -7
So, (5x + 3) = 5(2x + 4)/2 – 7
Now, ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = ʃ [{5(2x + 4)/2 – 7}/√(x2 + 4x + 10)]dx
= (5/2) * ʃ [(2x + 4)/√(x2 + 4x + 10)]dx - 7 * ʃ dx/√(x2 + 4x + 10)
Let I1 = ʃ [(2x + 4)/√(x2 + 4x + 10)]dx and I2 = ʃ dx/√(x2 + 4x + 10)
So, ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = 5I1/2 – 7I2 ……………..1
Now, consider I1 = ʃ [(2x + 4)/√(x2 + 4x + 10)]dx
Let x2 + 4x + 10 = t
=> (2x + 4)dx = dt
So, I1 = ʃ dt/√t
=> I1 = 2√t
=> I1 = 2√(x2 + 4x + 10)
Again consider I2 = ʃ dx/√(x2 + 4x + 10)
=> I2 = ʃ dx/√(x2 + 4x + 4 + 6)
=> I2 = ʃ dx/√{(x + 2)2 + 6}
=> I2 = ʃ dx/√{(x + 2)2 + (√6)2}
=> I2 = log{(x + 2) + (x2 + 4x + 10)}
From equation 1, we get
ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = 5[2√(x2 + 4x + 10)]/2 – 7 * log{(x + 2) + (x2 + 4x + 10)}
=> ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = 5 * √(x2 + 4x + 10) – 7 * log{(x + 2) + (x2 + 4x + 10)}
Integration by Partial Fractions
We know that a rational function is the ratio of two polynomials in the form P(x)/Q(x), where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If degree of P(x) is less than degree of Q(x), then it is called proper rational function otherwise it is improper. We can convert an improper rational function into proper rational function by division method.
So, if P(x)/Q(x) is improper, then P(x)/Q(x) = T(x) + P1(x)/Q(x)
Now to calculate ʃ [P(x)/Q(x)] dx where P(x)/Q(x) is a proper rational function, we first convert it into partial faction suing the method as shown in the given figure and then integrate using standard formula.
Problem: Find the integral
x/{(x - 1)(x - 2)(x - 3)}
Solution:
Let x/{(x - 1)(x - 2)(x - 3)} = A/(x - 1) + B/(x - 2) + C/(x - 3)
=> x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)
Equating the coefficients of x2, x and constant term, we get
A + B + C = 0
-5A – 4B – 3C = 1
6A + 3B + 2C = 0
Solving these equations, we get
A = 1/2, B = −2, and C = 3/2
So, x/{(x - 1)(x - 2)(x - 3)} = 1/2(x - 1) – 2/(x - 2) + 3/2(x - 3)
Now, ʃ [x/{(x - 1)(x - 2)(x - 3)}]dx = ʃ dx/2(x - 1) – ʃ dx/2(x - 2) + ʃ 3dx/2(x - 3)
=> ʃ [x/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * ʃ dx/(x - 1) – 2 ʃ dx/(x - 2) + (3/2) * ʃ dx/(x - 3)
=> ʃ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * log|x – 1| – 2 log|x – 2| + (3/2) * log|x – 3| + C
Problem: Find the integral
x/{(x2 + 1)(x - 1)}
Solution:
Let x/{(x2 + 1)(x - 1)} = (Ax + B)/(x2 + 1) + C/(x - 1)
=> x = (Ax + B)(x - 1) + C(x2 + 1)
=> x = Ax2 – Ax + Bx – B + Cx2 + C
Equating the coefficients of x2, x and constant terms, we get
A + C = 0
-A + B = 1
-B + C = 0
On solving these equations, we get
A = -1/2, B = 1/2 and C = 1/2
From equation 1, we get
x/{(x2 + 1)(x - 1)} = (-x/2 + 1/2)/(x2 + 1) + 1/2(x - 1)
x/{(x2 + 1)(x - 1)} = (-x/2)/(x2 + 1) + (1/2)/(x2 + 1) + 1/2(x - 1)
Now, ʃ [x/{(x2 + 1)(x - 1)} dx = (-1/2)ʃ [x/(x2 + 1)] + (1/2)ʃ [1/(x2 + 1)] + (1/2)ʃ dx/(x - 1)
= (-1/4)ʃ [2x/(x2 + 1)] dx + (1/2) * tan-1 x + (1/2) * log|x - 1| + C
Let x2 + 1 = t
=> 2x dx = dt
So, ʃ [2x/(x2 + 1)] = ʃ dt/t
= log|t|
= log|x2 + 1|
Now, ʃ [x/{(x2 + 1)(x - 1)}dx = (-1/4) log|x2 + 1| + (1/2)tan-1 x + (1/2)log|x - 1| + C
= (1/2)log|x - 1|- (1/4) log|x2 + 1| + (1/2)tan-1 x + C
Integration by Parts
Integration by parts means integration of product of two numbers. To find the integral of product of two functions, we use the following formula:
“The integral of the product of two functions = (first function) * (integral of the second function) –
Integral of [(differential coefficient of the first function) * (integral of the second function)]”
Let f(x) and g(x) are two functions. Assume f(x) is the first function and g(x) is the second function.
Now, ʃ [f(x) * g(x)] dx = f(x) * ʃ g(x) dx – ʃ {f’(x) * ʃ g(x) dx} dx
Problem: Find the integral
x * sin x
Solution:
Let I = ʃ x * sin x dx
Taking x as first function and sin x as second function and integrating by parts, we get
I = x * ʃ sin x dx - ʃ [dx/dx * ʃ sin x dx]dx
I = x * (-cos x) - ʃ 1 * (-cos x)]dx
I = -x * cos x + ʃ cos x dx
I = -x * cos x + sin x + C
Integral of the type ʃ ex[f(x) + f’(x)] dx
Let I = ʃ ex[f(x) + f’(x)] dx
= ʃ ex * f(x) dx + ʃ ex * f’(x) dx
=> I = I1 + ʃ ex * f’(x) dx, where I1 = ʃ ex * f(x) dx ………….1
Taking f(x) and first function and ex as second function in I1 and integrate, we get
I1 = f(x) * ex - ʃ f’(x) ex + C
From equation 1, we get
=> I = f(x) * ex - ʃ f’(x) ex + ʃ ex * f’(x) dx + C
=> I = ex * f(x) + C
Problem: Solve the integral
ex(sin x + cos x)
Solution:
Let I = ʃ ex(sin x + cos x) dx
Again let f(x) = sin x
=> f’(x) = cos x
So, I = ʃ ex{f(x) + f’(x)} dx
It is known that ʃ ex{f(x) + f’(x)} dx = ex f(x) + C
So, I = ex sin x + C
Integrals of some more types
There are some special types of integrals that will be solve using integration by parts:
(i) ʃ √(x2 – a2) dx (ii) ʃ √(x2 + a2) dx (iii) ʃ √(a2 – x2) dx
(i) Let I = ʃ √(x2 – a2) dx
=> I = ʃ 1 * √(x2 – a2) dx
Taking 1 as the second function and integrating by parts, we get
I = x * √(x2 – a2) - ʃ [(x/2) * 2x/√(x2 – a2)] dx
=> I = x * √(x2 – a2) - ʃ [x2/√(x2 – a2)] dx
=> I = x * √(x2 – a2) - ʃ [(x2 – a2 + a2)/√(x2 – a2)] dx
=> I = x * √(x2 – a2) - ʃ [(x2 – a2 )/√(x2 – a2)] dx - ʃ [a2/√(x2 – a2)] dx
=> I = x * √(x2 – a2) - ʃ √(x2 – a2) dx - a2 ʃ dx/√(x2 – a2)]
=> I = x * √(x2 – a2) - I - a2 log|x + √(x2 – a2)| + C
=> 2I = x * √(x2 – a2) - a2 log|x + √(x2 – a2)| + C
=> I = (x/2) * √(x2 – a2) – (a2 /2) * log|x + √(x2 – a2)| + C
=> ʃ √(x2 – a2) dx = (x/2) * √(x2 – a2) – (a2/2) * log|x + √(x2 – a2)| + C
Similarly, we can prove that
(ii) ʃ √(x2 + a2) dx = (x/2) * √(x2 + a2) + (a2/2) * log|x + √(x2 + a2)| + C
(iii) ʃ √(a2 – x2) dx = (x/2) * √(a2 – x2) + (a2/2) * sin-1 (x/a) + C
Problem: Find the integral
(i) √(1 – 4x2) (ii) √(x2 + 4x + 6)
Solution:
(i) Let I = √(1 – 4x2) = √{12 – (2x)2}
Again let 2x = t
=> 2 dx = dt
So, I = (1/2) √(12 – t2)
We know that ʃ √(a2 – x2) dx = (x/2) * √(a2 – x2) + (a2/2) * sin-1(x/a) + C
So, I = (1/2)[(t/2) * √(1 – t2) + (1/2) * sin-1 t] + C
=> I = (t/4) * √(1 – t2) + (1/4) * sin-1 t + C
=> I = (2x/4) * √(1 – 4x2) + (1/4) * sin-1 2x + C
=> I = (x/2) * √(1 – 4x2) + (1/4) * sin-1 2x + C
(ii) Let I = ʃ √(x2 + 4x + 6) dx
=> I = ʃ √(x2 + 4x + 4 + 2) dx
=> I = ʃ √{(x2 + 4x + 4) + 2} dx
=> I = ʃ √{(x + 2)2 + (√2)2} dx
We know that ʃ √(x2 + a2) dx = (x/2) * √(x2 + a2) + (a2/2) * log|x + √(x2 + a2)| + C
So, I = {(x + 2)/2} * √(x2 + 4x + 6) + (2/2) * log|(x + 2) + √(x2 + 4x + 6)| + C
=> I = {(x + 2)/2} * √(x2 + 4x + 6) + log|(x + 2) + √(x2 + 4x + 6)| + C
Definite Integral
A definite integral is denoted by aʃb f(x) dx, where a is called the lower limit of the integral and b is called the upper limit of the integral. The definite integral is introduced either as the limit of a sum or if it has an anti derivative F in the interval [a, b], then its value is the difference between the values of F at the end points, i.e., F(b) - F(a).
Definite integral as the limit of a sum:
The definite integral aʃb f(x) dx is the area bounded by the curve y = f (x), the ordinates x = a, x = b and the x-axis. To find this area Let us consider the region PQRSP between this curve and the ordinates x = a and x = b as shown in the given figure.
Now, divide the interval [a, b] into n equal sub intervals denoted by [x0, x1], [x1, x2], ……..,[xr-1, xr], ………..
[xn-1, xn] where x0 = a, x1 = a + h, x2 = a + 2h, ……….,xr = a + rh, xn = b = a + nh or h = (b - a)/n
Again as n -> ∞, h -> 0
The region PRSQP under consideration is the sum of n sub regions, where each sub region is defined on sub intervals [xr – 1, xr], r = 1, 2, 3, …, n.
Now, to find the area of the region PQRSQ is calculated as
Limx->∞ Sn = area of the region PQRSQ = aʃb f(x) dx
= limh->0 h[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]
= (b - a) * limh->∞ (1/n)[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]
So, aʃb f(x) dx = (b - a) * limh->∞ (1/n)[f(a) + f(a + h) + ………….+ f{a + (n - 1)h}]
Where h = (b - a)/n as n -> ∞
This is known as the definition of definite integral as the limit of sum.
Problem: Evaluate the definite integral as limit of sums
ʃ05 (x + 1) dx
Solution:
We know that
ʃab f(x) dx = (b - a)limn->∞ (1/n)[f(a) + f(a + h) + …………..+ f{a + (n - 1)h}], where h = (b - a)/n
Here a = 0, b = 5 and f(x) = x + 1
So, h = (5 - 0)/n = 5/n
Now, ʃ05 (x + 1) dx = (5 - 0)* limn->∞ (1/n)[f(0) + f(5/h) + …………..+ f{5(n - 1)/n}]
= 5 * limn->∞ (1/n)[1 + (5/n + 1) + …………..+ {1 + 5(n - 1)/n}]
= 5 * limn->∞ (1/n)[(1 + 1 + 1 + …….n times) + {5/n + 2 * 5/n + …………..+ 5(n - 1)/n}]
= 5 * limn->∞ (1/n)[(1 + 1 + …….n times) + (5/n){1 + 2 + 3 +…………..+ (n - 1)}]
= 5 * limn->∞ (1/n)[n + (5/n) * n(n - 1)/2]
= 5 * limn->∞ (1/n)[n + 5(n - 1)/2]
= 5 * limn->∞ (n/n)[1 + 5(1 – 1/n)/2]
= 5 * [1 + 5(1 – 1/∞)/2]
= 5 * [1 + 5(1 – 0)/2]
= 5 * [1 + 5/2]
= 5 * (7/2)
= 35/2
Fundamental Theorem of Calculus
There are some fundamental theorems which are going to discuss here one by one.
Area function
Let x be the given point in [a, b] as shown in the figure.
Then aʃb f(x) dx represents the area of the shaded region. The function A(x) is represented as Area function and is calculated as
A(x) = aʃb f(x) dx
According to this definition, two fundamental theorems are stated as:
First fundamental theorem of integral calculus
Theorem 1: Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function.
Then A′(x) = f (x), for all x ∈ [a, b].
Second fundamental theorem of integral calculus
Theorem 2: Let f be continuous function defined on the closed interval [a, b] and F be an anti derivative of f. Then aʃb f(x) dx = [F(x) a]b = F(b) – F(a), where F(x) is the indefinite integral of f(x).
Problem: Evaluate the following definite integrals
(a) ʃ-11 (x + 1) dx (b) ʃ21 dx/x
Solution:
(a) Let I = ʃ-11 (x + 1) dx
Now, ʃ (x + 1) dx = x2/2 + x = F(x)
By second fundamental theorem of calculus, we get
I = F(1) - F(-1)
= (1/2 + 1) – (1/2 - 1)
= 1/2 + 1 – 1/2 + 1
= 2
So, ʃ-11 (x + 1) dx = 2
(b) Let I = ʃ21 dx/x
Now, ʃ dx/x = log|x| = F(x)
By second fundamental theorem of calculus, we get
I = F(3) - F(2)
= log|3| - log|2|
= log 3 – log 2
= log(3/2)
So, ʃ21 dx/x = log(3/2)
Evaluation of Definite Integrals by Substitution
To evaluate aʃb f(x) dx by substitution, the steps could be as follows:
- Consider the integral without limits and substitute, y = f (x) or x = g(y) to reduce the given integral to a known form.
- Integrate the new integrand with respect to the new variable without mentioning the constant of integration.
- Resubstitute for the new variable and write the answer in terms of the original variable.
- Find the values of answers obtained in (3) at the given limits of integral and find the difference of the
values at the upper and lower limits.
Problem: Evaluate the following integrals using substitution method
(a) ʃ01 [x/(x2 + 1)] dx (b) ʃ0π/2 [√(sin ф) * cos5 ф] dф
Solution:
(a) Let x2 + 1 = t
=> 2x dx = dt
=> x dx = dt/2
When x = 0, t = 1 and when x = 1, t = 2
So, ʃ01 [x/(x2 + 1)] dx = (1/2)ʃ12 dt/t
= (1/2)[log t]12
= (1/2)[log 2 – log 1]
= (log 2)/2
(b) Let I = ʃ0π/2 [√(sin ф) * cos5 ф] dф
= ʃ0π/2 [√(sin ф) * cos4 ф * cos ф] dф
Let sin ф = t
=> cos ф dф = dt
When ф = 0, t = 0 and when ф = π/2, t = 1
So, I = ʃ01 [√t * (1 – t2)2] dt
= ʃ01 [√t * (1 + t4 - 2t2)] dt
= ʃ01 [t1/2 + t9/2 - 2t5/2] dt
= [t3/2/(3/2) + t11/2/(11/2) - 2t7/2/(7/2)]01
= 1/(3/2) + 1/(11/2) - 2/(7/2)
= 2/3 + 2/11 – 4/7
= (154 + 42 - 132)/231
= 64/231
Some Properties of Definite Integrals
There are some properties of definite integral which are very useful for calculating definite integral very
easily.
(i) aʃb f(x) dx = aʃb f(t) dt
(ii) aʃb f(x) dx = -bʃa f(t) dt
In particular, aʃa f(x) dx = 0
(iii) aʃb f(x) dx = aʃc f(x) dx + cʃb f(x) dx
(iv) aʃb f(x) dx = aʃb f(a + b - x) dx
(v) 0ʃa f(x) dx = 0ʃa f(a - x) dx
(vi) 0ʃ2a f(x) dx = 0ʃa f(x) dx + 0ʃa f(2a - x) dx
(vii) 0ʃ2a f(x) dx = 2 * 0ʃa f(x) dx, if f(2a - x) = f(x)
= 0, if f(2a - x) = -f(x)
(vii) -aʃa f(x) dx = 2 * 0ʃa f(x) dx, if f is an even function i.e. f(-x) = f(x)
= 0, if f is an odd function i.e. f(-x) = -f(x)
Problem: Evaluate the following integrals
(a) ʃ0π/2 cos2 x dx (b) ʃ-π/2π/2 sin2 x dx
Solution:
(a) Let I = ʃ0π/2 cos2 x dx …………..1
=> I = ʃ0π/2 cos2 (π/2 - x) dx [Since ʃ0a f(x) dx = ʃ0a f(a - x) dx]
=> I = ʃ0π/2 sin2 x dx …………..2
Adding equation 1and 2, we get
2I = I = ʃ0π/2 (sin2 x + cos2 x) dx
=> 2I = I = ʃ0π/2 1 dx
=> 2I = I = [x]0π/2
=> 2I = π/2
=> I = π/4
So, ʃ0π/2 cos2 x dx = π/4
(b) Let I = ʃ-π/2π/2 sin2 x dx
Since sin2 (−x) = {sin(−x)}2 = (−sin x)2 = sin2 x, therefore, sin2 x is an even function.
It is known that if f(x) is an even function, then ʃ-aa f(x) dx = 2 * ʃ0a f(x) dx
So, I = 2 * ʃ0π/2 sin2 x dx
=> I = 2 * ʃ0π/2 {(1 – cos 2x)/2} dx
=> I = ʃ0π/2 (1 – cos 2x) dx
=> I = [x – sin 2x /2]0π/2
=> I = [π/2 – sin (2 * π/2) /2]
=> I = [π/2 – (sin π)/2]
=> I = π/2
So, ʃ-π/2π/2 sin2 x dx = π/2
Integrals
