Introduction

Integral calculus is that part of the calculus which deals a change of function in macro level, That is, all micro level changes are 'integrated' and studied. This is opposite or the inverse of differential calculus.

The inverse operation of differentiation of a function is the integration. The result of integration is called integral and some times it is also termed as 'anti derivative' becuse the operation is inverse to the operation of differentiation.

An integral can also be described as summation. In fact, the integration symbol $\int$ is an elongated 'S', the letter indicating a summation. If the interval of the summation.or integration is not specified and if it is supposed to be general, then the integral is known as indefinite integral. But if the integration is carried out over a particular interval, then the integral is called as definite integral.

We shall try to grasp the concept about integrals, both indefinite and definite through some integral problems.

Let us start with the problems on basic concepts. Since we said that an integral is an anti derivative, let us start with that concept.Integral calculus is that part of the calculus which deals a change of function in macro level, That is, all micro level changes are 'integrated' and studied. This is opposite or the inverse of differential calculus.

The inverse operation of differentiation of a function is the integration. The result of integration is called integral and some times it is also termed as 'anti derivative' becuse the operation is inverse to the operation of differentiation.

An integral can also be described as summation. In fact, the integration symbol $\int$ is an elongated 'S', the letter indicating a summation. If the interval of the summation.or integration is not specified and if it is supposed to be general, then the integral is known as indefinite integral. But if the integration is carried out over a particular interval, then the integral is called as definite integral.

We shall try to grasp the concept about integrals, both indefinite and definite through some integral problems.

Problem 1: Evaluate the integral $\int$ cos dx

The question is expressed as $\int$ cos dx. It means the integral of cos x. The term 'dx' indicates that the integration has to be done with respect to the independent variable 'x'.

Now we defined an integral as anti derivative. That is, the answer to this question is some function of 'x' whose derivative is cos x. We know that derivative of sin x is cos x and hence we can say sin x is the anti derivative of cos x. In other words, the 'integral of cos x' is, sin x.

Now let us see what is the derivative of sin x + C, where 'C' is a constant. The derivative is still 'cos x'. Therefore, the integral of cos x could, in more general, be sinx + C. The constant C is better known as constant of integration.

Thus, $\int$ cos x dx = sin x + C

Problem 2: Evaluate the integral $\int$ [x

^{2}+ (1/x)]dx

The question is based the concept of rules of integration. As per the first rule of integration, integral of sum or difference of two functions is equal to the sum or difference of integrals.

Therefore, $\int$ [x

^{2}+ (1/x)]dx = $\int$ x

^{2 }dx + $\int$ (1/x) dx

We know the derivative of x

^{3}is 3x

^{2}or the derivative of (1/3)x

^{3}is x

^{2}. Therefore, in general, the integral of x

^{2}is, (1/3)x

^{3 }+ C

_{1}, where C

_{1}is the constant of integration. Similarly, applying the anti derivative concept we can say that the general integral of (1/x) is ln x + C

_{2}, where C

_{2}is another constant of integration. Therefore,

$\int$ [x

^{2}+ (1/x)]dx = (1/3)x

^{3 }+ ln x + C, where C = C

_{1 }+ C

_{2}, the net constant of integration.

*It is not necessary to find the anti derivative of a function every time or in every case. A table of standard integrals framed on the same principle is readily available and one can directly refer and complete the process of integration.*

Problem 3: Evaluate the integral $\int$ [x / (2 + x

^{2})]dx

This question is based on an important integration technique. We may not have any clue to integrate the function as it is nor we know of any function whose derivative is, [x / (2 + x

^{2})].

Let us see what happens when we bring in another variable 'u' to substitute for 2 + x

^{2}. That is, let 2 + x

^{2}= u or 2x dx = du or x dx = (1/2) du.

Substituting these in the original integral,

$\int$ [x / (2 + x

^{2})] = $\int$[(1/2) du]/[u] = (1/2)$\int$ du/[u]

Now, the right side is a standard integral which yields the answer as (1/2) ln u + C.

Now, we can undo the substitution and write the final answer in terms of original variable. That is,

$\int$ [x / (2 + x

^{2})]dx = (1/2) ln (2 + x

^{2}) + C

Problem 4: Evaluate the integral $\int$ [sin

^{2 }x] dx

In some cases the identities help us to simplify and integrate the functions which look hard at the first look. The function sin

^{2}x van not be integrated as it is. But there is a trigonometry which relates sin

^{2}x with a single degree trigonometric ratio, which is,

cos 2x = 1 - 2 sin

^{2}x, or sin

^{2}x = (1/2)(1 - cos 2x)

Therefore, $\int$ [sin

^{2 }x] dx = $\int$ (1/2)(1 - cos 2x) dx = (1/2) $\int$ dx - (1/2)$\int$ cos 2x dx], or,

$\int$ [sin

^{2 }x] dx = (x/2) - (sin 2x)/4] + C

Problem 5: Evaluate the integral $\int$dx/ [1 + x

^{2}]

This is another case of trigonometric substitution. The term (1 + x

^{2}) gets reduced to a single term by using the trigonometric identity

1 + tan

^{2}u = sec

^{2}u.

So let us substitute x = tan u, and hence dx = sec

^{2}u du

Therefore, $\int$dx/ [1 + x

^{2}] = $\int$[sec

^{2}u du}/ [1 + tan

^{2}u] = $\int$[sec

^{2}u du}/ [sec

^{2 }u] = $\int$ du = u + C

Now let us undo the substitution.

Since x = tan u, tan u = x or, u = tan

^{-1}x

Therefore, $\int$dx/ [1 + x

^{2}] = tan

^{-1}x + C

Problem 6: Evaluate the integral $\int$ (sin x * x)dx

This is an integration of a product. This type integration is done by parts and there is a formula for the same. It says,

**$\int$u dv = u* v - $\int$v du**

The formula can verbally remembered as ' integral of two functions = first function * integral of the second function - integral of [integral of the second function * differential coefficient of the first function]

It is important to identify which is the first function and which is the second function, rather than blindly considering the order inwhich they appear. For example, in the given problem it will be easier if 'x' is considered as the first function because its derivative is 1 and hence the second term in the formula becomes simple.

Hence, let u = x and hence du = dx

dv = sin x and hence v = -cos x

Therefore, $\int$(sin x * x) = $\int$ (x * sin x) = -x * cos x - $\int$-cosx dx = -x * cos x + sin x + C

Thus, $\int$(sin x * x) = sin x - x * cos x + C

Problem 7: Evaluate the integral $\int$ (cos x * e

^{x})dx

The integration of the above function is a little tricky. It is a product of two functions and obviously we need to do the integration by parts. But we know that the successive integrals or differential coefficients of e

^{x}is only e

^{x}. In case of cos x, the integrals and differential coefficients alternate between sin x and cos x (of course, with changes of signs). Hence in the process of integration by parts, we are likely to get back the given integral itself after a few operations. But we are going to use that to solve the problem!

Let u = e

^{x}and dv = cos x.

Hence, du = e

^{x}and v = sin x

$\int$ (cos x * ex)dx = e

^{x}* sin x - $\int$ sin x * e

^{x}dx

Now once again integrating the integral on the right side by parts (assuming u = e

^{x}and dv = sin x),

$\int$ (cos x * e

^{x})dx = e

^{x}* sin x - [-e

^{x}* cos x - $\int$ -cos x * e

^{x}dx] = e

^{x}* sin x + e

^{x}* cos x - $\int$ -cos x * e

^{x}dx

The last term on the right side is nothing but the given integral. Let it be I. So,

I = e

^{x}* sin x + e

^{x}* cos x - I

or, 2I = e

^{x}(sin x + cos x)

or,

*I = (e*

^{x}/ 2)( sin x + cos x)

Thus, $\int$ (cos x * ex)dx = (e

^{x}/ 2)( sin x + cos x)

Problem 8: Evaluate the integral $\int \frac{dx}{x^2 + 5x + 6}$

This integration is done by a method called 'integration by using partial fraction'. Let us first consider the denominator x

^{2}+ 5x + 6, of the rational function on the right side. It can be factored as (x + 3)(x + 2). Hence,

$\frac{1}{x^2 + 5x + 6}$ = $\frac{1}{(x + 3)(x + 2)}$

The right side can be decomposed into sum of two fractions called partial fractions in such a way,

$\frac{1}{(x + 3)(x + 2)}$ = $\frac{A}{x + 3}$ + $\frac{B}{x + 2}$, where A and B are constants. Now simplifying the right side under a common denominator,

$\frac{1}{x^2 + 5x + 6}$ = $\frac{A(x + 2)+ B(x + 3)}{(x + 3)(x + 2)}$ = $\frac{(A + B)x + (2A + 3B) }{x^2 + 5x + 6}$

Equating the coefficients of the numerator,

A + B = 1

2A + 3B = 0

Solving for A and B, we get, A = 3 and B = -2

Therefore, the integral now becomes as,

$\int \frac{dx}{x^2 + 5x + 6}$ = $\int \frac{3}{x + 3}$ - $\int \frac{2}{x + 2}$ = 3ln(x+3)- 2ln (x +2)= ln$\frac{(x+3)^3}{(x+2)^2}$

Thus, $\int \frac{dx}{x^2 + 5x + 6}$ = ln$\frac{(x+3)^3}{(x+2)^2}$

Problem 9: Find the area of the segment of the parabola y

^{2}= x, from origin to a point x = 4.

This is a integration problem based in definite integral. Let us draw a diagram to study the problem statement.

The equation of the parabola shown above is y

^{2}= x

Consider a thin strip AB inside the parabola at a distance 'x' from the origin and of thickness dx. The thickness is so thin that the strip can be considered as a rectangle.

The area of the strip given by dA = 2y*dx.

From the given equation of the parabola, y = √x.

Therefore, dA = 2√x*dx

Now, the area A of the segment of the parabola from origin to x = 4 is given by the following definite integral

A = $\int_{0}^{4}dA$ = $\int_{0}^{4}2*\sqrt{x} dx$ = 2 $\int_{0}^{4}\sqrt{x} dx$ = 2*(2/3)$[x^{3/2}]^4_0$ = (32/3) sq.units