Sexy definite and indefinite integral


Let’s encounter the integral in the problem of measuring the area of a plane region bounded by curved lines. The integral is an area. Let there be are given a function f(x), which is continuous and positive in an interval, and two values a and b (with a < b) in that interval. We think of the function as being represented by a curve and consider the area of the region which is bounded above by the curve, at the sides by the straight lines x = a and x = b and below by the portion of the x-axis between the points a and b.
We call this area the definite integral of the function f(x) between the limits a and b. We are, in general, unable to measure areas with curved boundaries, but we can measure polygons with straight sides by dividing them into rectangles and triangles. We subdivide the part of the x-axis between a and b into n equal parts and erect at each point of sub-division the ordinate up to the curve; the area is thus divided into n strips.

Eventually, as the x-axis dimension () of each rectangle approaches zero, the aggreate area between the function line at the top of the sub-intervals, and the top of the rectangles become smaller and smaller, making the area of the rectangles a better and better approximation of the area under the function. The following slide-show demonstrates the concept. As you tab through the four frames, notice the tan portion (representing the area under the rectangles) getting larger and larger.

Let f: D → R be a function defined on a subset D of the real line R. Let I = [a, b] be a closed interval contained in D, and let P = {[x0, x1), [x1, x2), … [xn-1, xn]} be a partition of I, where a = x0 < x1 < x2 … < xn = b.

The Riemann sum of f over I with partition P is defined as

where xi-1yixi. The choice of yi in this interval is arbitrary. If yi = xi-1 for all i, then S is called a left Riemann sum. If yi = xi, then S is called a right Riemann sum. If yi = (xi+xi-1)/2, then S is called a middle Riemann sum. The average of the left and right Riemann sum is the trapezoidal sum.

If it is given that

where vi is the supremum of f over [xi-1, xi], then S is defined to be an upper Riemann sum. Similarly, if vi is the infimum of f over [xi−1, xi], then S is a lower Riemann sum. Any Riemann sum on a given partition (that is, for any choice of yi between xi-1 and xi) is contained between the lower and the upper Riemann sums. A function is defined to be Riemann integrable if the lower and upper Riemann sums get ever closer as the partition gets finer and finer.

We have considered the definite integral as a number given by an area and have represented it as a limiting value.

Let f(x) be a function which is positive and continuous in the interval a x b (of length ba). We think of the interval as being sub-divided by (n—1) points x1, x2, ··· , xn-1 into n equal or unequal sub-intervals and, in addition, we let x0=a, xn=b. In each interval, we choose a perfectly arbitrary point, which may be within the interval or at either end; suppose that in the first interval we choose the point x1, in the second one the point x2, ··· and in the n-th interval the point xn. Instead of the continuous function f(x), we now consider a discontinuous function which has the constant value f(x1) in the first sub-interval, the constant value f(x2) in the second sub-interval, ··· , the constant value f(xn) in the n-th sub-interval.

This step-function defines a series of rectangles, the sum of the areas of which is given by

We shall call this limiting value the definite integral of the function f(x), the integrand, between the limits a and b; as we have already mentioned, we shall consider it as the definition of the area under the curve y = f(x) for a<x<b. If f(x) is continuous in a< x < b, its definite integral between the limits a and b exists.

The definition of the integral as the limit of a sum led Leibniz to express the integral by the symbol

\int_a^b f(x) dx

The integral symbol is a modification of a summation sign which has the shape of a long S.

We assume: (1) that the function f(x) is positive throughout the interval, and (2) that b > a.

Conventions: We are thus led to the relation:

\int_a^b f'(x) dx =f(b)-f(a)

\int_a^a f(x) dx=0

\int_a^b cf(x) dx =c \int_a^b f(x) dx

We obtain an important fundamental rule by considering the function cf(x), where c is a constant.

During evaluation of the integral, it does not matter whether we use the letter x or any other letter to denote the abscissae of the co-ordinate system, the independent variable.

Computing integrals

Most of techniques, that have been developed to evaluate integrals, rewrite one integral as a different one which is hopefully more tractable. Techniques include:

    • Integration by substitution:

1)  Write down the given integral: \int f(x) dx

2) Come up with a substitution u = u(x).

3) Ideally you may want to find the inverse function of u(x), meaning that you will find x = x(u).

4) Differentiate to find dx = x‘(u) du.

5) Back to the given integral and make the appropriate substitutions


6) Check after algebraic simplifications that the new integral is easier than the initial one. Otherwise, go back to step 2 and come up with another substitution.
Do not forget that the answer to \int f(x) dx  is a function of x. Therefore once you have finished doing all your calculations, you should substitute back to the initial variable x.

If you are given a definite integral \int_a^b f(x) dx, nothing will change except in step 5 you will have to replace a and b also, that is


In this case, you will never have to go back to the initial variable x.




Solution. It is clear that once we develop the (x^2+5)^{75}  through the binomial formula, we will get a polynomial function easy to integrate. But it is clear that this will take a lot of time with big possibility of doing mistakes !!
Let us consider the substitution u=x^2+5  (the reason behind is the presence of x in the integral since the derivative of  x^2+5 is 2x). Indeed, we have du = 2x dx and therefore


We may check that the new integral is easier to handle since




which does not complete the answer since the indefinite integral


is a function of x not of u. Therefore, we have to go back and replace u by u(x):


    • Integration by parts:

1) Write the given integral \int f(x) g(x) dx  where you identify the two functions f(x) and g(x). Note that if you are given only one function, then set the second one to be the constant function g(x)=1.

2) Introduce the intermediary functions u(x) and v(x) as:


Then you need to make one derivative (of f(x)) and one integration (of g(x)) to get


Note that at this step, you have the choice whether to differentiate f(x) or g(x). We will discuss this in little more details later.

3) Use the formula


4) Take care of the new integral tex2html_wrap_inline218 .For the definite integral tex2html_wrap_inline220 , we have two ways to go:

a)  Evaluate the indefinite integral displaymath222 which givesdisplaymath224 b) Use the above steps describing Integration by Parts directly on the given definite integral. This is how it goes:

(i) Write down the given definite integral \int_a^b f(x) dx where you identify the two functions f(x) and g(x);

(ii) Introduce the intermediary functions u(x) and v(x) as: displaymath204

Then you need to make one derivative (of f(x)) and one integration (of g(x)) to get


(iii) Use the formula


  (iv) Take care of the new integral \int_a^b v(x) du .




Let us follow the steps

This is an indefinite integral involving one function. The second needed function is g(x) = 1. Since the derivative of this function is 0, the only choice left is to differentiate the other function tex2html_wrap_inline30 .

2)We have


which gives


We have the formula




we get


Changing the order of integration

Integration by trigonometric substitution:

The familiar trigonometric identities


may be used to eliminate radicals from integrals. Specially when these integrals involve tex2html_wrap_inline29 and tex2html_wrap_inline31.

For tex2html_wrap_inline33 set tex2html_wrap_inline35 . In this case we talk about sine-substitution.2For tex2html_wrap_inline37 set tex2html_wrap_inline39 . In this case we talk about tangent-substitution.3For tex2html_wrap_inline41 set tex2html_wrap_inline43 . In this case we talk about secant-substitution.

The expressions tex2html_wrap_inline45 and tex2html_wrap_inline47 should be seen as a constant plus-minus a square of a function. In this case, x represents a function and a a constant.




Solution. First let us complete the square for tex2html_wrap_inline35 . We get


which suggests the secant-substitution tex2html_wrap_inline39 .

Hence we have

tex2html_wrap_inline41 and


Note that for x=0, we have  tex2html_wrap_inline47 which gives t=0 and for x=3, we have tex2html_wrap_inline53 which gives tex2html_wrap_inline55 . Therefore, we have


Using the trigonometric identities (you will find them at the end of this page), we get


The technique of integration related to the powers of the secant-function gives




which implies


One would check easily that


  • Integration by partial fractions:

A rational function is by definition the quotient of two polynomials. For example


are all rational functions. Remember in the definition of a rational function, you will not see \sqrt x  neither ln(x)  or |x| for example. Note that integration by parts will not be enough to help integrate a rational function. Therefore, a new technique is needed to do the job. This technique is called decomposition of rational functions into a sum of partial fractions (in short Partial Fraction Decomposition).
Let us summarize the practical steps how to integrate the rational function


1) If degree(P) >= degree(Q) , perform polynomial long-division. Otherwise go to step 2.

2)Factor the denominator Q(x) into irreducible polynomials: linear and irreducible quadratic polynomials.

3)Find the partial fraction decomposition.

4)Integrate the result of step 3.

The main difficulty encountered in general when using this technique is in dealing with step 2 and step 3. Therefore, it is highly recommended to do a serious review of partial decomposition technique before adventuring into integrating fractional functions.




Solution. Since the degree of the numerator is higher than the denominator, we should perform the long-division. We get


which implies


We concentrate on the fraction tex2html_wrap_inline165 .

The partial decomposition technique gives


This gives x + 2 = A(x-1) + B(x+1). If we substitute x=1, we get B = 3/2 and we substitute x=-1, we get A = -1/2. Therefore, we have


Now we are in position to perform the desired integration. Indeed, we have

displaymath87Since we have


we get


  • Integration by reduction formulae
  • Integration using parametric derivatives
  • Integration using Euler’s formula
  • Differentiation under the integral sign
  • Contour Integration


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s