Definite Integration-Topics in IB Mathematics

Definite Integration

In the previous post, we discussed indefinite integration. Now we shall discuss definite integration

► Definite Integration- We already know that   \int {f\left( x \right){\rm{ }}dx = g\left( x \right) + c}    \leftarrow  this c here is an integral constant. we are not sure about its value. This c is the reason we call this process indefinite integration. But suppose we do our integration between certain limits like:-

\int\limits_a^b {f(x)dx = \left[ {g(x) + c} \right]} _a^b   here a \to  lower limit while b \to  higher limit

\int\limits_a^b {f(x)dx = \left[ {g(b) + c} \right]} - \left[ {g(a) + c} \right]


You can clearly see that this function is independent of ‘c’. Means we can be sure about its value so this type of integration is called  Definite Integration.

►Definite Integration of a function f(x) is possible in [a,b] if f(x) is continuous in the given interval

►If f(x), the integrand, is not continuous for a given value of x then it doesn’t mean that g(x), the integral, is also discontinuous for that value of x.

► Definite integration of a function between given limits like     \int\limits_a^b {f\left( x \right)dx} \Rightarrow         Algebraic sum of areas bounded by the given curve f(x) and given lines x=a and x=b. That’s why the answer for definite integration problems is a single number.

► If \int\limits_a^b {f\left( x \right)dx} = 0 that shows a few things:-

(i) The lines between which area is bounded are co-incident(a=b)

(ii) Area covered above the x-axis=Area covered below the x-axis that means positive part of area and negative part of area is equal

(iii) there must be at least one solution/root to f(x) between x=a and x=b(this is something we study in ROLE’S THEOREM in detail)

► If given function f(x) is not continuous at x=c then we should write

\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^{{c^ - }} {f(x)dx} + \int\limits_{{c^ + }}^a {f(x)dx}

► If given function f(x) > or <0 in any given interval (a,b) then  \int\limits_a^b {f\left( x \right)dx} \Rightarrow  >0 or <0 in given interval (a,b)

► If given function f(x)  \ge  g(x) in the given interval (a,b) then    \int\limits_a^b {f(x)dx \ge } \int\limits_a^b {g(x) \ge } dx 

in the given interval

► If we integrate the given function f(x) in the given interval (a,b) then

\int\limits_a^b {f(x)dx \le } \left| {\int\limits_a^b {g(x) \ge } dx} \right| \le \int\limits_a^b {\left| {f(x)} \right|dx}

<img src="definite integration.jpg" alt="definite integration">

Some More Properties of Definite Integration:-

1.   We can interchange the limits on any definite integral, all that we need to do is tack a minus sign onto the integral when we do.

2. .  If the upper and lower limits are the equal then integration of function will be zero
3.  , where c is any constant/any real number

4.   that means definite integration is a distributive process

5.    here c is a number lying somewhere between a and b

6.    If we don’t change the integrand and the limits, then change in the variable will not affect the answer

7.(a) If f(x) is an odd function i.e. f(x) = – f(-x) then   \int\limits_{ - a}^a {f(x)} = 0

(b)  If f(x) is an even function i.e. f(x) = f(-x) then     \int\limits_{ - a}^a {f(x)} = 2x


8.   \int\limits_a^b {f(x)} dx = \int\limits_a^b {f(a + b - x)} dx   in particular    \int\limits_0^a {f(x)} dx = \int\limits_0^a {f(a - x)} dx

9.   \int\limits_0^a {f(x)} dx = 2\int\limits_0^{a/2} {f(x)} dx

10. \int\limits_{ma}^{na} {f(x)} dx = (n - m)\int\limits_0^a {f(x)} dx where f(a) is periodic with period ‘a’.

Walli’s Formula: 

\int\limits_0^{\pi /2} {{{\sin }^m}x.{{\cos }^n}x} dx\frac{{\left[ {(n - 1)(n - 3)(n - 5)............1or2} \right]\left[ {(m - 1)(m - 3)(m - 5)........1or2} \right]}}{{(m + n)(m + n - 2)(m + n - 4).......................1or2}}k

Where K =\pi /2   if both m and n are even   (m, n \in  N)

= 1 in case the function is odd


Leibnitz’s Rule- If f(x) is a continuous functuion and u(x) & v(x) are differentiable in the interval [a,b] then,

\frac{d}{{dx}}\int\limits_{u(x)}^{v(x)} {f(t)dt = f\left\{ {v(x)} \right\}\frac{d}{{dx}}} \left\{ {v(x)} \right\} - f\left\{ {u(x)} \right\}\frac{d}{{dx}}\left\{ {u(x)} \right\}

This rule is used when at least one of the limits is a function.

Here is a very detailed Pdf for definite integration download it solve the questions

Here are the links to articles written on calculus so far, you should read them for better understanding of calculus


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