Integration of Binomial Expansion

Edited By Komal Miglani | Updated on Feb 11, 2025 10:38 AM IST

An expression with two terms is called the binomial expansion. In the case of higher degree expression, it is difficult to calculate it manually. In these cases, Binomial theorem can be used to calculate it. Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial theorem is proved using the concept of mathematical induction. Binomial coefficients are the coefficients of the terms in the Binomial expansion. Apart from Mathematics, Binomial theorem is also used in statistical and financial data analysis.

JEE Main: Study Materials | High Scoring Topics | Preparation Guide

JEE Main: Syllabus | Sample Papers | Mock Tests | PYQs

Integration of Binomial Expansion

This article is about the integration in Binomial coefficients which falls under the broader category of Binomial Theorem and its applications. It is one of the important topics for competitive exams.

Binomial Theorem

Statement:

If $n$ is any positive integer, then

$(a + b)^{n} = (\binom{n}{0}) a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n - 1}) a b^{n - 1} + (\binom{n}{n}) b^{n}$

Proof:

The proof is obtained by applying the principle of mathematical induction.

Let the given statement be:

$P (n) : (a + b)^{n} = (\binom{n}{0}) a^{n} + (\binom{n}{1}) a^{n - 1} b + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n - 1}) a b^{n - 1} + (\binom{n}{n}) b^{n}$

For $n = 1$ , we have:

$P (1) : (a + b)^{1} = (\binom{1}{0}) a^{1} + (\binom{1}{1}) b^{1} = a + b$

Thus, $P (1)$ is true.

Suppose $P (k)$ is true for some positive integer $k$ , i.e.,

$(a + b)^{k} = (\binom{k}{0}) a^{k} + (\binom{k}{1}) a^{k - 1} b + (\binom{k}{2}) a^{k - 2} b^{2} + \dots + (\binom{k}{k}) b^{k}$

We shall prove that $P (k + 1)$ is also true, i.e.,

$(a + b)^{k + 1} = (\binom{k + 1}{0}) a^{k + 1} + (\binom{k + 1}{1}) a^{k} b + (\binom{k + 1}{2}) a^{k - 1} b^{2} + \dots + (\binom{k + 1}{k + 1}) b^{k + 1}$

Now,

$(a + b)^{k + 1} = (a + b) (a + b)^{k}$

$= (a + b) [(\binom{k}{0}) a^{k} + (\binom{k}{1}) a^{k - 1} b + (\binom{k}{2}) a^{k - 2} b^{2} + \dots + (\binom{k}{k - 1}) a b^{k - 1} + (\binom{k}{k}) b^{k}]$

[from (1)]

$= (\binom{k}{0}) a^{k + 1} + (\binom{k}{1}) a^{k} b + (\binom{k}{2}) a^{k - 1} b^{2} + \dots + (\binom{k}{k - 1}) a^{2} b^{k - 1} + (\binom{k}{k}) a b^{k}$

$+ (\binom{k}{0}) a^{k} b + (\binom{k}{1}) a^{k - 1} b^{2} + (\binom{k}{2}) a^{k - 2} b^{3} + \dots + (\binom{k}{k - 1}) a b^{k} + (\binom{k}{k}) b^{k + 1}$

[by actual multiplication]

$= (\binom{k}{0}) a^{k + 1} + ((\binom{k}{1}) + (\binom{k}{0})) a^{k} b + ((\binom{k}{2}) + (\binom{k}{1})) a^{k - 1} b^{2} + \dots + ((\binom{k}{k}) + (\binom{k}{k - 1})) a b^{k} + (\binom{k}{k}) b^{k + 1}$

[grouping like terms]

$= (\binom{k + 1}{0}) a^{k + 1} + (\binom{k + 1}{1}) a^{k} b + (\binom{k + 1}{2}) a^{k - 1} b^{2} + \dots + (\binom{k + 1}{k}) a b^{k} + (\binom{k + 1}{k + 1}) b^{k + 1}$

(by using $(\binom{k + 1}{0}) = 1$ , $(\binom{k}{r}) + (\binom{k}{r - 1}) = (\binom{k + 1}{r})$ , and $(\binom{k}{k}) = 1 = (\binom{k + 1}{k + 1})$ )

Thus, it has been proved that $P (k + 1)$ is true whenever $P (k)$ is true. Therefore, by the principle of mathematical induction, $P (n)$ is true for every positive integer $n$ .

Integration of Binomial Expansion

Limits for Integration:

If the numbers occur as the denominator of the binomial coefficient, then this method is applicable.

S. No.	Conditions	Limits of integration
1	If the binomial series contains all positive sign terms	$0$ to $1$
2	If the binomial series contains alternate sign $(+ a n d -)$	$- 1$ to $0$
3	If the binomial series contains odd coefficients $(C 0, C 2, C 4, . . . . .)$	$- 1$ to $1$
4	If the binomial series contains even coefficients $(C 1, C 3, C 5, . . . . .)$	subtract (2) from (1) then divide by 2

For Example,

$C_{0} + 2^{2} \cdot \frac{C_{1}}{2} + 2^{3} \cdot \frac{C_{2}}{3} + \dots + 2^{n + 1} \cdot \frac{C_{n}}{n + 1} = \frac{3^{n + 1} - 1}{n + 1}$

Proof:

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$

As each term has $+ sign$ and each term also has powers of 2 , so we will integrate it from $0$ to $2$

$\int_{0}^{2} (1 + x)^{n} d x = \int_{0}^{2} [C_{0} + C_{1} x + \dots + C_{n} x^{n}] d x$

${[\frac{(1 + x)^{n + 1}}{n + 1}]}_{0}^{2} = {[C_{0} x + C_{1} \cdot \frac{x^{2}}{2} + C_{2} \cdot \frac{x^{3}}{3} + \dots C_{n} \cdot \frac{x^{n + 1}}{n + 1}]}_{0}^{2}$

$\Rightarrow \frac{3^{n + 1} - 1}{n + 1} = C_{0} \cdot 2 + 2^{2} \cdot \frac{C_{1}}{2} + 2^{3} \cdot \frac{C_{2}}{3} + \dots + 2^{n + 1} \cdot \frac{C_{n}}{n + 1}$

Solved Examples Based on Integration in Binomial Expansion

Example 1: If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ be binomial coefficients in the expansion of $(1 + x)^{n}$

Find the value of $C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \dots + (- 1)^{n} \frac{C_{n}}{n + 1}$ .

1) $\frac{n}{n - 1}$

2) $\frac{n}{n + 1}$

3) $\frac{1}{n - 1}$

4) $\frac{1}{n + 1}$

Solution:

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$

Integrating within limits $- 1$ to $0$ , then we get,

$\int_{- 1}^{0} (1 + x)^{n} d x = \int_{- 1}^{0} (C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}) d x$

$\Rightarrow {[\frac{(1 + x)^{n + 1}}{n + 1}]}_{- 1}^{0} = {[C_{0} x + \frac{C_{1} x^{2}}{2} + \frac{C_{2} x^{3}}{3} + \dots + \frac{C_{n} x^{n + 1}}{n + 1}]}_{- 1}^{0}$

$\Rightarrow \frac{1 - 0}{n + 1} = 0 - (- C_{0} + \frac{C_{1}}{2} - \frac{C_{2}}{3} + \dots + (- 1)^{n + 1} \frac{C_{n}}{n + 1})$

$\Rightarrow \frac{1}{n + 1} = C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \dots + (- 1)^{n + 2} \frac{C_{n}}{n + 1}$

$\Rightarrow \frac{1}{n + 1} = C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \dots + (- 1)^{n} \frac{C_{n}}{n + 1}$

$∵ [(- 1)^{n + 2} = (- 1)^{n} (- 1)^{2} = (- 1)^{n}]$

Hence, $C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \dots + (- 1)^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1}$

Hence, the answer is option (4).

Example 2: The value of $C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1}$ is.

1) $\frac{2^{n} - 1}{n + 1}$

2) $\frac{2^{n + 1}}{n + 1}$

3) $\frac{2^{n + 1} - 1}{n + 1}$

4) None of these

Solution:

This binomial series contains all positive sign terms

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots \dots + C_{n} x^{n}$

Integrating between limits $0$ and $1$ , we get :

$\int_{0}^{1} (1 + x)^{n} d x = \int_{0}^{1} (C_{0} d x + \int_{0}^{1} C_{1} x d x + \int_{0}^{1} C_{2} x^{2} d x + \dots + \int_{0}^{1} C_{n} x^{n}) d x$

${[\frac{(1 + x)^{n + 1}}{n + 1}]}_{0}^{1} = {\frac{C_{0} x}{1} |}_{0}^{1} + {C_{1} \frac{x^{2}}{2} |}_{0}^{1} + {C_{2} \cdot \frac{x^{3}}{3} |}_{0}^{1} + \dots + {C_{n} \cdot \frac{x^{n + 1}}{n + 1} |}_{0}^{1}$

$\Rightarrow \frac{2^{n + 1}}{n + 1} - \frac{1}{n + 1} = C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1}$

$\Rightarrow C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}$

Hence, the answer is option (3).

Example 3: The sum to $(n + 1)$ terms of the series $\frac{C_{0}}{2} + \frac{C_{1}}{3} + \frac{C_{2}}{4} + \frac{C_{3}}{5} + \dots$ is:

1) $\frac{1}{n + 1}$

2) $\frac{1}{n + 2}$

3) $\frac{1}{n (n + 1)}$

4) none of these

Solution:

As we have learned

Result of Binomial Theorem -

$c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + - - - - - \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}$

Take $\int_{0}^{1} (1 + x)^{n} d x = \int_{o}^{1} \sum_{r = 0}^{n}^{n} c_{r} x^{r} d x$

We have

$(1 - x)^{n} = c_{0} - c_{1} x + c_{2} x^{2} - c_{3} x^{3} + \dots$

$x (1 - x)^{n} = c_{0} x - c_{1} x^{2} + c_{2} x^{3} - c_{3} x^{4} + \dots$

$\int_{0}^{1} x (1 - x)^{n} d x = \int_{0}^{1} x (1 - x)^{n} d x = \int_{0}^{1} (1 - t) t^{n} (- 1) d t [put 1 - x = t]$

$= [Put 1 - x = t]$

$= \int_{0}^{1} (t^{n} - t^{n + 1}) d t = {| \frac{t^{n + 1}}{n + 1} - \frac{t^{n + 2}}{n + 2} |}_{0}^{1}$

$= \frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{(n + 1) (n + 2)}$

$Integrating R.H.S of (1) we get {(\frac{C_{0} x^{2}}{2} - \frac{C_{1} x^{3}}{3} + \frac{C_{2} x^{4}}{4} -) |}_{0}^{1}$

$= \frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \dots . = \frac{1}{(n + 1) (n + 2)}$

$Thus, \frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \dots . =$

Hence, the answer is option (4).

xample 4: If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ be binomial coefficients in the expansion of $(1 + x)^{n}$ Find the value of $3 C_{0} + 3^{2} \frac{C_{1}}{2} + \frac{3^{3} C_{2}}{3} + \frac{3^{4} C_{3}}{4} + \dots + \frac{3^{n + 1} C_{n}}{n + 1}$ .

1) $\frac{4^{n} - 1}{n + 1}$

2) $\frac{4^{n + 1} + 1}{n + 1}$

3) $\frac{4^{n + 1} - 1}{n + 1}$

4) $\frac{4^{n} - 1}{n + 1}$

Solution:

$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{n} x^{n}$

Integrating within limits $0$ to $3$ , then we get,

$\int_{0}^{3} (1 + x)^{n} d x = \int_{0}^{3} (C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{n} x^{n}) d x$

$\Rightarrow {[\frac{(1 + x)^{n + 1}}{n + 1}]}_{0}^{3} = [C_{0} x + \frac{C_{1} x^{2}}{2} + \frac{C_{2} x^{3}}{3} + \frac{C_{3} x^{4}}{4} + \dots . + \frac{C_{n} x^{n + 1}}{n + 1}]$

$\Rightarrow \frac{4^{n + 1} - 1}{n + 1} = 3 C_{0} + \frac{3^{2} C_{1}}{2} + \frac{3^{3} C_{2}}{3} + \frac{3^{4} C_{3}}{4} + \dots + \frac{3^{n + 1} C_{n}}{n + 1}$

Hence,

$3 C_{0} + \frac{3^{2} C_{1}}{2} + \frac{3^{3} C_{2}}{3} + \frac{3^{4} C_{3}}{4} + \dots + \frac{3^{n + 1} C_{n}}{n + 1} = \frac{4^{n + 1} - 1}{n + 1}$

Hence, the answer is option (3).

Example 5: $\frac{C_{0}}{1} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1} =$

1) $\frac{2^{n}}{n + 1}$

2) $\frac{2^{n} - 1}{n + 1}$

3) $\frac{2^{n + 1} - 1}{n + 1}$

4) None of these

Solution:

Consider the expansion $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots . + C_{n} x^{n}$

Integrating both sides of (i) within limits $0$ to $1$ we get.

$\int_{0}^{1} (1 + x)^{n} d x = \int_{0}^{1} C_{0} + \int_{0}^{1} C_{1} x + \int_{0}^{1} C_{2} x^{2} + \dots \dots + \int_{0}^{1} C_{n} x^{n} d x$

${[\frac{(1 + 2)^{n + 1}}{n + 1}]}_{0}^{1} = C_{0} [x]_{0}^{1} + C_{1} {[\frac{x^{2}}{2}]}_{0}^{1} + \dots \dots + C_{n} {[\frac{x^{n + 1}}{n + 1}]}_{0}^{1}$

$\frac{2^{n + 1}}{n + 1} - \frac{1}{n + 1} - C_{0} [1] + C_{1} \frac{1}{2} + C_{2} \frac{1}{3} + \dots \dots + C_{n} \cdot \frac{1}{n + 1}; C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \dots + \frac{C_{n}}{n + 1} - \frac{2^{n + 1} - 1}{n + 1}$