Let a_n = \int\limits_ - 1^n ( 1 + x \over 2 + x^2 \over 3 + \,\,.....\,\, + \,\,x^n - 1 \over n )dx for every n \in N. Then the sum of all the elements of the set n \in N : an \in (2, 30) is \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q43 with Solution

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Correct answer: 5

Write the integral in summation form

Given

a_n=\int_{-1}^{n}\left(1+\frac{x}{2}+\frac{x^2}{3}+\cdots+\frac{x^{n-1}}{n}\right)dx

we rewrite the integrand as

\sum_{k=0}^{n-1} \frac{x^k}{k+1}.

So,

a_n=\int_{-1}^{n} \sum_{k=0}^{n-1}\frac{x^k}{k+1}\,dx.

Integrate term-by-term

Since

\int \frac{x^k}{k+1}\,dx=\frac{x^{k+1}}{(k+1)^2},

we get

a_n=\sum_{k=0}^{n-1}\left[\frac{x^{k+1}}{(k+1)^2}\right]_{-1}^{n}.

Let $j=k+1$ . Then $j=1,2,\dots,n$ . Hence

a_n=\sum_{j=1}^{n}\frac{n^j-(-1)^j}{j^2}.

We need those $n\in\mathbb N$ for which $a_n\in(2,30)$

So check small values of $n$ .

For $n=1$

a_1=\int_{-1}^{1}1\,dx=2.

But interval is open $(2,30)$ , so $2\notin(2,30)$ . Thus, $n=1$ is not included.

For $n=2$

Integrand is

1+\frac{x}{2}.

Then

a_2=\int_{-1}^{2}\left(1+\frac{x}{2}\right)dx =\left[x+\frac{x^2}{4}\right]_{-1}^{2}.

Evaluate:

\left(2+1\right)-\left(-1+\frac14\right)=3-\left(-\frac34\right)=\frac{15}{4}=3.75.

a_2\in(2,30).

Thus, $n=2$ is included.

For $n=3$

Using the formula,

a_3=\sum_{j=1}^{3}\frac{3^j-(-1)^j}{j^2}.

Compute each term:

\frac{3-(-1)}{1^2}=4,

\frac{9-1}{2^2}=\frac84=2,

\frac{27-(-1)}{3^2}=\frac{28}{9}.

Hence

a_3=4+2+\frac{28}{9}=6+\frac{28}{9}=\frac{82}{9}>9.

a_3\in(2,30).

Thus, $n=3$ is included.

For $n=4$

It is enough to show $a_4>30$ . Using the formula,

a_4=\sum_{j=1}^{4}\frac{4^j-(-1)^j}{j^2}.

Already the first three terms give

\frac{4-(-1)}{1}=5,

\frac{16-1}{4}=\frac{15}{4},

\frac{64-(-1)}{9}=\frac{65}{9}.

a_4>5+\frac{15}{4}+\frac{65}{9}.

Now,

5+\frac{15}{4}+\frac{65}{9}=\frac{180+135+260}{36}=\frac{575}{36}>15.

Including the 4th term,

\frac{256-1}{16}=\frac{255}{16}>15.

Hence

a_4>15+15=30.

Therefore $a_4\notin(2,30)$ .

For all $n\ge 4$ , $a_n>30$

For $n\ge 4$ , all terms

\frac{n^j-(-1)^j}{j^2}>0

are positive, and in particular the first term is

\frac{n-(-1)}{1^2}=n+1\ge 5.

Also $a_n$ grows very rapidly with $n$ , so since $a_4>30$ , certainly for $n\ge 4$ we have $a_n>30$ .

Thus the only values of $n$ for which $a_n\in(2,30)$ are

n=2,\;3.

Required sum

2+3=5.

Therefore, the required integer is

\boxed{5}.

More from Definite Integration

View written solutionFree

Correct answer: 5

Write the integral in summation form

Given

a_n=\int_{-1}^{n}\left(1+\frac{x}{2}+\frac{x^2}{3}+\cdots+\frac{x^{n-1}}{n}\right)dx

we rewrite the integrand as

\sum_{k=0}^{n-1} \frac{x^k}{k+1}.

So,

a_n=\int_{-1}^{n} \sum_{k=0}^{n-1}\frac{x^k}{k+1}\,dx.

Integrate term-by-term

Since

\int \frac{x^k}{k+1}\,dx=\frac{x^{k+1}}{(k+1)^2},

we get

a_n=\sum_{k=0}^{n-1}\left[\frac{x^{k+1}}{(k+1)^2}\right]_{-1}^{n}.

Let $j=k+1$ . Then $j=1,2,\dots,n$ . Hence

a_n=\sum_{j=1}^{n}\frac{n^j-(-1)^j}{j^2}.

We need those $n\in\mathbb N$ for which $a_n\in(2,30)$

So check small values of $n$ .

For $n=1$

a_1=\int_{-1}^{1}1\,dx=2.

But interval is open $(2,30)$ , so $2\notin(2,30)$ . Thus, $n=1$ is not included.

For $n=2$

Integrand is

1+\frac{x}{2}.

Then

a_2=\int_{-1}^{2}\left(1+\frac{x}{2}\right)dx =\left[x+\frac{x^2}{4}\right]_{-1}^{2}.

Evaluate:

\left(2+1\right)-\left(-1+\frac14\right)=3-\left(-\frac34\right)=\frac{15}{4}=3.75.

a_2\in(2,30).

Thus, $n=2$ is included.

For $n=3$

Using the formula,

a_3=\sum_{j=1}^{3}\frac{3^j-(-1)^j}{j^2}.

Compute each term:

\frac{3-(-1)}{1^2}=4,

\frac{9-1}{2^2}=\frac84=2,

\frac{27-(-1)}{3^2}=\frac{28}{9}.

Hence

a_3=4+2+\frac{28}{9}=6+\frac{28}{9}=\frac{82}{9}>9.

a_3\in(2,30).

Thus, $n=3$ is included.

For $n=4$

It is enough to show $a_4>30$ . Using the formula,

a_4=\sum_{j=1}^{4}\frac{4^j-(-1)^j}{j^2}.

Already the first three terms give

\frac{4-(-1)}{1}=5,

\frac{16-1}{4}=\frac{15}{4},

\frac{64-(-1)}{9}=\frac{65}{9}.

a_4>5+\frac{15}{4}+\frac{65}{9}.

Now,

5+\frac{15}{4}+\frac{65}{9}=\frac{180+135+260}{36}=\frac{575}{36}>15.

Including the 4th term,

\frac{256-1}{16}=\frac{255}{16}>15.

Hence

a_4>15+15=30.

Therefore $a_4\notin(2,30)$ .

For all $n\ge 4$ , $a_n>30$

For $n\ge 4$ , all terms

\frac{n^j-(-1)^j}{j^2}>0

are positive, and in particular the first term is

\frac{n-(-1)}{1^2}=n+1\ge 5.

Also $a_n$ grows very rapidly with $n$ , so since $a_4>30$ , certainly for $n\ge 4$ we have $a_n>30$ .

Thus the only values of $n$ for which $a_n\in(2,30)$ are

n=2,\;3.

Required sum

2+3=5.

Therefore, the required integer is

\boxed{5}.

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