Let [.] denote the greatest integer function. Then the value of integral from 0 to 3 of (e^x + e^-x) / ([x]!) dx is :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths IntegralsQuestion #204

Let

[.]

denote the greatest integer function. Then the value of

\int_{0}^{3} \frac{e^{x} + e^{- x}}{[x]!} d x

is :

Clue 1 of 3: Greatest Integer Function

The greatest integer function $[x]$ is a step function, which means its value is constant over intervals of the form $[n, n + 1)$ . This property is crucial for splitting the integral.

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● Video Walkthrough

Step 1: Split the integral based on the greatest integer function✦ Active

The greatest integer function $[x]$ changes its value at integer points. For the interval $[0, 3]$ , we need to consider the intervals $[0, 1)$ , $[1, 2)$ , and $[2, 3)$ . The value of $[x]!$ will be constant within these intervals:

- For $x \in [0, 1)$ , $[x] = 0$ , so $[x]! = 0! = 1$ .

- For $x \in [1, 2)$ , $[x] = 1$ , so $[x]! = 1! = 1$ .

- For $x \in [2, 3)$ , $[x] = 2$ , so $[x]! = 2! = 2$ .

The integral can be written as:

I = \int_{0}^{1} \frac{e^{x} + e^{- x}}{0!} d x + \int_{1}^{2} \frac{e^{x} + e^{- x}}{1!} d x + \int_{2}^{3} \frac{e^{x} + e^{- x}}{2!} d x

I = \int_{0}^{1} (e^{x} + e^{- x}) d x + \int_{1}^{2} (e^{x} + e^{- x}) d x + \frac{1}{2} \int_{2}^{3} (e^{x} + e^{- x}) d x

Step 2: Evaluate each definite integral○ Expand

The antiderivative of $e^{x} + e^{- x}$ is $e^{x} - e^{- x}$ . Evaluate each part:

I_{1} = [e^{x} - e^{- x}]_{0}^{1} = (e^{1} - e^{- 1}) - (e^{0} - e^{0}) = e - \frac{1}{e}

I_{2} = [e^{x} - e^{- x}]_{1}^{2} = (e^{2} - e^{- 2}) - (e^{1} - e^{- 1}) = e^{2} - \frac{1}{e^{2}} - e + \frac{1}{e}

I_{3} = \frac{1}{2} [e^{x} - e^{- x}]_{2}^{3} = \frac{1}{2} ((e^{3} - e^{- 3}) - (e^{2} - e^{- 2})) = \frac{1}{2} (e^{3} - \frac{1}{e^{3}} - e^{2} + \frac{1}{e^{2}})

💡 Teacher's Secret Hint

Remember that $e^{0} = 1$ and $0! = 1$ .

Step 3: Sum the results of the integrals○ Expand

Add the results from the three parts:

I = (e - \frac{1}{e}) + (e^{2} - \frac{1}{e^{2}} - e + \frac{1}{e}) + \frac{1}{2} (e^{3} - \frac{1}{e^{3}} - e^{2} + \frac{1}{e^{2}})

Combine like terms:

I = (e - e) + (- \frac{1}{e} + \frac{1}{e}) + (e^{2} - \frac{e^{2}}{2}) + (- \frac{1}{e^{2}} + \frac{1}{2 e^{2}}) + \frac{e^{3}}{2} - \frac{1}{2 e^{3}}

I = 0 + 0 + \frac{e^{2}}{2} - \frac{1}{2 e^{2}} + \frac{e^{3}}{2} - \frac{1}{2 e^{3}}

I = \frac{1}{2} (e^{2} - \frac{1}{e^{2}} + e^{3} - \frac{1}{e^{3}})

I = \frac{1}{2} (e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}})

This matches option 2.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Greatest Integer Function

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