Why do we compare with 2 instead of general ?

Because on , is identically 1, so the expression is constant and equals 2. We only need to find where .

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

[x]

denote the greatest integer

\leq x

. Consider the function

f(x) = \max\{x^2, 1 + [x]\}

. Then the value of the integral

\int_0^2 f(x)dx

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{5+4\sqrt{2}}{3}

\frac{4+5\sqrt{2}}{3}

\frac{1+5\sqrt{2}}{3}

\frac{8+4\sqrt{2}}{3}

Step-by-step solutionView

Correct answer

Splitting the integral across the integer intervals and determining where $x^2$ exceeds $1 + [x]$ yields $\frac{5+4\sqrt{2}}{3}$ .

Option analysis

Why each option works or fails

A ·

\frac{5+4\sqrt{2}}{3}

None. This is the correct evaluation. Correct: On $[0, 1)$ , $f(x) = 1$ ; on $[1, \sqrt{2})$ , $f(x) = 2$ ; and on $[\sqrt{2}, 2]$ , $f(x) = x^2$ . Integrating these gives $1 + 2(\sqrt{2}-1) + \frac{8 - 2\sqrt{2}}{3} = \frac{5 + 4\sqrt{2}}{3}$ .

B ·

\frac{4+5\sqrt{2}}{3}

Swapping the constant and radical coefficients during the combination of the constant terms and radical terms. Carefully collect like terms: the constants sum to $1 - 2 + \frac{8}{3} = \frac{5}{3}$ , and the $\sqrt{2}$ terms sum to $2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$ .

C ·

\frac{1+5\sqrt{2}}{3}

Arithmetic sign error when evaluating the definite integral of $x^2$ or combining $-1$ with $\frac{8}{3}$ . Check the evaluation of $[x^3/3]_{\sqrt{2}}^2 = \frac{8 - 2\sqrt{2}}{3}$ ; adding this to $-1 + 2\sqrt{2}$ gives $\frac{5 + 4\sqrt{2}}{3}$ , not $\frac{1 + 5\sqrt{2}}{3}$ .

D ·

\frac{8+4\sqrt{2}}{3}

Forgetting to subtract the lower limit contribution when integrating over $[1, \sqrt{2})$ or miscomputing the constant term as $\frac{8}{3}$ without adjusting for $1 - 2 = -1$ . Ensure the integral $\int_1^{\sqrt{2}} 2\,dx = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2$ , so the total constant contribution is $1 - 2 + \frac{8}{3} = \frac{5}{3}$ .

Reviewed route

Solution

StepWorking

01given

$f(x) = \max\{x^2, 1 + [x]\}$ on $[0, 2]$ , where $[x]$ is the greatest integer $\leq x$ .

02goal

Evaluate $I = \int_0^2 f(x)\,dx$ .

03approach

Analyze $1 + [x]$ on natural sub-intervals $[0, 1)$ , $[1, 2)$ , and $\{2\}$ , and compare it with $x^2$ to find where $x^2$ overtakes $1 + [x]$ .

04execute

For $x \in [0, 1)$ , $[x] = 0 \implies 1 + [x] = 1$ . Since $x^2 < 1$ , $f(x) = 1$ . For $x \in [1, 2)$ , $[x] = 1 \implies 1 + [x] = 2$ . Comparing $x^2$ and $2$ : $x^2 \leq 2$ for $x \in [1, \sqrt{2}]$ , so $f(x) = 2$ . For $x \in (\sqrt{2}, 2)$ , $x^2 > 2$ , so $f(x) = x^2$ . At $x = 2$ , isolated point does not change the integral. Thus, $f(x) = 1$ on $[0, 1)$ , $2$ on $[1, \sqrt{2}]$ , and $x^2$ on $(\sqrt{2}, 2]$ .

05execute

Compute $\int_0^2 f(x)\,dx = \int_0^1 1\,dx + \int_1^{\sqrt{2}} 2\,dx + \int_{\sqrt{2}}^2 x^2\,dx = 1(1 - 0) + 2(\sqrt{2} - 1) + \left[\frac{x^3}{3}\right]_{\sqrt{2}}^2 = 1 + 2\sqrt{2} - 2 + \frac{8 - 2\sqrt{2}}{3} = -1 + 2\sqrt{2} + \frac{8 - 2\sqrt{2}}{3} = \frac{-3 + 6\sqrt{2} + 8 - 2\sqrt{2}}{3} = \frac{5 + 4\sqrt{2}}{3}$ .

✓verify

Approximate values: $\frac{5 + 4(1.414)}{3} = \frac{10.656}{3} \approx 3.55$ . Areas: $\int_0^1 1 = 1$ , $\int_1^{\sqrt{2}} 2 \approx 2(0.414) = 0.828$ , $\int_{\sqrt{2}}^2 x^2 \approx \frac{8 - 2.828}{3} = \frac{5.172}{3} \approx 1.724$ . Total $\approx 1 + 0.828 + 1.724 = 3.552$ . Matches perfectly.

Hints that build this answer step by step

How should the domain $[0, 2]$ be partitioned to evaluate $f(x) = \max\{x^2, 1 + [x]\}$ ?
Partition into $[0, 1)$ , $[1, \sqrt{2})$ , and $[\sqrt{2}, 2]$ based on where $[x]$ changes and where $x^2 = 1 + [x]$ .
What is the piecewise expression for $f(x)$ on each sub-interval?
$f(x) = 1$ on $[0, 1)$ , $f(x) = 2$ on $[1, \sqrt{2})$ , and $f(x) = x^2$ on $[\sqrt{2}, 2]$ .
What is the sum of the integrals $\int_0^1 1\,dx + \int_1^{\sqrt{2}} 2\,dx + \int_{\sqrt{2}}^2 x^2\,dx$ ?
$\frac{5 + 4\sqrt{2}}{3}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we compare

x^2

with 2 instead of general

1+[x]

Because on $[1, 2)$ , $[x]$ is identically 1, so the expression $1 + [x]$ is constant and equals 2. We only need to find where $x^2 \geq 2$ .

Answer

Splitting the integral across the integer intervals and determining where $x^2$ exceeds $1 + [x]$ yields $\frac{5+4\sqrt{2}}{3}$ .

Why each option works or fails

A · correct: $\frac{5+4\sqrt{2}}{3}$ - None. This is the correct evaluation. Correct: On $[0, 1)$ , $f(x) = 1$ ; on $[1, \sqrt{2})$ , $f(x) = 2$ ; and on $[\sqrt{2}, 2]$ , $f(x) = x^2$ . Integrating these gives $1 + 2(\sqrt{2}-1) + \frac{8 - 2\sqrt{2}}{3} = \frac{5 + 4\sqrt{2}}{3}$ .
B: $\frac{4+5\sqrt{2}}{3}$ - Swapping the constant and radical coefficients during the combination of the constant terms and radical terms. Carefully collect like terms: the constants sum to $1 - 2 + \frac{8}{3} = \frac{5}{3}$ , and the $\sqrt{2}$ terms sum to $2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$ .
C: $\frac{1+5\sqrt{2}}{3}$ - Arithmetic sign error when evaluating the definite integral of $x^2$ or combining $-1$ with $\frac{8}{3}$ . Check the evaluation of $[x^3/3]_{\sqrt{2}}^2 = \frac{8 - 2\sqrt{2}}{3}$ ; adding this to $-1 + 2\sqrt{2}$ gives $\frac{5 + 4\sqrt{2}}{3}$ , not $\frac{1 + 5\sqrt{2}}{3}$ .
D: $\frac{8+4\sqrt{2}}{3}$ - Forgetting to subtract the lower limit contribution when integrating over $[1, \sqrt{2})$ or miscomputing the constant term as $\frac{8}{3}$ without adjusting for $1 - 2 = -1$ . Ensure the integral $\int_1^{\sqrt{2}} 2\,dx = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2$ , so the total constant contribution is $1 - 2 + \frac{8}{3} = \frac{5}{3}$ .

Step-by-step solution

given: $f(x) = \max\{x^2, 1 + [x]\}$ on $[0, 2]$ , where $[x]$ is the greatest integer $\leq x$ .
goal: Evaluate $I = \int_0^2 f(x)\,dx$ .
approach: Analyze $1 + [x]$ on natural sub-intervals $[0, 1)$ , $[1, 2)$ , and $\{2\}$ , and compare it with $x^2$ to find where $x^2$ overtakes $1 + [x]$ .
execute: For $x \in [0, 1)$ , $[x] = 0 \implies 1 + [x] = 1$ . Since $x^2 < 1$ , $f(x) = 1$ . For $x \in [1, 2)$ , $[1, \sqrt{2})$ . Comparing $x^2$ and $2$ : $x^2 \leq 2$ for $x \in [1, \sqrt{2}]$ , so $f(x) = 2$ . For $x \in (\sqrt{2}, 2)$ , $x^2 > 2$ , so $f(x) = x^2$ . At $x = 2$ , isolated point does not change the integral. Thus, $f(x) = 1$ on $[0, 1)$ , $2$ on $[1, \sqrt{2}]$ , and $x^2$ on $(\sqrt{2}, 2]$ .
execute: Compute $\int_0^2 f(x)\,dx = \int_0^1 1\,dx + \int_1^{\sqrt{2}} 2\,dx + \int_{\sqrt{2}}^2 x^2\,dx = 1(1 - 0) + 2(\sqrt{2} - 1) + \left[\frac{x^3}{3}\right]_{\sqrt{2}}^2 = 1 + 2\sqrt{2} - 2 + \frac{8 - 2\sqrt{2}}{3} = -1 + 2\sqrt{2} + \frac{8 - 2\sqrt{2}}{3} = \frac{-3 + 6\sqrt{2} + 8 - 2\sqrt{2}}{3} = \frac{5 + 4\sqrt{2}}{3}$ .
verify: Approximate values: $\frac{5 + 4(1.414)}{3} = \frac{10.656}{3} \approx 3.55$ . Areas: $\int_0^1 1 = 1$ , $\int_1^{\sqrt{2}} 2 \approx 2(0.414) = 0.828$ , $\int_{\sqrt{2}}^2 x^2 \approx \frac{8 - 2.828}{3} = \frac{5.172}{3} \approx 1.724$ . Total $\approx 1 + 0.828 + 1.724 = 3.552$ . Matches perfectly.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

Splitting the integral across the integer intervals and determining where $x^2$ exceeds $1 + [x]$ yields $\frac{5+4\sqrt{2}}{3}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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