Logo
Jeeify
MCQMediumJEE 2026Definite Integrals

JEE Mathematics 2026 Question with Solution

The value of the integral ∫−π2π21[x]+4 dx,\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{[x]+4}\,dx, where [⋅][\cdot] denotes the greatest integer function, is

  • A

    160(π−7)\dfrac{1}{60}(\pi-7)

  • B

    160(21π−1)\dfrac{1}{60}(21\pi-1)

  • C

    760(3π−1)\dfrac{7}{60}(3\pi-1)

  • D

    760(π−3)\dfrac{7}{60}(\pi-3)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

∫−π2π21[x]+4 dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{[x]+4}\,dx

where [x][x] denotes the greatest integer less than or equal to xx.

Find: The value of the integral and the correct option.

For the greatest integer function, split the interval at integer points inside [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right].

Since

−π2≈−1.57andπ2≈1.57,-\frac{\pi}{2}\approx -1.57 \quad \text{and} \quad \frac{\pi}{2}\approx 1.57,

the interval is divided as

[−π2,−1), [−1,0), [0,1), [1,π2].\left[-\frac{\pi}{2},-1\right),\ \left[-1,0\right),\ \left[0,1\right),\ \left[1,\frac{\pi}{2}\right].

Hence,

  • on [−π2,−1)\left[-\frac{\pi}{2},-1\right), [x]=−2[x]=-2, so the integrand is 12\frac{1}{2},
  • on [−1,0)\left[-1,0\right), [x]=−1[x]=-1, so the integrand is 13\frac{1}{3},
  • on [0,1)\left[0,1\right), [x]=0[x]=0, so the integrand is 14\frac{1}{4},
  • on [1,π2]\left[1,\frac{\pi}{2}\right], [x]=1[x]=1, so the integrand is 15\frac{1}{5}.

Therefore,

∫−π2π21[x]+4 dx=∫−π2−112 dx+∫−1013 dx+∫0114 dx+∫1π215 dx\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{[x]+4}\,dx = \int_{-\frac{\pi}{2}}^{-1} \frac{1}{2}\,dx + \int_{-1}^{0} \frac{1}{3}\,dx + \int_{0}^{1} \frac{1}{4}\,dx + \int_{1}^{\frac{\pi}{2}} \frac{1}{5}\,dx

Now evaluate each part:

∫−π2−112 dx=12(−1+π2)\int_{-\frac{\pi}{2}}^{-1} \frac{1}{2}\,dx = \frac{1}{2}\left(-1+\frac{\pi}{2}\right) ∫−1013 dx=13\int_{-1}^{0} \frac{1}{3}\,dx = \frac{1}{3} ∫0114 dx=14\int_{0}^{1} \frac{1}{4}\,dx = \frac{1}{4} ∫1π215 dx=15(π2−1)\int_{1}^{\frac{\pi}{2}} \frac{1}{5}\,dx = \frac{1}{5}\left(\frac{\pi}{2}-1\right)

Adding,

12(π2−1)+13+14+15(π2−1)\frac{1}{2}\left(\frac{\pi}{2}-1\right)+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\left(\frac{\pi}{2}-1\right) =(π2−1)(12+15)+13+14= \left(\frac{\pi}{2}-1\right)\left(\frac{1}{2}+\frac{1}{5}\right)+\frac{1}{3}+\frac{1}{4} =710(π2−1)+712= \frac{7}{10}\left(\frac{\pi}{2}-1\right)+\frac{7}{12} =7π20−710+712= \frac{7\pi}{20}-\frac{7}{10}+\frac{7}{12} =21π−760= \frac{21\pi-7}{60} =760(3π−1)= \frac{7}{60}(3\pi-1)

Therefore, the value of the integral is 760(3π−1)\dfrac{7}{60}(3\pi-1) and the correct option is C.

The solution's intermediate integrands contain a mismatch, but the final result and option selection clearly give C.](streamdown:incomplete-link)

Piecewise Evaluation of Greatest Integer Function

Given: An integral involving the greatest integer function.

Find: The exact value by evaluating the integrand on each integer subinterval.

The key idea is that [x][x] remains constant between consecutive integers. So first identify the integer breakpoints inside the interval [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right], namely −1,0,1-1,0,1.

Then write the four constant values:

[x]=−2,x∈[−π2,−1)[x]=−1,x∈[−1,0)[x]=0,x∈[0,1)[x]=1,x∈[1,π2]\begin{aligned} [x] &= -2, && x \in \left[-\frac{\pi}{2},-1\right) \\ [x] &= -1, && x \in \left[-1,0\right) \\ [x] &= 0, && x \in \left[0,1\right) \\ [x] &= 1, && x \in \left[1,\frac{\pi}{2}\right] \end{aligned}

Hence,

1[x]+4={12,x∈[−π2,−1)13,x∈[−1,0)14,x∈[0,1)15,x∈[1,π2]\frac{1}{[x]+4}= \begin{cases} \frac{1}{2}, & x \in \left[-\frac{\pi}{2},-1\right) \\ \frac{1}{3}, & x \in \left[-1,0\right) \\ \frac{1}{4}, & x \in \left[0,1\right) \\ \frac{1}{5}, & x \in \left[1,\frac{\pi}{2}\right] \end{cases}

Now integrate constant functions over their interval lengths:

I=12(−1+π2)+13(1)+14(1)+15(π2−1)=12(π2−1)+13+14+15(π2−1)=710(π2−1)+712=7π20−710+712=21π−42+3560=21π−760=760(3π−1)\begin{aligned} I &= \frac{1}{2}\left(-1+\frac{\pi}{2}\right)+\frac{1}{3}(1)+\frac{1}{4}(1)+\frac{1}{5}\left(\frac{\pi}{2}-1\right) \\ &= \frac{1}{2}\left(\frac{\pi}{2}-1\right)+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\left(\frac{\pi}{2}-1\right) \\ &= \frac{7}{10}\left(\frac{\pi}{2}-1\right)+\frac{7}{12} \\ &= \frac{7\pi}{20}-\frac{7}{10}+\frac{7}{12} \\ &= \frac{21\pi-42+35}{60} \\ &= \frac{21\pi-7}{60} \\ &= \frac{7}{60}(3\pi-1) \end{aligned}

So the correct option is C.](streamdown:incomplete-link)

Common mistakes

  • A common mistake is taking [x][x] as the nearest integer instead of the greatest integer less than or equal to xx. For example, on x∈[−π2,−1)x \in \left[-\frac{\pi}{2},-1\right), [x][x] is −2-2, not −1-1. Always use the floor definition carefully.](streamdown:incomplete-link)

  • Another mistake is not splitting the interval at the integer points −1,0,1-1,0,1. Since [x][x] changes value only at integers, failing to break the interval there gives an incorrect integrand over parts of the domain.

  • A frequent error is copying the piecewise denominators incorrectly, such as writing [x]+4=3,4,5,6[x]+4=3,4,5,6 on the four subintervals. Check each interval separately by first finding [x][x] and then adding 44.

Practice more Definite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Definite Integrals

Related questions