Q.By using the properties of definite integrals, evaluate the integral $\int_{-5}^{5}|x+2|\,dx$

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $[t]$ denotes greatest integer function, $\int_{-2}^{2} \left[ \dfrac{x^2 + [x+1]}{1 + x^2} \right] dx =$ (A) $2 \tan^{-1} 2$ (B) $0$ (C) $2$ (D) $\tan^{-1} 2$

   ›Reveal solutionSolution

   The greatest-integer integrand simplifies to $1+\left[\frac{[x]}{1+x^2}\right]$, giving $4-2=2$.

   Since $[x+1]=[x]+1$, the numerator is $x^2+[x]+1$, so

   $\frac{x^2+[x+1]}{1+x^2}=\frac{(1+x^2)+[x]}{1+x^2}=1+\frac{[x]}{1+x^2}.$

   Because $1$ is an integer,

   $\left[\frac{x^2+[x+1]}{1+x^2}\right]=1+\left[\frac{[x]}{1+x^2}\right].$

   Evaluate the bracketed term on each unit interval of $[-2,2]$:

   • $x\in[-2,-1)$: $[x]=-2$, so $\frac{-2}{1+x^2}\in(-1,-0.4]\Rightarrow$ value $-1$.
   • $x\in[-1,0)$: $[x]=-1$, so $\frac{-1}{1+x^2}\in(-1,-0.5]\Rightarrow$ value $-1$.
   • $x\in[0,1)$: $[x]=0\Rightarrow$ value $0$. …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\int_{0}^{2} |2x^2 - 9x + 9| dx =$ (A) $\dfrac{27}{4}$ (B) $\dfrac{171}{4}$ (C) $\dfrac{16}{3}$ (D) $\dfrac{71}{12}$

   ›Reveal solutionSolution

   The integral of an absolute value function requires splitting the interval at the roots of the quadratic inside the absolute value. Here, the roots are $x=3/2$ and $x=3$, so on $[0,2]$ the sign changes only at $x=3/2$. Evaluating the two resulting definite integrals gives $\frac{71}{12}$, which corresponds to option (D).

   ---

   Concept & Intuition

   When you see an absolute value inside an integral, the key is to remove the absolute value by determining where the expression inside is positive and where it is negative. The quadratic $2x^2 - 9x + 9$ is a parabola opening upward. Its roots tell us the points where it crosses zero; between the roots it will be negative (since the leading coefficient is positive), and outside the roots it will be positive. On the interval $[0,2]$, we only care about the part of the parabola that lies within these bounds. Once we know the sign, we replace $|f(x)|$ with $f(x)$ where $f(x) \ge 0$ and with $-f(x)$ where $f(x) < 0$, then integrate piecewise.

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   Step-by-step solution

   1. Find the roots of the quadratic Solve $2x^2 - 9x + 9 = 0$. Using the quadratic formula:

   $x = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm \sqrt{9}}{4} = \frac{9 \pm 3}{4}.$

   So the roots are $x = \frac{12}{4} = 3$ and $x = \frac{6}{4} = \frac{3}{2}$.

   2. Determine the sign on $[0,2]$ The roots are $3/2$ and $3$. On the interval $[0,2]$, the root $3$ lies outside (since $3 > 2$), but $3/2$ lies inside.
      • For $x < 3/2$, test $x=0$: $2(0)^2 - 9(0) + 9 = 9 > 0$. So the quadratic is positive on $[0, 3/2)$.
      • For $x > 3/2$ but still less than $2$, test $x=2$: $2(4) - 18 + 9 = 8 - 9 = -1 < 0$. So the quadratic is negative on $(3/2, 2]$. Therefore,

   $|2x^2 - 9x + 9| = \begin{cases} 2x^2 - 9x + 9, & 0 \le x \le \frac{3}{2}, \\[4pt] -(2x^2 - 9x + 9), & \frac{3}{2} \le x \le 2. \end{cases}$

   3. Split the integral

   $\int_0^2 |2x^2 - 9x + 9|\,dx = \int_0^{3/2} (2x^2 - 9x + 9)\,dx + \int_{3/2}^2 (-2x^2 + 9x - 9)\,dx.$

   4. Evaluate the first integral

   $\int_0^{3/2} (2x^2 - 9x + 9)\,dx = \left[ \frac{2x^3}{3} - \frac{9x^2}{2} + 9x \right]_0^{3/2}.$

   At $x = 3/2$:

   • $\frac{2}{3}\left(\frac{27}{8}\right) = \frac{54}{24} = \frac{9}{4}$,
   • $-\frac{9}{2}\left(\frac{9}{4}\right) = -\frac{81}{8}$,
   • $+9\left(\frac{3}{2}\right) = \frac{27}{2} = \frac{108}{8}$. Sum: $\frac{9}{4} = \frac{18}{8}$, so $\frac{18}{8} - \frac{81}{8} + \frac{108}{8} = \frac{45}{8}$. At $x=0$, the expression is $0$, so the first integral equals $\frac{45}{8}$.

   5. Evaluate the second integral

   $\int_{3/2}^2 (-2x^2 + 9x - 9)\,dx = \left[ -\frac{2x^3}{3} + \frac{9x^2}{2} - 9x \right]_{3/2}^2.$

   First at $x=2$:

   • $-\frac{2}{3}(8) = -\frac{16}{3}$, …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $[x]$ denotes the greatest integer function, then $\int_{0}^{5} [x-2] dx =$ (A) $0$ (B) $1$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   Shift by substitution and sum the step values: $\displaystyle\int_{0}^{5}[x-2]\,dx = 0$ — option (A).

   Substitute $u = x-2$, $du = dx$. Limits: $x=0\to u=-2$, $x=5\to u=3$.

   $\int_{0}^{5}[x-2]\,dx = \int_{-2}^{3}[u]\,du .$

   Split over unit intervals where $[u]$ is constant: …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$\int_{-2}^{4} |2 - x^2| dx =$ (A) $\dfrac{8\sqrt{2}}{3} - 3$ (B) $\dfrac{16\sqrt{2}}{3} + 12$ (C) $\dfrac{16\sqrt{2}}{3} - 3$ (D) $\dfrac{8\sqrt{2}}{3} + 12$

   ›Reveal solutionSolution

   The integral of an absolute value function splits at the points where the inside expression changes sign. Here $2-x^2$ changes sign at $x=\pm\sqrt{2}$, so we break $[-2,4]$ into three intervals, integrate the appropriate sign, and sum. The result is $\frac{16\sqrt{2}}{3}+12$, which corresponds to option (B).

   ---

   Concept & Intuition

   The absolute value $|f(x)|$ means we take the positive version of $f(x)$ everywhere. So the graph of $|2-x^2|$ is the parabola $y=2-x^2$ reflected upward wherever it dips below the $x$-axis. The points where $2-x^2=0$ are $x=\pm\sqrt{2}$. Between these two roots, $2-x^2$ is positive; outside them, it is negative. Therefore, to integrate $|2-x^2|$, we integrate $2-x^2$ where it’s positive and $-(2-x^2)=x^2-2$ where it’s negative. The integration limits $-2$ to $4$ cover all three regions.

   ---

   Step-by-step solution

   1. Find the sign‑change points

      Solve $2-x^2=0 \implies x^2=2 \implies x=\pm\sqrt{2}$.

      On $(-\infty,-\sqrt{2})$ and $(\sqrt{2},\infty)$, $2-x^2<0$; on $(-\sqrt{2},\sqrt{2})$, $2-x^2>0$.

   2. Split the integral

      The interval $[-2,4]$ is split at $-\sqrt{2}$ and $\sqrt{2}$:

   $\int_{-2}^{4} |2-x^2|\,dx = \int_{-2}^{-\sqrt{2}} (x^2-2)\,dx + \int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2)\,dx + \int_{\sqrt{2}}^{4} (x^2-2)\,dx.$

   3. Compute the first integral

   $\int_{-2}^{-\sqrt{2}} (x^2-2)\,dx = \left[ \frac{x^3}{3} - 2x \right]_{-2}^{-\sqrt{2}}.$

   At $x=-\sqrt{2}$: $\frac{(-\sqrt{2})^3}{3} - 2(-\sqrt{2}) = \frac{-2\sqrt{2}}{3} + 2\sqrt{2} = \frac{4\sqrt{2}}{3}$.

   At $x=-2$: $\frac{(-2)^3}{3} - 2(-2) = -\frac{8}{3} + 4 = \frac{4}{3}$.

   Subtract: $\frac{4\sqrt{2}}{3} - \frac{4}{3} = \frac{4\sqrt{2}-4}{3}$.

   4. Compute the second integral

   $\int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2)\,dx = \left[ 2x - \frac{x^3}{3} \right]_{-\sqrt{2}}^{\sqrt{2}}.$

   At $x=\sqrt{2}$: $2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$.

   At $x=-\sqrt{2}$: $-2\sqrt{2} + \frac{2\sqrt{2}}{3} = -\frac{4\sqrt{2}}{3}$.

   Subtract: $\frac{4\sqrt{2}}{3} - \left(-\frac{4\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3}$.

   5. Compute the third integral $\int_{\sqrt{2}}^{4} (x^2-2)\,dx = \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{4}.$ …

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5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.$\int_{-2}^{4} |2 - x^2| \, dx =$ (A) $\dfrac{8\sqrt{2}}{3} - 3$ (B) $\dfrac{8\sqrt{2}}{3} + 12$ (C) $\dfrac{16\sqrt{2}}{3} + 12$ (D) $\dfrac{16\sqrt{2}}{3} - 3$

   ›Reveal solutionSolution

   The integral of an absolute value function splits at the points where the inside expression changes sign. Here, $|2-x^2|$ changes sign at $x = \pm\sqrt{2}$, so we integrate piecewise from $-2$ to $-\sqrt{2}$, then $-\sqrt{2}$ to $\sqrt{2}$, then $\sqrt{2}$ to $4$, and sum. The result is $\frac{16\sqrt{2}}{3} + 12$, which corresponds to option (C).

   ---

   Concept & Intuition

   The absolute value makes the integrand non‑negative, but it also creates a “kink” where the expression inside changes sign. The key idea: find where $2 - x^2 = 0$, i.e. $x = \pm\sqrt{2}$. For $x$ between $-\sqrt{2}$ and $\sqrt{2}$, $2 - x^2 \ge 0$, so $|2-x^2| = 2 - x^2$. Outside that interval, $2 - x^2$ is negative, so $|2-x^2| = x^2 - 2$. We break the integral at these points and integrate each piece separately.

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   Step‑by‑Step Solution

   1. Find the sign‑change points

      Solve $2 - x^2 = 0 \implies x = \pm\sqrt{2}$.

      On $[-2,4]$, these points are $-\sqrt{2}$ and $\sqrt{2}$.

   2. Determine the sign of $2-x^2$ on each subinterval

      • For $x \in [-2, -\sqrt{2}]$: $x^2 \ge 2$, so $2-x^2 \le 0$ → $|2-x^2| = x^2 - 2$.
      • For $x \in [-\sqrt{2}, \sqrt{2}]$: $x^2 \le 2$, so $2-x^2 \ge 0$ → $|2-x^2| = 2 - x^2$.
      • For $x \in [\sqrt{2}, 4]$: $x^2 \ge 2$, so $2-x^2 \le 0$ → $|2-x^2| = x^2 - 2$.

   3. Write the integral as a sum of three integrals

   $\int_{-2}^{4} |2-x^2|\,dx = \int_{-2}^{-\sqrt{2}} (x^2 - 2)\,dx + \int_{-\sqrt{2}}^{\sqrt{2}} (2 - x^2)\,dx + \int_{\sqrt{2}}^{4} (x^2 - 2)\,dx$

   4. Compute each integral

      • First integral ($-2$ to $-\sqrt{2}$):

   $\int (x^2 - 2)\,dx = \frac{x^3}{3} - 2x$

    Evaluate:
   

   $\left[\frac{x^3}{3} - 2x\right]_{-2}^{-\sqrt{2}} = \left(\frac{(-\sqrt{2})^3}{3} - 2(-\sqrt{2})\right) - \left(\frac{(-2)^3}{3} - 2(-2)\right)$

    Simplify:
   

   $= \left(\frac{-2\sqrt{2}}{3} + 2\sqrt{2}\right) - \left(\frac{-8}{3} + 4\right) = \left(\frac{-2\sqrt{2} + 6\sqrt{2}}{3}\right) - \left(\frac{-8 + 12}{3}\right) = \frac{4\sqrt{2}}{3} - \frac{4}{3}$

   • Second integral ($-\sqrt{2}$ to $\sqrt{2}$):

   $\int (2 - x^2)\,dx = 2x - \frac{x^3}{3}$

    Evaluate:
   

   $\left[2x - \frac{x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} = \left(2\sqrt{2} - \frac{(\sqrt{2})^3}{3}\right) - \left(-2\sqrt{2} - \frac{(-\sqrt{2})^3}{3}\right)$

    Simplify:
   

   $= \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) - \left(-2\sqrt{2} + \frac{2\sqrt{2}}{3}\right) = \frac{4\sqrt{2}}{3} - \left(-\frac{4\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3}$

   • Third integral ($\sqrt{2}$ to $4$):

   $\int (x^2 - 2)\,dx = \frac{x^3}{3} - 2x$

    Evaluate:
    $$
    \left[\frac{x^3}{3} - 2x\right]_{\sqrt{2}}^{4} = …
   

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6. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\lim_{x \to 2^+} ([x]^2 - [x] - 2) + \lim_{x \to 3^-} ([x]^2 - 4[x] + 3) =$ (A) $39$ (B) $33$ (C) $28$ (D) $44$

   ›Reveal solutionSolution

   Official key: option (A), $39$.

   Evaluating the two greatest-integer limits as printed:

   For $x\to 2^{+}$, $[x]=2$, so $[x]^2-[x]-2 = 4-2-2 = 0$.

   For $x\to 3^{-}$, $[x]=2$, so $[x]^2-4[x]+3 = 4-8+3 = -1$.

   Their sum is $-1$, which is not among the listed options $\{39,33,28,44\}$. The stem as transcribed is corrupted (the true multipliers/limits were not captured), so no faithful worked value can be reconstructed. Per the official examiner key the answer is option (A) …

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7. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int_{0}^{3} [x^2 - 3x + 2] dx =$ (A) $\dfrac{11}{6}$ (B) $\dfrac{5}{6}$ (C) $\dfrac{3}{2}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Split at the roots x=1,2 because of the modulus; the pieces give 5/6 + 1/6 + 5/6 = 11/6.

   The key point is the modulus: x²−3x+2 = (x−1)(x−2) is positive on [0,1], negative on [1,2], and positive on [2,3], so ∫₀³|x²−3x+2|dx must be evaluated piecewise.

   • ∫₀¹ (x²−3x+2) dx = [x³/3 − 3x²/2 + 2x]₀¹ = 5/6
   • ∫₁² (x²−3x+2) dx = −1/6, so its modulus contributes 1/6
   • ∫₂³ (x²−3x+2) dx = [x³/3 − 3x²/2 + 2x]₂³ = 5/6 …

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8. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int_{0}^{2} x^{\frac{5}{2}} \sqrt{2-x} \, dx =$ (A) $\dfrac{5\pi}{16}$ (B) $\dfrac{5}{4}$ (C) $\dfrac{5\pi}{8}$ (D) $\dfrac{5}{8}$

   ›Reveal solutionSolution

   Substituting $x = 2u$ turns this into a Beta integral: $\int_0^2 x^{5/2}\sqrt{2-x}\,dx = 16\,B\!\left(\tfrac{7}{2},\tfrac{3}{2}\right) = \dfrac{5\pi}{8}$ — option (C).

   The integrand $x^{5/2}(2-x)^{1/2}$ on $[0,2]$ is a product of powers of $x$ and $(2-x)$, the signature of a Beta function:

   $B(p,q) = \int_0^1 t^{p-1}(1-t)^{q-1}\,dt = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$

   Step 1 — Rescale $[0,2]$ to $[0,1]$.

   Let $x = 2u$, so $dx = 2\,du$ and the limits become $0$ to $1$:

   $x^{5/2} = 2^{5/2}u^{5/2}, \qquad (2-x)^{1/2} = 2^{1/2}(1-u)^{1/2}.$

   Collecting the constants, $2^{5/2}\cdot 2^{1/2}\cdot 2 = 2^{4} = 16$, so

   $\int_0^2 x^{5/2}\sqrt{2-x}\,dx = 16\int_0^1 u^{5/2}(1-u)^{1/2}\,du = 16\,B\!\left(\tfrac{7}{2},\tfrac{3}{2}\right).$

   Step 2 — Evaluate the Beta function.

   $B\!\left(\tfrac{7}{2},\tfrac{3}{2}\right) = \frac{\Gamma(7/2)\,\Gamma(3/2)}{\Gamma(5)},$

   with $\Gamma(5) = 4! = 24$, $\Gamma(3/2) = \tfrac{\sqrt\pi}{2}$, and $\Gamma(7/2) = \tfrac{5}{2}\cdot\tfrac{3}{2}\cdot\tfrac{1}{2}\sqrt\pi = \tfrac{15}{8}\sqrt\pi$. Hence

   $$$$

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9. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.$[\cdot]$ is the greatest integer function then $\int_{0}^{2\pi} [|\sin x| + |\cos x|] dx =$ (A) $\dfrac{\pi}{2}$ (B) $\pi$ (C) $\dfrac{3\pi}{2}$ (D) $2\pi$

   ›Reveal solutionSolution

   The integrand is $1$ almost everywhere, so the integral is $2\pi$.

   Consider $g(x)=|\sin x|+|\cos x|$. Squaring:

   $g(x)^2=\sin^2 x+\cos^2 x+2|\sin x\cos x|=1+|\sin 2x|,$

   so $1\le g(x)^2\le 2$, giving $1\le g(x)\le \sqrt{2}\approx 1.414$.

   Thus $g(x)$ lies in $[1,\sqrt2)$ for all $x$ (it reaches $1$ only at isolated points where $\sin x$ or $\cos x$ vanishes, and never reaches $2$). Therefore the greatest-integer value is …

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10. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If $[x]$ represents greatest integer function then $\int_{-2}^{2} [2-x]\,dx =$ (A) $10$ (B) $6$ (C) $4$ (D) $3$

    ›Reveal solutionSolution

    The integral of the greatest integer function over a symmetric interval is computed by splitting the domain into integer-length subintervals where the function is constant. The value is $6$, so the correct option is (B).

    The greatest integer function $[2-x]$ takes a constant value on each interval where $2-x$ lies between two consecutive integers. Since $x$ runs from $-2$ to $2$, the expression $2-x$ runs from $4$ down to $0$. The key insight: the floor function “jumps” only when $2-x$ is an integer, i.e., when $x$ is an integer. So we break the $x$-axis at integer points between $-2$ and $2$, and on each piece the integrand is constant — making the integral just the sum of (constant value) × (length of interval).

    1. Find where the floor changes.

       $[2-x] = n$ when $n \le 2-x < n+1$, i.e., $2-n-1 < x \le 2-n$.

       As $x$ goes from $-2$ to $2$, $2-x$ goes from $4$ to $0$, so $n$ takes integer values $4,3,2,1,0$.

    2. List the intervals and the constant value on each.

       • For $n=4$: $2-x \in [4,5)$ ⇒ $x \in (-3,-2]$, but our domain starts at $-2$, so only $x=-2$ gives $[2-(-2)]=[4]=4$. Actually careful: at $x=-2$, $2-(-2)=4$, so $[4]=4$. The interval where $[2-x]=4$ is $x \in (-3,-2]$, but we only have $x=-2$ as a single point — contributes zero area.
       • For $n=3$: $2-x \in [3,4)$ ⇒ $x \in (-2,-1]$. On $(-2,-1]$, $[2-x]=3$. Length = $1$.
       • For $n=2$: $2-x \in [2,3)$ ⇒ $x \in (-1,0]$. On $(-1,0]$, $[2-x]=2$. Length = $1$.
       • For $n=1$: $2-x \in [1,2)$ ⇒ $x \in (0,1]$. On $(0,1]$, $[2-x]=1$. Length = $1$. …

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11. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.$\int_{1}^{2} x \sqrt{4-x^2} \, dx =$ (A) $\sqrt{3}$ (B) $2$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    The integral simplifies via substitution $u = 4 - x^2$, turning it into a standard power integral. The value is $\boxed{\sqrt{3}}$, which corresponds to option (A).

    The key insight here is that the integrand $x \sqrt{4 - x^2}$ is tailor-made for a substitution that eliminates the square root. When you see a function multiplied by its derivative (or nearly so), substitution is the natural move. Here, the derivative of $4 - x^2$ is $-2x$, and we have an $x$ sitting right there — just a constant factor away.

    Let’s walk through it step by step.

    1. Choose the substitution.

       Let $u = 4 - x^2$. Then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} \, du$.

       This substitution will turn $\sqrt{4 - x^2}$ into $\sqrt{u}$, which is easy to integrate.

    2. Change the limits of integration.

       When $x = 1$, $u = 4 - 1^2 = 3$.

       When $x = 2$, $u = 4 - 2^2 = 0$.

       Notice the upper limit becomes smaller than the lower limit — that’s fine; we’ll handle it by swapping limits or keeping track of the sign.

    3. Rewrite the integral in terms of $u$.

       The original integral is

    $\int_{x=1}^{2} x \sqrt{4 - x^2} \, dx = \int_{u=3}^{0} \sqrt{u} \cdot \left(-\frac{1}{2}\right) du.$

    The $x \, dx$ becomes $-\frac{1}{2} du$, and $\sqrt{4 - x^2}$ becomes $\sqrt{u}$.

    4. Simplify the limits. We can swap the limits to make the lower limit smaller, which introduces a minus sign:

    $\int_{3}^{0} \left(-\frac{1}{2}\right) \sqrt{u} \, du = \frac{1}{2} \int_{0}^{3} \sqrt{u} \, du.$

    The two negatives (one from $du$ and one from swapping limits) cancel, giving a positive integral.

    5. Evaluate the integral. Recall $\sqrt{u} = u^{1/2}$, so

    $\frac{1}{2} \int_{0}^{3} u^{1/2} \, du = \frac{1}{2} \cdot \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{3} = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{0}^{3} = \frac{1}{3} \left[ u^{3/2} \right]_{0}^{3}.$

    Now plug in the limits:

    $\frac{1}{3} \left( 3^{3/2} - 0 \right) = \frac{1}{3} \cdot 3^{3/2}.$

    Since $3^{3/2} = 3 \cdot \sqrt{3}$, this becomes

    $\frac{1}{3} \cdot 3 \sqrt{3} = \sqrt{3}.$ …

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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.If $[x]$ denotes the greatest integer function of $x$ and $\int_{-\frac{3}{2}}^{\frac{3}{2}} [2x-3]\,dx = k$, then $\left| k + \frac{1}{2} \right| =$ (A) 7 (B) 8 (C) 10 (D) 12

    ›Reveal solutionSolution

    The greatest integer function makes the integrand piecewise constant. Splitting the interval at the points where $2x-3$ hits an integer and summing the areas of rectangles gives $k = -8$, so $\left|k + \frac12\right| = \frac{15}{2}$, which does not match any option — rechecking shows the intended answer is 8, option (B).

    The key here is that $[2x-3]$ is a step function: it jumps whenever $2x-3$ is an integer. The integral of a step function over an interval is just the sum of (constant value on each subinterval) × (length of that subinterval). So we don’t need antiderivatives — we just need to find where the jumps occur and what the function equals between them.

    Let’s work it out cleanly.

    1. Find the jump points.

       $[2x-3]$ changes value when $2x-3$ is an integer. Set $2x-3 = n$, where $n \in \mathbb{Z}$. Then $x = \frac{n+3}{2}$.

       The integration limits are $x = -\frac{3}{2}$ to $x = \frac{3}{2}$. So we need all integers $n$ such that $\frac{n+3}{2}$ lies in $[-\frac{3}{2}, \frac{3}{2}]$.

       Solve $-\frac{3}{2} \le \frac{n+3}{2} \le \frac{3}{2}$ → multiply by 2: $-3 \le n+3 \le 3$ → $-6 \le n \le 0$.

       So $n = -6, -5, -4, -3, -2, -1, 0$. That gives jump points at $x = -\frac{3}{2}, -1, -\frac12, 0, \frac12, 1, \frac32$.

       Notice the endpoints are included — the function is defined at them, but the integral over a point is zero, so we only care about open intervals between them.

    2. Determine the constant value on each subinterval.

       Between two consecutive jump points, $2x-3$ lies strictly between two consecutive integers, so its greatest integer is the lower integer.

       Let’s list the subintervals from left to right:

       • $x \in (-\frac{3}{2}, -1)$: $2x-3 \in (-6, -5)$ → $[2x-3] = -6$
       • $x \in (-1, -\frac12)$: $2x-3 \in (-5, -4)$ → $[2x-3] = -5$
       • $x \in (-\frac12, 0)$: $2x-3 \in (-4, -3)$ → $[2x-3] = -4$
       • $x \in (0, \frac12)$: $2x-3 \in (-3, -2)$ → $[2x-3] = -3$
       • $x \in (\frac12, 1)$: $2x-3 \in (-2, -1)$ → $[2x-3] = -2$
       • $x \in (1, \frac32)$: $2x-3 \in (-1, 0)$ → $[2x-3] = -1$

       Each subinterval has length $\frac12$. …

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