Q.Evaluate the definite integral $\int_{1}^{4}\left[|x-1|+|x-2|+|x-3|\right]dx$

Showing the 12 most recent of 30 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.$\displaystyle\int_{2}^{4} \{|x-2| + |x-3|\}\,dx =$ (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   Break the integral at $x=3$, the point where the second absolute value flips sign; the two pieces integrate to $1$ and $2$, totalling $3$.

   Concept and Intuition

   For definite integrals with absolute values, identify all the points inside the range where the expressions inside the absolute values change sign, split the interval there, and remove the absolute values piecewise.

   Step-by-Step Solution

   1. On $[2,4]$, $|x-2|=x-2$ throughout since $x\ge2$. But $|x-3|$ changes sign at $x=3$.
   2. For $2\le x\le3$: $|x-3|=3-x$. Sum: $(x-2)+(3-x)=1$.
   3. For $3\le x\le4$: $|x-3|=x-3$. Sum: $(x-2)+(x-3)=2x-5$.
   4. Compute $\displaystyle\int_2^3 1\,dx = [x]_2^3 = 1$. …

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2. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_1^5 (|x-3| + |1-x|)\,dx =$ (A) $4$ (B) $8$ (C) $12$ (D) $24$

   ›Reveal solutionSolution

   Splitting $[1,5]$ at $x=3$ and removing the absolute values gives a total of $12$.

   Concept and Intuition

   To integrate a sum of absolute values, split the domain at each point where an inner expression changes sign, then integrate the resulting piecewise-linear (constant-slope) function directly.

   Step-by-Step Solution

   1. On $[1,5]$: $|1-x| = x-1$ throughout (since $x\ge1$).
   2. $|x-3| = 3-x$ for $x\in[1,3]$, and $=x-3$ for $x\in[3,5]$.
   3. On $[1,3]$: integrand $= (3-x)+(x-1) = 2$. Integral $=2\times(3-1)=4$. …

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3. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int_{-3}^{3}|2-x|\,dx=$ (A) $12$ (B) $16$ (C) $13$ (D) $25$

   ›Reveal solutionSolution

   Split the absolute-value integral at the point where the expression inside changes sign, then integrate each piece.

   Concept and Intuition

   $|2-x|$ equals $2-x$ when $x<2$ and $x-2$ when $x>2$. Since the sign-change point $x=2$ lies inside $[-3,3]$, the integral must be split there before integrating.

   Step-by-Step Solution

   1. $\displaystyle\int_{-3}^{3}|2-x|\,dx=\int_{-3}^{2}(2-x)\,dx+\int_{2}^{3}(x-2)\,dx$.
   2. First piece: $\int_{-3}^2(2-x)dx=\left[2x-\dfrac{x^2}{2}\right]_{-3}^{2}=(4-2)-(-6-4.5)=2-(-10.5)=12.5$.
   3. Second piece: $\int_2^3(x-2)dx=\left[\dfrac{x^2}{2}-2x\right]_2^3=(4.5-6)-(2-4)=(-1.5)-(-2)=0.5$. …

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4. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\displaystyle\int_0^{2\pi} |x\sin x|\,dx = k\pi$, then $k=$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Splitting the interval where $\sin x$ changes sign and integrating $x\sin x$ by parts on each piece gives total $4\pi$, so $k=4$.

   Concept and Intuition

   The factor $x>0$ throughout $(0,2\pi)$ doesn't change sign, so $x\sin x$ has the same sign as $\sin x$: positive on $(0,\pi)$, negative on $(\pi,2\pi)$. To handle the absolute value, split the integral at $x=\pi$ and flip the sign of the integrand on the second piece.

   Step-by-Step Solution

   1. $|x\sin x|=x\sin x$ on $(0,\pi)$ (both factors non-negative there) and $|x\sin x|=-x\sin x$ on $(\pi,2\pi)$ (since $\sin x<0$ there but $x>0$).
   2. Antiderivative (integration by parts): $\displaystyle\int x\sin x\,dx=-x\cos x+\sin x+C$.
   3. Evaluate on $(0,\pi)$: $[-x\cos x+\sin x]_0^\pi=(-\pi\cos\pi+\sin\pi)-(0+0)=(-\pi(-1)+0)-0=\pi$.
   4. Evaluate on $(\pi,2\pi)$: $[-x\cos x+\sin x]_\pi^{2\pi}=(-2\pi\cos2\pi+\sin2\pi)-(-\pi\cos\pi+\sin\pi)=(-2\pi(1)+0)-(-\pi(-1)+0)=-2\pi-\pi=-3\pi$. …

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5. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.$\displaystyle\int_{\pi/4}^{5\pi/4} \left(|\cos t|\sin t + |\sin t|\cos t\right)dt =$ (A) 0 (B) 1 (C) $1/2$ (D) $\sqrt{3}/2$

   ›Reveal solutionSolution

   Breaking the integral at the sign-change points of $\sin t$ and $\cos t$ shows the middle piece vanishes and the two outer pieces exactly cancel, giving $0$.

   Concept and Intuition

   Absolute values force us to split the integration range wherever $\sin t$ or $\cos t$ changes sign, since $|\cos t|$ and $|\sin t|$ are piecewise expressions of $\cos t$ and $\sin t$ (with a sign flip) on each sub-interval.

   Step-by-Step Solution

   1. On $[\pi/4,\pi/2]$: $\cos t\ge0,\ \sin t\ge0$, so $|\cos t|\sin t+|\sin t|\cos t = 2\sin t\cos t=\sin2t$. $\displaystyle\int_{\pi/4}^{\pi/2}\sin2t\,dt = \left[-\tfrac12\cos2t\right]_{\pi/4}^{\pi/2} = \left(-\tfrac12\cos\pi\right)-\left(-\tfrac12\cos\tfrac\pi2\right) = \tfrac12-0=\tfrac12$.
   2. On $[\pi/2,\pi]$: $\cos t\le0,\ \sin t\ge0$, so $|\cos t|=-\cos t,\ |\sin t|=\sin t$; integrand $=-\cos t\sin t+\sin t\cos t=0$. Contribution: $0$.
   3. On $[\pi,5\pi/4]$: $\cos t\le0,\ \sin t\le0$, so $|\cos t|=-\cos t,\ |\sin t|=-\sin t$; integrand $=-\cos t\sin t-\sin t\cos t=-\sin2t$. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} |x\cos 2x|\, dx =$ (A) $\pi$ (B) $\pi - 2$ (C) $\pi + \dfrac{1}{4}$ (D) $\pi - \dfrac{1}{4}$

   ›Reveal solutionSolution

   Split the interval at the zeros of $\cos2x$ (at $x=\pi/4,3\pi/4$) and flip the sign of the middle piece to handle the absolute value, then use integration by parts on $x\cos2x$. Answer: $\pi$.

   Concept and Intuition

   To integrate $|f(x)|$, first find where $f$ changes sign inside the interval, then integrate $f$ (or $-f$) piecewise so the result is always non-negative on each piece. Here $x\ge0$ throughout, so the sign of $x\cos2x$ tracks the sign of $\cos2x$ alone.

   Step-by-Step Solution

   1. $\cos2x=0$ at $2x=\pi/2,3\pi/2\Rightarrow x=\pi/4,3\pi/4$ inside $[0,\pi]$. $\cos2x>0$ on $[0,\pi/4)$ and $(3\pi/4,\pi]$; $\cos2x<0$ on $(\pi/4,3\pi/4)$.
   2. So $\displaystyle\int_0^\pi|x\cos2x|dx=\int_0^{\pi/4}x\cos2x\,dx-\int_{\pi/4}^{3\pi/4}x\cos2x\,dx+\int_{3\pi/4}^{\pi}x\cos2x\,dx$.
   3. By parts: $\displaystyle\int x\cos2x\,dx=\frac{x\sin2x}{2}+\frac{\cos2x}{4}+C=F(x)$.
   4. Evaluate: $F(0)=0+\frac14=\frac14$; $F(\pi/4)=\frac{(\pi/4)(1)}{2}+0=\frac\pi8$; $F(3\pi/4)=\frac{(3\pi/4)(-1)}{2}+0=-\frac{3\pi}{8}$; $F(\pi)=0+\frac14=\frac14$.
   5. Total $=[F(\pi/4)-F(0)]-[F(3\pi/4)-F(\pi/4)]+[F(\pi)-F(3\pi/4)]$ …

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7. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.$\int_2^5 \sqrt{x + 2\sqrt{x-1}} + \sqrt{x - 2\sqrt{x-1}}\,dx =$ (A) $16/3$ (B) $32/3$ (C) $28/3$ (D) $4/3$

   ›Reveal solutionSolution

   This tests simplifying nested square roots via a substitution that reveals them as perfect squares, before integrating.

   Concept and Intuition

   Expressions of the form $\sqrt{x\pm2\sqrt{x-1}}$ simplify beautifully once you set $t=\sqrt{x-1}$, since $x = t^2+1$ turns $x\pm2\sqrt{x-1}$ into $(t\pm1)^2$ — a perfect square whose root is just $|t\pm1|$.

   Step-by-Step Solution

   1. Let $t=\sqrt{x-1}$, so $x=t^2+1$. Then $x+2\sqrt{x-1} = t^2+1+2t=(t+1)^2$ and $x-2\sqrt{x-1}=t^2+1-2t=(t-1)^2$.
   2. So the integrand becomes $|t+1|+|t-1|$.
   3. As $x$ ranges over $[2,5]$, $t=\sqrt{x-1}$ ranges over $[1,2]$. On this range, $t+1>0$ and $t-1\ge0$, so $|t+1|+|t-1| = (t+1)+(t-1)=2t = 2\sqrt{x-1}$.
   4. So the original integral reduces to $\displaystyle\int_2^5 2\sqrt{x-1}\,dx$. …

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8. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.$\displaystyle\int_{1/e}^{e^2} \left|\dfrac{\log_e x}{x}\right|\,dx =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{5}{2}$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   Since $\log x$ changes sign at $x=1$ within the interval $[1/e,e^2]$, the absolute value forces splitting the integral there; substituting $u=\log x$ turns each piece into a trivial polynomial integral, totaling $5/2$.

   Concept and Intuition

   $\left|\dfrac{\log x}{x}\right|$ is not simply $\dfrac{\log x}{x}$ throughout $[1/e,e^2]$ because $\log x$ is negative on $(1/e,1)$ and positive on $(1,e^2)$. So the absolute value must be resolved by splitting the integral at the sign-change point $x=1$, and on each sub-interval the substitution $u=\log x$ (with $du=dx/x$) makes the integral immediate.

   Step-by-Step Solution

   1. Note $\log x<0$ for $x\in(1/e,1)$ and $\log x>0$ for $x\in(1,e^2)$, with $\log x=0$ at $x=1$.
   2. Split: $\displaystyle\int_{1/e}^{e^2}\left|\frac{\log x}{x}\right|dx = \int_{1/e}^{1}\left(-\frac{\log x}{x}\right)dx + \int_{1}^{e^2}\frac{\log x}{x}dx$.
   3. Substitute $u=\log x$, $du=dx/x$. Limits: $x=1/e\Rightarrow u=-1$; $x=1\Rightarrow u=0$; $x=e^2\Rightarrow u=2$. …

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9. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $[.]$ represents greatest integer function, then $\displaystyle\int_{3\pi/4}^{\pi} \left[\sin x + \left[\dfrac{4x}{\pi}\right]\right] dx =$ (A) $\pi/4$ (B) $\pi/2$ (C) $3\pi/4$ (D) $\pi$

   ›Reveal solutionSolution

   Over $[3\pi/4,\pi]$ the greatest-integer term $[4x/\pi]$ is constantly $3$ and $\sin x$ stays in $[0,1)$, so the whole integrand is the constant $3$, giving $3\pi/4$.

   Concept and Intuition

   Greatest-integer-function integrals are solved by identifying the sub-intervals on which the bracketed quantity is constant, then integrating that constant over each piece. Here two floor-type effects combine, but they both turn out to be constant on the whole interval, dramatically simplifying the problem.

   Step-by-Step Solution

   1. For $x\in[3\pi/4,\pi]$: at $x=3\pi/4$, $4x/\pi=3$; at $x=\pi$, $4x/\pi=4$. So $4x/\pi$ ranges over $[3,4]$, and $\left[\dfrac{4x}{\pi}\right]=3$ for all $x\in[3\pi/4,\pi)$ (equals $4$ only at the single point $x=\pi$, which doesn't affect the integral).
   2. For $x\in[3\pi/4,\pi]$, $\sin x$ decreases from $\sin(3\pi/4)=\tfrac{\sqrt2}{2}\approx0.707$ down to $\sin\pi=0$, so $\sin x\in[0,\tfrac{\sqrt2}{2}]\subset[0,1)$.
   3. Since $[4x/\pi]=3$ is an integer, $[\sin x + 3] = 3+[\sin x]$. Because $0\le\sin x<1$ throughout, $[\sin x]=0$. …

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10. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.The area enclosed by the curves $y = x|x|$, $x = -1$ and $x = 1$ is ______ sq. units. (A) $\dfrac{3}{2}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{5}{3}$ (D) $\dfrac{7}{3}$

    ›Reveal solutionSolution

    Split the piecewise curve $y=x|x|$ at $x=0$; each half contributes an area of $1/3$, giving a total of $2/3$.

    Concept and Intuition

    $y=x|x|$ is an odd, S-shaped curve: it behaves like $y=x^2$ (above the axis) for $x>0$ and like $y=-x^2$ (below the axis) for $x<0$. The AREA enclosed (as opposed to the signed integral, which would cancel to zero) must be computed by taking the magnitude on each piece.

    Step-by-Step Solution

    1. For $x\in[0,1]$: $y=x\cdot x = x^2 \ge 0$, so the region lies above the axis; area $=\int_0^1 x^2\,dx = \dfrac13$.
    2. For $x\in[-1,0]$: $y = x\cdot(-x) = -x^2 \le 0$, so the region lies below the axis; area (magnitude) $=\int_{-1}^0 x^2\,dx = \dfrac13$. …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $f(x) = \text{Max}\{x^3 - 4, x^4 - 4\}$, and $g(x) = \text{Min}\{x^2, x^3\}$, then $\displaystyle\int_{-1}^1 (f(x) - g(x))\,dx =$ (A) $-\dfrac{151}{20}$ (B) $\dfrac{9}{20}$ (C) $\dfrac{131}{22}$ (D) $-\dfrac{67}{9}$

    ›Reveal solutionSolution

    This tests locating where two power functions cross on $[-1,1]$ to convert a max/min piecewise definition into an ordinary integral, then a clean simplification kills $g$'s contribution entirely.

    Concept and Intuition

    For $\max$/$\min$ of two functions, find where their difference changes sign — that tells you which one is larger on each sub-interval. Here $x^4-x^3=x^3(x-1)$ and $x^2-x^3=x^2(1-x)$ each factor cleanly, revealing that $g(x)=\min(x^2,x^3)=x^3$ on the entire interval $[-1,1]$ — a nice simplification that removes the need to split $g$ at all.

    Step-by-Step Solution

    1. Simplify $g$: $x^2-x^3=x^2(1-x)\ge0$ for all $x\in[-1,1]$ (since $x^2\ge0$ and $1-x\ge0$ there), so $x^3\le x^2$ throughout, meaning $g(x)=x^3$ on all of $[-1,1]$.
    2. So $\int_{-1}^1 g(x)\,dx=\int_{-1}^1 x^3\,dx=0$ (odd function over a symmetric interval).
    3. Simplify $f$: $x^4-x^3=x^3(x-1)$. For $x\in[-1,0)$: $x^3<0,\ x-1<0\Rightarrow$ product $>0\Rightarrow x^4>x^3$, so $\max=x^4-4$. For $x\in(0,1)$: $x^3>0,\ x-1<0\Rightarrow$ product $<0\Rightarrow x^4<x^3$, so $\max=x^3-4$.
    4. $\int_{-1}^0(x^4-4)\,dx=\big[\tfrac{x^5}{5}-4x\big]_{-1}^0=0-(-\tfrac15+4)=\tfrac15-4=-\tfrac{19}{5}$. …

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12. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If [x] is the greatest integer not exceeding x, then $\displaystyle\int_{-0.5}^{1.5} x^2[x]\,dx=$ (A) $\dfrac{4.5}{4}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{3.5}{4}$ (D) $\dfrac{2.375}{2}$

    ›Reveal solutionSolution

    Splitting the interval according to where the floor function $[x]$ is constant ($-1$, then $0$, then $1$), the integral evaluates to $3/4$.

    Concept and Intuition

    The greatest integer function $[x]$ is piecewise constant, jumping at every integer. To integrate any expression containing $[x]$, we must break the interval of integration at each integer point inside it and replace $[x]$ by its constant value on each piece.

    Step-by-Step Solution

    1. The interval is $[-0.5, 1.5]$. The integers inside/bounding it are $-1, 0, 1$ (relevant jump points at $x=0$ and $x=1$). So split as:

    $[-0.5,0)\cup[0,1)\cup[1,1.5]$

    with $[x]=-1,\ 0,\ 1$ respectively on each piece.

    2. Compute each piece:

    Piece 1 ($x\in[-0.5,0)$, $[x]=-1$):

    $\int_{-0.5}^0 x^2\cdot(-1)\,dx = -\left[\frac{x^3}{3}\right]_{-0.5}^{0} = -\left(0 - \frac{(-0.5)^3}{3}\right) = -\left(0+\frac{0.125}{3}\right) = -\frac{1}{24}$

    Piece 2 ($x\in[0,1)$, $[x]=0$):

    $\int_0^1 x^2\cdot 0\,dx = 0$

    Piece 3 ($x\in[1,1.5]$, $[x]=1$): …

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