Q.Evaluate $\int_{-1}^2 \left| x^3 - x \right|\, dx$

Showing the 12 most recent of 15 on this concept.

1. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.The value of $\displaystyle\int_{-1}^{2} (x - 2|x|)\, dx$ is equal to (A) $\dfrac{-1}{2}$ (B) $\dfrac{-3}{2}$ (C) $\dfrac{-5}{2}$ (D) $\dfrac{-7}{2}$ (E) $\dfrac{-9}{2}$

   ›Reveal solutionSolution

   The integral equals $-\frac72$.

   Concept and Intuition

   The absolute value forces a split of the domain at $x=0$, where $|x|$ changes formula.

   Step-by-Step Solution

   1. For $x<0$: $|x|=-x$, so $x-2|x|=x+2x=3x$.
   2. For $x\geq0$: $|x|=x$, so $x-2|x|=x-2x=-x$.
   3. $\int_{-1}^{0}3x\,dx=3\left[\frac{x^2}{2}\right]_{-1}^0=3\left(0-\frac12\right)=-\frac32$.
   4. $\int_{0}^{2}(-x)\,dx=-\left[\frac{x^2}{2}\right]_0^2=-2$.
   5. Sum $=-\frac32-2=-\frac72$.

   Common Mistakes

   • Not splitting the integral at $x=0$.
   • Using $|x|=x$ on the negative part.

   ✓Final answer

   The correct option is (D) — $\dfrac{-7}{2}$.

   ANSWER: D

2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\displaystyle\int_0^3 \left|3x^2 - 3\right|\, dx$ is equal to (A) $16$ (B) $18$ (C) $20$ (D) $22$ (E) $24$

   ›Reveal solutionSolution

   Split at $x=1$ where $x^2-1$ changes sign; the two pieces give $3\cdot\dfrac{22}{3}=22$.

   Write $\int_0^3|3x^2-3|\,dx=3\int_0^3|x^2-1|\,dx$. Since $x^2-1<0$ on $[0,1]$ and $>0$ on $[1,3]$:

   $\int_0^1(1-x^2)dx=\left[x-\tfrac{x^3}{3}\right]_0^1=1-\tfrac13=\tfrac23,$

   $\int_1^3(x^2-1)dx=\left[\tfrac{x^3}{3}-x\right]_1^3=(9-3)-\left(\tfrac13-1\right)=6+\tfrac23=\tfrac{20}{3}.$

   Sum $=\tfrac23+\tfrac{20}{3}=\tfrac{22}{3}$; multiplying by $3$ gives $22$.

   ✓Final answer

   The correct option is (D).

3. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.The value of $\displaystyle\int_{0}^{3} |x-2|\,dx$ is equal to (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{2}$ (C) $\dfrac{5}{2}$ (D) $\dfrac{2}{5}$ (E) $\dfrac{9}{2}$

   ›Reveal solutionSolution

   Split the absolute value at $x=2$ and integrate each piece.

   For $0\le x\le2$, $|x-2|=2-x$; for $2\le x\le3$, $|x-2|=x-2$:

   $\int_0^2(2-x)dx=\left[2x-\frac{x^2}{2}\right]_0^2=4-2=2,$

   $\int_2^3(x-2)dx=\left[\frac{x^2}{2}-2x\right]_2^3=\left(\frac92-6\right)-(2-4)=-\frac32+2=\frac12.$

   Total $=2+\frac12=\frac52$.

   ✓Final answer

   The correct option is (C).

4. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The value of $\int_{-2}^{1} \frac{|x|}{x}\, dx$ is equal to (A) $-1$ (B) $-2$ (C) $1$ (D) $2$ (E) $3$

   ›Reveal solutionSolution

   Split at $0$ since $|x|/x=-1$ for $x<0$ and $+1$ for $x>0$.

   For $x<0$, $\frac{|x|}{x}=-1$; for $x>0$, $\frac{|x|}{x}=+1$.

   $\int_{-2}^{0}(-1)\,dx=-2$ and $\int_{0}^{1}(1)\,dx=1$.

   Sum $=-2+1=-1$.

   ✓Final answer

   The correct option is (A).

5. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.$\displaystyle\int_{-2}^{2} |x+3|\, dx =$ (A) $14$ (B) $16$ (C) $8$ (D) $10$ (E) $12$

   ›Reveal solutionSolution

   $x+3>0$ throughout $[-2,2]$, so drop the absolute value and integrate.

   For $x\in[-2,2]$, $x+3\ge 1>0$, hence $|x+3|=x+3$.

   $\int_{-2}^{2}(x+3)\,dx=\left[\frac{x^2}{2}+3x\right]_{-2}^{2}=(2+6)-(2-6)=8-(-4)=12.$

   ✓Final answer

   The correct option is (E).

6. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.$\displaystyle\int_{-2}^{4}x^2|x|\,dx$ is equal to (A) $72$ (B) $68$ (C) $64$ (D) $48$ (E) $37$

   ›Reveal solutionSolution

   $x^2|x|=-x^3$ for $x<0$ and $x^3$ for $x>0$; the two parts give $4+64=68$.

   Split the integral at $x=0$ where $|x|$ changes form:

   $\int_{-2}^{4}x^2|x|\,dx=\int_{-2}^{0}x^2(-x)\,dx+\int_{0}^{4}x^2(x)\,dx=\int_{-2}^{0}(-x^3)\,dx+\int_{0}^{4}x^3\,dx.$

   Evaluate each:

   $\int_{-2}^{0}(-x^3)\,dx=\left[-\frac{x^4}{4}\right]_{-2}^{0}=0-\left(-\frac{16}{4}\right)=4,$

   $\int_{0}^{4}x^3\,dx=\left[\frac{x^4}{4}\right]_{0}^{4}=\frac{256}{4}=64.$

   Total $=4+64=68$.

   ✓Final answer

   The correct option is (B).

7. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The value of $\int_{-1}^{1}|x-3|\,dx$ is equal to (A) $5$ (B) $6$ (C) $-5$ (D) $-6$ (E) $0$

   ›Reveal solutionSolution

   Since $x-3<0$ on $[-1,1]$, replace $|x-3|$ with $3-x$ and integrate.

   For $x\in[-1,1]$, $x-3<0$, so $|x-3|=3-x$. Then

   $\int_{-1}^{1}(3-x)\,dx=\left[3x-\dfrac{x^{2}}{2}\right]_{-1}^{1}=\left(3-\dfrac{1}{2}\right)-\left(-3-\dfrac{1}{2}\right)=\dfrac{5}{2}+\dfrac{7}{2}=6.$

   ✓Final answer

   The correct option is (B).

8. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \begin{cases} x & \text{if } x \le 1 \\ -x + 2 & \text{if } x > 1 \end{cases}$ Then $\int_{0}^{2} f(x)\, dx =$ (A) $\dfrac{\pi}{2}$ (B) $1$ (C) $2$ (D) $4$ (E) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   $\int_0^2 f(x)\,dx=1$.

   Concept and Intuition

   Split the integral at the junction $x=1$ where the definition changes.

   Step-by-Step Solution

   1. $\int_0^1 x\,dx=\left[\dfrac{x^2}{2}\right]_0^1=\dfrac12$.
   2. $\int_1^2(-x+2)\,dx=\left[-\dfrac{x^2}{2}+2x\right]_1^2=(2)-(1.5)=\dfrac12$.
   3. Total $=\dfrac12+\dfrac12=1$.

   Common Mistakes

   • Using one formula over the whole interval.
   • Arithmetic slip in the second integral's evaluation.

   ✓Final answer

   The correct option is (B) — $1$.

   ANSWER: B

9. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.The value of $\int_1^3 [x - 1]\,dx$, where $[x]$ denotes the greatest integer function in $x$, is equal to (A) $-2$ (B) $-1$ (C) $0$ (D) $1$ (E) $2$

   ›Reveal solutionSolution

   Split at $x=2$: $[x-1]=0$ on $[1,2)$ and $=1$ on $[2,3)$, giving $0+1=1$.

   For $x\in[1,2)$, $x-1\in[0,1)$ so $[x-1]=0$. For $x\in[2,3)$, $x-1\in[1,2)$ so $[x-1]=1$. Thus

   $\int_1^3[x-1]\,dx=\int_1^2 0\,dx+\int_2^3 1\,dx=0+1=1.$

   ✓Final answer

   The correct option is (D).

10. KEAM 2025Set eng-2025-04274 marksMCQ
    Q.If $[x]$ is the greatest integer less than or equal to $x$, then $\int_{-3}^{3} [x]\,dx =$ (A) $-3$ (B) $-6$ (C) $-4$ (D) $-2$ (E) $0$

    ›Reveal solutionSolution

    Integrating the step function $[x]$ over $[-3,3]$ gives $-3-2-1+0+1+2=-3$.

    The greatest-integer function $[x]$ is constant on each unit interval:

    $[x]=-3\ \text{on}\ [-3,-2),\ -2\ \text{on}\ [-2,-1),\ -1\ \text{on}\ [-1,0),$

    $[x]=0\ \text{on}\ [0,1),\ 1\ \text{on}\ [1,2),\ 2\ \text{on}\ [2,3).$

    Each interval has width $1$, so the integral is the sum of these constant values:

    $\int_{-3}^{3}[x]\,dx=(-3)+(-2)+(-1)+0+1+2=-3.$

    ✓Final answer

    The correct option is (A).

11. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.If $f(x) = \begin{cases} \cos x & \text{for } x \geq 0 \\ 2x & \text{for } x < 0 \end{cases}$, then the value of $\displaystyle\int_{-2}^{\pi/2} f(x)\, dx$ is equal to (A) $2$ (B) $-2$ (C) $-3$ (D) $3$ (E) $0$

    ›Reveal solutionSolution

    The integral equals $-3$.

    Concept and Intuition

    The piecewise function splits the integral at $x=0$, using $2x$ below and $\cos x$ above.

    Step-by-Step Solution

    1. Split at $0$: $\int_{-2}^{\pi/2}f=\int_{-2}^{0}2x\,dx+\int_{0}^{\pi/2}\cos x\,dx$.
    2. $\int_{-2}^{0}2x\,dx=[x^2]_{-2}^0=0-4=-4$.
    3. $\int_{0}^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=1-0=1$.
    4. Total $=-4+1=-3$.

    Common Mistakes

    • Applying $\cos x$ on the negative part where $2x$ is required.
    • Arithmetic slip on $[x^2]_{-2}^0=-4$.

    ✓Final answer

    The correct option is (C) — $-3$.

    ANSWER: C

12. KEAM 2024Set eng-2024-06054 marksMCQ
    Q.If $\left[x^2\right]$ is the greatest integer less than or equal to $x^2$, then $\int_{0}^{\sqrt{2}} \left[x^2\right]\,dx =$ (A) $\sqrt{2}$ (B) $2$ (C) $\sqrt{2} - 1$ (D) $\sqrt{2} + 1$ (E) $2\sqrt{2} + 1$

    ›Reveal solutionSolution

    Split at $x=1$: the integrand is 0 up to 1 and 1 from 1 to $\sqrt2$, giving area $\sqrt2-1$.

    For $0\le x<1$, $x^2\in[0,1)$ so $[x^2]=0$.

    For $1\le x<\sqrt2$, $x^2\in[1,2)$ so $[x^2]=1$.

    $\displaystyle\int_0^{\sqrt2}[x^2]\,dx=\int_0^1 0\,dx+\int_1^{\sqrt2}1\,dx=0+(\sqrt2-1)=\sqrt2-1$.

    ✓Final answer

    The correct option is (C).