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

Showing the 12 most recent of 30 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.$\displaystyle\int_{-1}^{1} \dfrac{|x|}{x}\, dx =$ (A) $1$ (B) $-1$ (C) $0$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   $|x|/x$ is simply $\pm1$ depending on the sign of $x$, an odd function, so its integral over a symmetric interval is zero.

   Concept and Intuition

   The function $f(x)=|x|/x = \operatorname{sgn}(x)$ takes the constant value $-1$ on $(-1,0)$ and $+1$ on $(0,1)$. This makes $f$ an odd function about $x=0$ (aside from the removable point at $x=0$ itself, which doesn't affect the integral), so integrating over the symmetric interval $[-1,1]$ gives exactly zero — the negative area on the left cancels the positive area on the right.

   Step-by-Step Solution

   1. For $x\in(-1,0)$: $|x|=-x$, so $|x|/x = -x/x = -1$.
   2. For $x\in(0,1)$: $|x|=x$, so $|x|/x = x/x = 1$.
   3. $\displaystyle\int_{-1}^1\dfrac{|x|}{x}dx = \int_{-1}^0(-1)\,dx + \int_0^1(1)\,dx$.
   4. $\displaystyle\int_{-1}^0(-1)\,dx = -1\cdot[0-(-1)] = -1$.
   5. $\displaystyle\int_0^1(1)\,dx = 1\cdot[1-0] = 1$.
   6. Sum: $-1+1 = 0$.

   Common Mistakes

   • Treating $|x|/x$ as $|x|$ (forgetting the sign flip for negative $x$) and getting a nonzero answer.
   • Ignoring that the function is undefined exactly at $x=0$, which is a single point and doesn't affect the definite integral's value.

   ✓Final answer

   The correct option is (C) — $0$.

   ANSWER: C

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 $=2\times(3-1)=4$.
   4. On $[3,5]$: integrand $=(x-3)+(x-1)=2x-4$. Integral $=\left[x^2-4x\right]_3^5=(25-20)-(9-12)=5+3=8$.
   5. Total $=4+8=12$.

   Common Mistakes

   • Forgetting to split at $x=3$ and integrating $|x-3|$ with a single fixed sign over the whole interval.
   • Arithmetic slip evaluating $[x^2-4x]$ at the endpoints.

   ✓Final answer

   The correct option is (C) — $12$.

   ANSWER: C

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: $\int_2^3(x-2)dx=\left[\dfrac{x^2}{2}-2x\right]_2^3=(4.5-6)-(2-4)=(-1.5)-(-2)=0.5$.
   4. Total $=12.5+0.5=13$.

   Common Mistakes

   • Integrating $|2-x|$ as if it were $2-x$ throughout the whole interval, ignoring the sign change at $x=2$.
   • Arithmetic slip evaluating the quadratic antiderivative at the negative limit $x=-3$.

   ✓Final answer

   The correct option is (C) — $13$.

   ANSWER: C

4. 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 $\displaystyle\int_2^3 1\,dx = [x]_2^3 = 1$.
   5. Compute $\displaystyle\int_3^4(2x-5)\,dx = [x^2-5x]_3^4 = (16-20)-(9-15) = -4-(-6)=2$.
   6. Add: $1+2=3$.

   Common Mistakes

   • Forgetting to split the interval at $x=3$ and integrating $|x-3|$ as if it had one sign throughout.
   • Arithmetic slip evaluating $x^2-5x$ at the limits.

   ✓Final answer

   The correct option is (C) — 3.

   ANSWER: C

5. 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: $x=1/e\Rightarrow u=-1$; $x=1\Rightarrow u=0$; $x=e^2\Rightarrow u=2$.
   4. First piece: $\displaystyle\int_{-1}^{0}(-u)\,du = \left[-\frac{u^2}{2}\right]_{-1}^{0} = 0-\left(-\frac12\right)=\frac12$.
   5. Second piece: $\displaystyle\int_{0}^{2}u\,du = \left[\frac{u^2}{2}\right]_0^2 = 2$.
   6. Total $=\dfrac12+2=\dfrac52$.

   Common Mistakes

   • Integrating $\log x/x$ directly over the whole interval without splitting at $x=1$, which gives the wrong (too small, possibly even sign-cancelled) value.
   • Sign error in the first piece — forgetting the extra minus sign needed since $\log x$ is negative there.

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{2}$.

   ANSWER: B

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 $=[F(\pi/4)-F(0)]-[F(3\pi/4)-F(\pi/4)]+[F(\pi)-F(3\pi/4)]$ $=\left(\frac\pi8-\frac14\right)-\left(-\frac{3\pi}{8}-\frac\pi8\right)+\left(\frac14+\frac{3\pi}{8}\right)$ $=\left(\frac\pi8-\frac14\right)+\frac\pi2+\left(\frac14+\frac{3\pi}8\right)$ $=\frac\pi8+\frac\pi2+\frac{3\pi}8=\pi$ (the $-\frac14,+\frac14$ terms cancel).

   Common Mistakes

   • Forgetting to flip the sign of the middle integral (over $(\pi/4,3\pi/4)$) where $\cos2x<0$.
   • Sign slips in evaluating $F$ at $3\pi/4$ where $\sin(3\pi/2)=-1$.

   ✓Final answer

   The correct option is (A) — $\pi$.

   ANSWER: A

7. 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. $\int_{-1}^0(x^4-4)\,dx=\big[\tfrac{x^5}{5}-4x\big]_{-1}^0=0-(-\tfrac15+4)=\tfrac15-4=-\tfrac{19}{5}$.
   5. $\int_0^1(x^3-4)\,dx=\big[\tfrac{x^4}{4}-4x\big]_0^1=\tfrac14-4=-\tfrac{15}{4}$.
   6. $\int_{-1}^1(f-g)\,dx=\int f\,dx-0=-\tfrac{19}{5}-\tfrac{15}{4}=-\tfrac{76}{20}-\tfrac{75}{20}=-\tfrac{151}{20}$.

   Common Mistakes

   • Assuming $\max(x^3,x^4)$ switches at $x=0$ and $x=1$ symmetric to $\min(x^2,x^3)$ without actually checking sign of the difference.
   • Forgetting the $-4$ constant carries through the max/min unchanged (since it's added to both branches equally).

   ✓Final answer

   The correct option is (A) — $-\dfrac{151}{20}$.

   ANSWER: A

8. 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 $=-\cos t\sin t-\sin t\cos t=-\sin2t$. $\displaystyle\int_{\pi}^{5\pi/4}(-\sin2t)\,dt = \left[\tfrac12\cos2t\right]_{\pi}^{5\pi/4} = \tfrac12\cos\tfrac{5\pi}2 - \tfrac12\cos2\pi = 0-\tfrac12=-\tfrac12$.
   4. Total: $\tfrac12+0-\tfrac12=0$.

   Common Mistakes

   • Not splitting the interval at $\pi/2$ and $\pi$ where the signs of $\sin t,\cos t$ change, leading to a wrong single antiderivative over the whole range.
   • Sign slips when writing $|\cos t|=-\cos t$ vs $\cos t$ on each piece.

   ✓Final answer

   The correct option is (A) — $0$.

   ANSWER: A

9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.$f(x)=\begin{cases}x^2, & 0\le x<1\\ \sqrt{x}, & 1\le x\le2\end{cases} \Rightarrow \int_0^2 f(x)\,dx=$ (A) $\dfrac{4\sqrt{2}-1}{3}$ (B) $\dfrac{4\sqrt{2}+1}{3}$ (C) $\dfrac{4\sqrt{2}-1}{6}$ (D) $\dfrac{4\sqrt{2}+1}{6}$

   ›Reveal solutionSolution

   Split the integral at the point where the piecewise definition changes ($x=1$) and integrate each piece separately.

   Concept and Intuition

   For a piecewise-defined function, $\int_a^b f = \int_a^c f + \int_c^b f$ whenever $c$ is the breakpoint — additivity of the definite integral over adjacent intervals.

   Step-by-Step Solution

   1. $\displaystyle\int_0^1 x^2\,dx = \left[\dfrac{x^3}{3}\right]_0^1=\dfrac13$.
   2. $\displaystyle\int_1^2 \sqrt{x}\,dx = \left[\dfrac{2}{3}x^{3/2}\right]_1^2=\dfrac23\left(2^{3/2}-1\right)=\dfrac23\left(2\sqrt2-1\right)=\dfrac{4\sqrt2-2}{3}$.
   3. Sum: $\dfrac13+\dfrac{4\sqrt2-2}{3}=\dfrac{1+4\sqrt2-2}{3}=\dfrac{4\sqrt2-1}{3}$.

   Common Mistakes

   • Forgetting that $2^{3/2}=2\sqrt2$, not $\sqrt2$ or $4\sqrt2$.
   • Using the wrong piece of the definition on the wrong sub-interval.

   ✓Final answer

   The correct option is (A) — $\dfrac{4\sqrt2-1}{3}$.

   ANSWER: A

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) $=\int_{-1}^0 x^2\,dx = \dfrac13$.
    3. Total enclosed area $= \dfrac13+\dfrac13 = \dfrac23$.

    Common Mistakes

    • Directly integrating $\int_{-1}^1 x|x|\,dx$ without splitting — since the function is odd, this signed integral is $0$, which is NOT the enclosed area.

    ✓Final answer

    The correct option is (B) — $\dfrac{2}{3}$.

    ANSWER: B

11. 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 $(\pi,2\pi)$: $[-x\cos x+\sin x]_\pi^{2\pi}=(-2\pi\cos2\pi+\sin2\pi)-(-\pi\cos\pi+\sin\pi)=(-2\pi(1)+0)-(-\pi(-1)+0)=-2\pi-\pi=-3\pi$.
    5. So $\displaystyle\int_\pi^{2\pi} x\sin x\,dx=-3\pi$, which means $\displaystyle\int_\pi^{2\pi}(-x\sin x)\,dx=3\pi$.
    6. Total: $\displaystyle\int_0^{2\pi}|x\sin x|\,dx=\pi+3\pi=4\pi$. So $k=4$.

    Common Mistakes

    • Not splitting the domain at the sign-change point of $\sin x$ and integrating $x\sin x$ over the whole range directly (which would give a much smaller, wrong value due to cancellation).
    • Arithmetic slip evaluating $\cos\pi=-1$, $\cos2\pi=1$.

    ✓Final answer

    The correct option is (D) — $4$.

    ANSWER: D

12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.The area of the region (in sq. units) enclosed between the curves $y = |x|$, $y = [x]$ and the ordinates $x = -1, x = 0, x = 1$ is (A) $2$ (B) $\dfrac{3}{2}$ (C) $3$ (D) $\dfrac{5}{2}$

    ›Reveal solutionSolution

    Split the region into $[-1,0]$ and $[0,1]$, using $[x]=-1$ on the first (non-integer negative $x$) and $[x]=0$ on the second, then integrate $|y_{\text{top}}-y_{\text{bottom}}|$ on each piece. Answer: $2$.

    Concept and Intuition

    The floor function $[x]$ is piecewise constant, jumping at each integer. Between $x=-1$ and $x=0$ (excluding the endpoint $0$), every value of $x$ is a non-integer negative number, so $[x]=-1$ throughout. Between $x=0$ and $x=1$ (excluding $1$), $[x]=0$ throughout. This lets us treat $[x]$ as two separate constants on the two sub-intervals and just integrate the vertical gap between $|x|$ and that constant.

    Step-by-Step Solution

    1. On $[-1,0]$: $|x|=-x$, ranging from $1$ (at $x=-1$) down to $0$ (at $x=0$). Also $[x]=-1$ for all $x\in[-1,0)$ (and at $x=-1$ itself, $[x]=-1$). Since $-x\ge -1$ throughout this interval, the enclosed strip has height $(-x)-(-1) = 1-x$.

    $\int_{-1}^0(1-x)\,dx = \left[x-\frac{x^2}{2}\right]_{-1}^0 = 0 - \left(-1-\frac12\right) = \frac32.$

    2. On $[0,1]$: $|x|=x$, ranging from $0$ to $1$. Also $[x]=0$ for all $x\in[0,1)$. Height $= x - 0 = x$.

    $\int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac12.$

    3. Total enclosed area $= \dfrac32 + \dfrac12 = 2$.

    Common Mistakes

    • Using $[x]=0$ on the whole interval $[-1,1]$ instead of correctly switching to $[x]=-1$ for negative non-integer $x$.
    • Sign errors when computing $(-x)-(-1)=1-x$ on the negative side.

    ✓Final answer

    The correct option is (A) — $2$.

    ANSWER: A