Q.Evaluate $\int_{-1}^{3/2} \left| x \sin(\pi x) \right|\, dx$

Showing the 12 most recent of 30 on this concept.

1. 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 $=[F(\pi/4)-F(0)]-[F(3\pi/4)-F(\pi/4)]+[F(\pi)-F(3\pi/4)]$ …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$[\ .\ ]$ represents greatest integer function, then $\displaystyle\int_{-1}^{1}(x[1+\sin\pi x]+1)\,dx =$ (A) $1$ (B) $2$ (C) $5/2$ (D) $3/2$

   ›Reveal solutionSolution

   Splitting the interval where $[1+\sin\pi x]$ equals $0$ versus $1$ reduces the integral to two easy pieces, totaling $\frac52$.

   Concept and Intuition

   The greatest integer function $[\,\cdot\,]$ is piecewise constant, so the key step is figuring out exactly which sub-intervals give which integer value of $1+\sin\pi x$, then integrating each piece separately as an ordinary polynomial.

   Step-by-Step Solution

   1. For $x\in(-1,0)$: $\sin\pi x\in(-1,0)$, so $1+\sin\pi x\in(0,1)$, giving $[1+\sin\pi x]=0$.
   2. For $x\in(0,1)$: $\sin\pi x\in(0,1]$, so $1+\sin\pi x\in(1,2]$; except exactly at $x=\frac12$ (where it equals $2$), $[1+\sin\pi x]=1$ throughout — and a single point doesn't affect the integral's value.
   3. Split the integral: $\displaystyle\int_{-1}^{1}(x[1+\sin\pi x]+1)\,dx = \int_{-1}^0(0\cdot x+1)\,dx + \int_0^1(1\cdot x+1)\,dx$.
   4. First piece: $\displaystyle\int_{-1}^0 1\,dx = 1$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $=-\cos t\sin t-\sin t\cos t=-\sin2t$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.$\displaystyle\int_{-\pi/2}^{\pi/2} \sin(x - [x])\,dx =$ (Here $[x]$ is the greatest integer function) (A) $0$ (B) $3(1 - \cos 1) + \sin 2 - \sin 1$ (C) $3(1 - \cos 1) + \cos 2 - \sin 1$ (D) $\cos 2 - \sin 2$

   ›Reveal solutionSolution

   This tests handling the greatest-integer function inside an integral by splitting the domain at every integer inside it, then integrating a plain shifted sine on each piece.

   Concept and Intuition

   $[x]$ (floor of $x$) is constant on each interval between consecutive integers, and jumps by $1$ at every integer. So over $[-\pi/2,\pi/2]\approx[-1.57,1.57]$, the integers $-1,0,1$ split the domain into four sub-intervals, and on each one $x-[x]$ is just $x$ shifted by a constant, so $\sin(x-[x])$ becomes an ordinary sine we can integrate directly.

   Step-by-Step Solution

   1. Break the domain: $[-\pi/2,-1],[-1,0],[0,1],[1,\pi/2]$, with $[x]=-2,-1,0,1$ respectively, so $x-[x]=x+2,\,x+1,\,x,\,x-1$.
   2. $I_1=\int_{-\pi/2}^{-1}\sin(x+2)\,dx=\big[-\cos(x+2)\big]_{-\pi/2}^{-1}=-\cos1+\cos(2-\tfrac{\pi}{2})=-\cos1+\sin2$ (using $\cos(2-\tfrac{\pi}{2})=\sin2$).
   3. $I_2=\int_{-1}^{0}\sin(x+1)\,dx=\big[-\cos(x+1)\big]_{-1}^{0}=-\cos1+\cos0=1-\cos1$.
   4. $I_3=\int_{0}^{1}\sin x\,dx=1-\cos1$. …

   🔒 Start your 14-day free trial to see the full solution →
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: $x=1/e\Rightarrow u=-1$; $x=1\Rightarrow u=0$; $x=e^2\Rightarrow u=2$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $(\pi,2\pi)$: $[-x\cos x+\sin x]_\pi^{2\pi}=(-2\pi\cos2\pi+\sin2\pi)-(-\pi\cos\pi+\sin\pi)=(-2\pi(1)+0)-(-\pi(-1)+0)=-2\pi-\pi=-3\pi$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $=2\times(3-1)=4$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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, $[\sin x]=0$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $(-x)-(-1) = 1-x$. $\int_{-1}^0(1-x)\,dx = \left[x-\frac{x^2}{2}\right]_{-1}^0 = 0 - \left(-1-\frac12\right) = \frac32.$ …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $\displaystyle\int_2^3 1\,dx = [x]_2^3 = 1$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.$\displaystyle\int_{0}^{\pi} \dfrac{\cos x}{\sqrt{1-\sin^2 x}}\, dx =$ (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $-\dfrac{\pi}{2}$ (D) $0$

    ›Reveal solutionSolution

    The integrand simplifies to $\pm1$ depending on the sign of $\cos x$, and the two halves of $[0,\pi]$ cancel exactly.

    Concept and Intuition

    $\sqrt{1-\sin^2x}=|\cos x|$, not simply $\cos x$, since square roots are always non-negative.

    Step-by-Step Solution

    1. $\dfrac{\cos x}{\sqrt{1-\sin^2x}} = \dfrac{\cos x}{|\cos x|} = \text{sign}(\cos x)$.
    2. On $[0,\pi/2)$, $\cos x>0$, so the integrand is $+1$.
    3. On $(\pi/2,\pi]$, $\cos x<0$, so the integrand is $-1$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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. $\displaystyle\int_{-1}^0(-1)\,dx = -1\cdot[0-(-1)] = -1$. …

    🔒 Start your 14-day free trial to see the full solution →