Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{4}|x-1|\,dx$

Showing the 12 most recent of 30 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $f(x) = \left|\dfrac{x}{3} - 3^{-x}\right|$ and $\displaystyle\int_0^5 f(x)\,dx = K + \dfrac{3^n + 1}{3^5 \log 3}$, then $nK =$ (A) $\dfrac{46}{3}$ (B) $\dfrac{23}{6}$ (C) $\dfrac{54}{5}$ (D) $\dfrac{28}{9}$

   ›Reveal solutionSolution

   This tests splitting a modulus integral at its sign-change point and matching the result to a given algebraic form; the answer is $nK=\dfrac{46}{3}$.

   Concept and Intuition

   $f(x)=\left|\dfrac{x}{3}-3^{-x}\right|$ is the absolute value of $g(x)=\dfrac{x}{3}-3^{-x}$. To integrate an absolute value, we must find where $g$ changes sign and split the integral there — inside each sub-interval $f$ equals $+g$ or $-g$ depending on the sign.

   Step-by-Step Solution

   1. Find the zero of $g(x)=\dfrac{x}{3}-3^{-x}$: try $x=1$: $\dfrac13 - 3^{-1}=\dfrac13-\dfrac13=0$. So $x=1$ is the crossing point.
   2. $g'(x)=\dfrac13+3^{-x}\log 3>0$ always, so $g$ is strictly increasing — it is negative on $[0,1)$ and positive on $(1,5]$, with exactly one crossing.
   3. So $\displaystyle\int_0^5 f\,dx=\int_0^1(-g)\,dx+\int_1^5 g\,dx$.
   4. Antiderivative: $\displaystyle\int g\,dx=\int\left(\frac{x}{3}-3^{-x}\right)dx=\frac{x^2}{6}+\frac{3^{-x}}{\log 3}=G(x)$.
   5. $\displaystyle\int_0^1(-g)\,dx = -(G(1)-G(0)) = -\left[\left(\frac16+\frac{1}{3\ln3}\right)-\left(0+\frac{1}{\ln3}\right)\right]=\frac{2}{3\ln3}-\frac16$.
   6. $\displaystyle\int_1^5 g\,dx = G(5)-G(1)=\left(\frac{25}{6}+\frac{3^{-5}}{\ln3}\right)-\left(\frac16+\frac{1}{3\ln3}\right)=4+\frac{\frac{1}{243}-\frac13}{\ln3}=4-\frac{80}{243\ln3}$. …

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2. 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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3. 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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4. 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 $(-x)-(-1) = 1-x$. $\int_{-1}^0(1-x)\,dx = \left[x-\frac{x^2}{2}\right]_{-1}^0 = 0 - \left(-1-\frac12\right) = \frac32.$ …

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5. 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. $I_3=\int_{0}^{1}\sin x\,dx=1-\cos1$. …

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6. 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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7. 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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8. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $[x]$ is the greatest integer function, then $\displaystyle\int_0^5 [x]\,dx =$ (A) $15$ (B) $2$ (C) $3$ (D) $10$

   ›Reveal solutionSolution

   The greatest-integer function is piecewise constant on each unit interval; summing the constant values times width $1$ gives $10$ — (D).

   Concept and Intuition

   $[x]$ (the floor function) takes the constant integer value $n$ throughout $[n,n+1)$. Integrating it over $[0,5]$ is therefore just a sum of rectangle areas, each of width $1$ and height equal to the integer part on that strip.

   Step-by-Step Solution

   1. On $[0,1)$: $[x]=0$, contributes $0\times1=0$.
   2. On $[1,2)$: $[x]=1$, contributes $1\times1=1$.
   3. On $[2,3)$: $[x]=2$, contributes $2\times1=2$.
   4. On $[3,4)$: $[x]=3$, contributes $3\times1=3$.
   5. On $[4,5)$: $[x]=4$, contributes $4\times1=4$.
   6. Sum: $0+1+2+3+4=10$.

   Common Mistakes …

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9. 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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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $f(x) = \begin{cases} \dfrac{6x^2+1}{4x^3+2x+3}, & 0 < x < 1 \\ x^2+1, & 1 \le x < 2 \end{cases}$ then $\displaystyle\int_0^2 f(x)\, dx =$ (A) $\dfrac{1}{2}\log 3 + \dfrac{10}{3}$ (B) $\dfrac{1}{2}\log 3 - \dfrac{10}{3}$ (C) $\dfrac{1}{2}\log 3 + \dfrac{13}{3}$ (D) $\dfrac{1}{2}\log 3 + \dfrac{20}{3}$

    ›Reveal solutionSolution

    Split the definite integral at the piecewise boundary $x=1$; the first piece is a disguised logarithmic derivative, the second is a simple polynomial integral.

    Concept and Intuition

    The numerator $6x^2+1$ is exactly half the derivative of the denominator $4x^3+2x+3$ (since $\frac{d}{dx}(4x^3+2x+3)=12x^2+2=2(6x^2+1)$), so that piece integrates directly to a logarithm — no partial fractions needed.

    Step-by-Step Solution

    1. $\displaystyle\int_0^2 f(x)dx=\int_0^1\frac{6x^2+1}{4x^3+2x+3}dx+\int_1^2(x^2+1)dx$.
    2. Since $\frac{d}{dx}(4x^3+2x+3)=12x^2+2=2(6x^2+1)$: $\int_0^1\frac{6x^2+1}{4x^3+2x+3}dx=\frac12\Big[\log(4x^3+2x+3)\Big]_0^1=\frac12\big[\ln9-\ln3\big]=\frac12\ln3.$ …

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11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $\displaystyle\int_1^n [x]\, dx = 120$, then $n =$ (A) $15$ (B) $16$ (C) $14$ (D) $12$

    ›Reveal solutionSolution

    The floor function is constant on each unit interval, turning the integral into a simple triangular-number sum; solve the resulting quadratic for $n$.

    Concept and Intuition

    On each interval $[k,k+1)$, $[x]=k$, so integrating the floor function from $1$ to an integer $n$ just adds up $k\times1$ for $k=1,\dots,n-1$ — a sum of consecutive integers, which is a well-known triangular number.

    Step-by-Step Solution

    1. $\displaystyle\int_1^n[x]\,dx=\sum_{k=1}^{n-1}\int_k^{k+1}k\,dx=\sum_{k=1}^{n-1}k=\frac{(n-1)n}{2}$.
    2. Set equal to $120$: $\dfrac{n(n-1)}{2}=120 \Rightarrow n^2-n-240=0$. …

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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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