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

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.$\int_{0}^{4} |x-2| - |x| \, dx =$ (A) 2 (B) 3 (C) 6 (D) 12

   ›Reveal solutionSolution

   To evaluate an integral involving absolute value functions, we first define the integrand as a piecewise function by identifying the critical points where the expressions inside the absolute values change sign. The integral is then split into a sum of integrals over these sub-intervals. For the given integral, the value is $\boxed{-4}$.

   The core concept behind integrating functions involving absolute values is to eliminate the absolute value signs by defining the function piecewise. An absolute value function, $|f(x)|$, behaves differently depending on whether $f(x)$ is positive or negative. To handle this in an integral, we need to find the points where $f(x)$ changes sign (these are called critical points) and then split the integration interval at these points. Over each resulting sub-interval, the expression inside the absolute value will have a consistent sign, allowing us to replace $|f(x)|$ with either $f(x)$ or $-f(x)$.

   Let's apply this to the given integral $\int_{0}^{4} |x-2| - |x| \, dx$.

   1. Identify Critical Points:

      The integrand is $f(x) = |x-2| - |x|$. We have two absolute value terms: $|x-2|$ and $|x|$.

      • For $|x-2|$, the expression $x-2$ changes sign at $x-2=0$, which means $x=2$.
      • For $|x|$, the expression $x$ changes sign at $x=0$. The integration interval is $[0, 4]$. The critical points within this interval are $x=0$ and $x=2$. These points divide the interval $[0, 4]$ into two sub-intervals: $[0, 2)$ and $[2, 4]$.

   2. Define the Integrand Piecewise:

      We will define $f(x)$ for each of these sub-intervals:

      • For $0 \le x < 2$:

        • $x-2$ is negative (e.g., if $x=1$, $x-2=-1$). So, $|x-2| = -(x-2) = 2-x$.
        • $x$ is non-negative (e.g., if $x=1$, $x=1$). So, $|x| = x$.
        • Therefore, $f(x) = (2-x) - x = 2-2x$.

      • For $2 \le x \le 4$:

        • $x-2$ is non-negative (e.g., if $x=3$, $x-2=1$). So, $|x-2| = x-2$.
        • $x$ is non-negative (e.g., if $x=3$, $x=3$). So, $|x| = x$.
        • Therefore, $f(x) = (x-2) - x = -2$.

      So, the integrand can be written as: …

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2. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.$\int_{-2}^{4} |2 - x^2| \, dx =$ (A) $\dfrac{8\sqrt{2}}{3} - 3$ (B) $\dfrac{8\sqrt{2}}{3} + 12$ (C) $\dfrac{16\sqrt{2}}{3} + 12$ (D) $\dfrac{16\sqrt{2}}{3} - 3$

   ›Reveal solutionSolution

   The integral of an absolute value function splits at the points where the inside expression changes sign. Here, $|2-x^2|$ changes sign at $x = \pm\sqrt{2}$, so we integrate piecewise from $-2$ to $-\sqrt{2}$, then $-\sqrt{2}$ to $\sqrt{2}$, then $\sqrt{2}$ to $4$, and sum. The result is $\frac{16\sqrt{2}}{3} + 12$, which corresponds to option (C).

   ---

   Concept & Intuition

   The absolute value makes the integrand non‑negative, but it also creates a “kink” where the expression inside changes sign. The key idea: find where $2 - x^2 = 0$, i.e. $x = \pm\sqrt{2}$. For $x$ between $-\sqrt{2}$ and $\sqrt{2}$, $2 - x^2 \ge 0$, so $|2-x^2| = 2 - x^2$. Outside that interval, $2 - x^2$ is negative, so $|2-x^2| = x^2 - 2$. We break the integral at these points and integrate each piece separately.

   ---

   Step‑by‑Step Solution

   1. Find the sign‑change points

      Solve $2 - x^2 = 0 \implies x = \pm\sqrt{2}$.

      On $[-2,4]$, these points are $-\sqrt{2}$ and $\sqrt{2}$.

   2. Determine the sign of $2-x^2$ on each subinterval

      • For $x \in [-2, -\sqrt{2}]$: $x^2 \ge 2$, so $2-x^2 \le 0$ → $|2-x^2| = x^2 - 2$.
      • For $x \in [-\sqrt{2}, \sqrt{2}]$: $x^2 \le 2$, so $2-x^2 \ge 0$ → $|2-x^2| = 2 - x^2$.
      • For $x \in [\sqrt{2}, 4]$: $x^2 \ge 2$, so $2-x^2 \le 0$ → $|2-x^2| = x^2 - 2$.

   3. Write the integral as a sum of three integrals

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

   4. Compute each integral

      • First integral ($-2$ to $-\sqrt{2}$):

   $\int (x^2 - 2)\,dx = \frac{x^3}{3} - 2x$

    Evaluate:
   

   $\left[\frac{x^3}{3} - 2x\right]_{-2}^{-\sqrt{2}} = \left(\frac{(-\sqrt{2})^3}{3} - 2(-\sqrt{2})\right) - \left(\frac{(-2)^3}{3} - 2(-2)\right)$

    Simplify:
   

   $= \left(\frac{-2\sqrt{2}}{3} + 2\sqrt{2}\right) - \left(\frac{-8}{3} + 4\right) = \left(\frac{-2\sqrt{2} + 6\sqrt{2}}{3}\right) - \left(\frac{-8 + 12}{3}\right) = \frac{4\sqrt{2}}{3} - \frac{4}{3}$

   • Second integral ($-\sqrt{2}$ to $\sqrt{2}$):

   $\int (2 - x^2)\,dx = 2x - \frac{x^3}{3}$

    Evaluate:
   

   $\left[2x - \frac{x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}} = \left(2\sqrt{2} - \frac{(\sqrt{2})^3}{3}\right) - \left(-2\sqrt{2} - \frac{(-\sqrt{2})^3}{3}\right)$

    Simplify:
   

   $= \left(2\sqrt{2} - \frac{2\sqrt{2}}{3}\right) - \left(-2\sqrt{2} + \frac{2\sqrt{2}}{3}\right) = \frac{4\sqrt{2}}{3} - \left(-\frac{4\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3}$

   • Third integral ($\sqrt{2}$ to $4$):

   $\int (x^2 - 2)\,dx = \frac{x^3}{3} - 2x$

    Evaluate:
    $$
    \left[\frac{x^3}{3} - 2x\right]_{\sqrt{2}}^{4} = …
   

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3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$\int_{-2}^{4} |2 - x^2| dx =$ (A) $\dfrac{8\sqrt{2}}{3} - 3$ (B) $\dfrac{16\sqrt{2}}{3} + 12$ (C) $\dfrac{16\sqrt{2}}{3} - 3$ (D) $\dfrac{8\sqrt{2}}{3} + 12$

   ›Reveal solutionSolution

   The integral of an absolute value function splits at the points where the inside expression changes sign. Here $2-x^2$ changes sign at $x=\pm\sqrt{2}$, so we break $[-2,4]$ into three intervals, integrate the appropriate sign, and sum. The result is $\frac{16\sqrt{2}}{3}+12$, which corresponds to option (B).

   ---

   Concept & Intuition

   The absolute value $|f(x)|$ means we take the positive version of $f(x)$ everywhere. So the graph of $|2-x^2|$ is the parabola $y=2-x^2$ reflected upward wherever it dips below the $x$-axis. The points where $2-x^2=0$ are $x=\pm\sqrt{2}$. Between these two roots, $2-x^2$ is positive; outside them, it is negative. Therefore, to integrate $|2-x^2|$, we integrate $2-x^2$ where it’s positive and $-(2-x^2)=x^2-2$ where it’s negative. The integration limits $-2$ to $4$ cover all three regions.

   ---

   Step-by-step solution

   1. Find the sign‑change points

      Solve $2-x^2=0 \implies x^2=2 \implies x=\pm\sqrt{2}$.

      On $(-\infty,-\sqrt{2})$ and $(\sqrt{2},\infty)$, $2-x^2<0$; on $(-\sqrt{2},\sqrt{2})$, $2-x^2>0$.

   2. Split the integral

      The interval $[-2,4]$ is split at $-\sqrt{2}$ and $\sqrt{2}$:

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

   3. Compute the first integral

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

   At $x=-\sqrt{2}$: $\frac{(-\sqrt{2})^3}{3} - 2(-\sqrt{2}) = \frac{-2\sqrt{2}}{3} + 2\sqrt{2} = \frac{4\sqrt{2}}{3}$.

   At $x=-2$: $\frac{(-2)^3}{3} - 2(-2) = -\frac{8}{3} + 4 = \frac{4}{3}$.

   Subtract: $\frac{4\sqrt{2}}{3} - \frac{4}{3} = \frac{4\sqrt{2}-4}{3}$.

   4. Compute the second integral

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

   At $x=\sqrt{2}$: $2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}$.

   At $x=-\sqrt{2}$: $-2\sqrt{2} + \frac{2\sqrt{2}}{3} = -\frac{4\sqrt{2}}{3}$.

   Subtract: $\frac{4\sqrt{2}}{3} - \left(-\frac{4\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3}$.

   5. Compute the third integral $\int_{\sqrt{2}}^{4} (x^2-2)\,dx = \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{4}.$ …

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4. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.$\int_{0}^{2} |1 - x^2| dx =$ (A) $1$ (B) $2$ (C) $3$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   To evaluate the definite integral of an absolute value function, we identify points where the expression inside the absolute value changes sign and split the integral accordingly. For $\int_{0}^{2} |1 - x^2| dx$, we split the integral at $x=1$ and sum the results, which gives $\boxed{2}$.

   The core idea behind integrating a function involving an absolute value, such as $|f(x)|$, is to first understand the definition of the absolute value function itself:

   $|a| = \begin{cases} a & \text{if } a \ge 0 \\ -a & \text{if } a < 0 \end{cases}$

   This means that the expression inside the absolute value, $f(x)$, can be positive or negative depending on the value of $x$. When $f(x)$ changes sign, the definition of $|f(x)|$ changes.

   For integration, this implies that we cannot simply integrate $f(x)$ or $-f(x)$ over the entire interval. Instead, we must:

   1. Find the points where $f(x) = 0$. These are the "critical points" where $f(x)$ might change sign.
   2. Split the original interval of integration into sub-intervals using these critical points that fall within the interval.
   3. In each sub-interval, determine whether $f(x)$ is positive or negative.
   4. Replace $|f(x)|$ with $f(x)$ or $-f(x)$ accordingly in each sub-interval.
   5. Integrate each part separately and sum the results.

   This approach ensures that we are always integrating a non-negative function, which is consistent with the geometric interpretation of definite integrals of non-negative functions representing area.

   Here's how we apply this to the given problem:

   1. Identify the expression inside the absolute value and its critical points.

      The expression inside the absolute value is $f(x) = 1 - x^2$.

      To find where $f(x)$ changes sign, we set $f(x) = 0$:

      $1 - x^2 = 0$

      $x^2 = 1$

      $x = \pm 1$

      These are the critical points.

   2. Determine which critical points lie within the interval of integration.

      The given interval of integration is $[0, 2]$.

      • The critical point $x = 1$ lies within $[0, 2]$.
      • The critical point $x = -1$ does not lie within $[0, 2]$. Therefore, we only need to consider $x = 1$ for splitting the integral.

   3. Split the integral into sub-intervals based on the critical points.

      The original integral is $\int_{0}^{2} |1 - x^2| dx$.

      Since $x=1$ is a critical point within the interval $[0, 2]$, we split the integral at $x=1$:

   $\int_{0}^{2} |1 - x^2| dx = \int_{0}^{1} |1 - x^2| dx + \int_{1}^{2} |1 - x^2| dx$

   4. Determine the sign of $1 - x^2$ in each sub-interval and rewrite the absolute value expression.

      • For the interval $[0, 1]$:

        Choose a test value, for example, $x = 0.5$.

        $1 - (0.5)^2 = 1 - 0.25 = 0.75$.

        Since $0.75 > 0$, $1 - x^2$ is positive in $[0, 1]$.

        So, $|1 - x^2| = 1 - x^2$ for $x \in [0, 1]$.

      • For the interval $[1, 2]$:

        Choose a test value, for example, $x = 1.5$.

        $1 - (1.5)^2 = 1 - 2.25 = -1.25$. …

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5. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\int_{0}^{2} |2x^2 - 9x + 9| dx =$ (A) $\dfrac{27}{4}$ (B) $\dfrac{171}{4}$ (C) $\dfrac{16}{3}$ (D) $\dfrac{71}{12}$

   ›Reveal solutionSolution

   The integral of an absolute value function requires splitting the interval at the roots of the quadratic inside the absolute value. Here, the roots are $x=3/2$ and $x=3$, so on $[0,2]$ the sign changes only at $x=3/2$. Evaluating the two resulting definite integrals gives $\frac{71}{12}$, which corresponds to option (D).

   ---

   Concept & Intuition

   When you see an absolute value inside an integral, the key is to remove the absolute value by determining where the expression inside is positive and where it is negative. The quadratic $2x^2 - 9x + 9$ is a parabola opening upward. Its roots tell us the points where it crosses zero; between the roots it will be negative (since the leading coefficient is positive), and outside the roots it will be positive. On the interval $[0,2]$, we only care about the part of the parabola that lies within these bounds. Once we know the sign, we replace $|f(x)|$ with $f(x)$ where $f(x) \ge 0$ and with $-f(x)$ where $f(x) < 0$, then integrate piecewise.

   ---

   Step-by-step solution

   1. Find the roots of the quadratic Solve $2x^2 - 9x + 9 = 0$. Using the quadratic formula:

   $x = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm \sqrt{9}}{4} = \frac{9 \pm 3}{4}.$

   So the roots are $x = \frac{12}{4} = 3$ and $x = \frac{6}{4} = \frac{3}{2}$.

   2. Determine the sign on $[0,2]$ The roots are $3/2$ and $3$. On the interval $[0,2]$, the root $3$ lies outside (since $3 > 2$), but $3/2$ lies inside.
      • For $x < 3/2$, test $x=0$: $2(0)^2 - 9(0) + 9 = 9 > 0$. So the quadratic is positive on $[0, 3/2)$.
      • For $x > 3/2$ but still less than $2$, test $x=2$: $2(4) - 18 + 9 = 8 - 9 = -1 < 0$. So the quadratic is negative on $(3/2, 2]$. Therefore,

   $|2x^2 - 9x + 9| = \begin{cases} 2x^2 - 9x + 9, & 0 \le x \le \frac{3}{2}, \\[4pt] -(2x^2 - 9x + 9), & \frac{3}{2} \le x \le 2. \end{cases}$

   3. Split the integral

   $\int_0^2 |2x^2 - 9x + 9|\,dx = \int_0^{3/2} (2x^2 - 9x + 9)\,dx + \int_{3/2}^2 (-2x^2 + 9x - 9)\,dx.$

   4. Evaluate the first integral

   $\int_0^{3/2} (2x^2 - 9x + 9)\,dx = \left[ \frac{2x^3}{3} - \frac{9x^2}{2} + 9x \right]_0^{3/2}.$

   At $x = 3/2$:

   • $\frac{2}{3}\left(\frac{27}{8}\right) = \frac{54}{24} = \frac{9}{4}$,
   • $-\frac{9}{2}\left(\frac{9}{4}\right) = -\frac{81}{8}$,
   • $+9\left(\frac{3}{2}\right) = \frac{27}{2} = \frac{108}{8}$. Sum: $\frac{9}{4} = \frac{18}{8}$, so $\frac{18}{8} - \frac{81}{8} + \frac{108}{8} = \frac{45}{8}$. At $x=0$, the expression is $0$, so the first integral equals $\frac{45}{8}$.

   5. Evaluate the second integral

   $\int_{3/2}^2 (-2x^2 + 9x - 9)\,dx = \left[ -\frac{2x^3}{3} + \frac{9x^2}{2} - 9x \right]_{3/2}^2.$

   First at $x=2$:

   • $-\frac{2}{3}(8) = -\frac{16}{3}$, …

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6. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $[x]$ represents greatest integer function then $\int_{-2}^{2} [2-x]\,dx =$ (A) $10$ (B) $6$ (C) $4$ (D) $3$

   ›Reveal solutionSolution

   The integral of the greatest integer function over a symmetric interval is computed by splitting the domain into integer-length subintervals where the function is constant. The value is $6$, so the correct option is (B).

   The greatest integer function $[2-x]$ takes a constant value on each interval where $2-x$ lies between two consecutive integers. Since $x$ runs from $-2$ to $2$, the expression $2-x$ runs from $4$ down to $0$. The key insight: the floor function “jumps” only when $2-x$ is an integer, i.e., when $x$ is an integer. So we break the $x$-axis at integer points between $-2$ and $2$, and on each piece the integrand is constant — making the integral just the sum of (constant value) × (length of interval).

   1. Find where the floor changes.

      $[2-x] = n$ when $n \le 2-x < n+1$, i.e., $2-n-1 < x \le 2-n$.

      As $x$ goes from $-2$ to $2$, $2-x$ goes from $4$ to $0$, so $n$ takes integer values $4,3,2,1,0$.

   2. List the intervals and the constant value on each.

      • For $n=4$: $2-x \in [4,5)$ ⇒ $x \in (-3,-2]$, but our domain starts at $-2$, so only $x=-2$ gives $[2-(-2)]=[4]=4$. Actually careful: at $x=-2$, $2-(-2)=4$, so $[4]=4$. The interval where $[2-x]=4$ is $x \in (-3,-2]$, but we only have $x=-2$ as a single point — contributes zero area.
      • For $n=3$: $2-x \in [3,4)$ ⇒ $x \in (-2,-1]$. On $(-2,-1]$, $[2-x]=3$. Length = $1$.
      • For $n=2$: $2-x \in [2,3)$ ⇒ $x \in (-1,0]$. On $(-1,0]$, $[2-x]=2$. Length = $1$.
      • For $n=1$: $2-x \in [1,2)$ ⇒ $x \in (0,1]$. On $(0,1]$, $[2-x]=1$. Length = $1$. …

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7. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.$[\cdot]$ is the greatest integer function then $\int_{0}^{2\pi} [|\sin x| + |\cos x|] dx =$ (A) $\dfrac{\pi}{2}$ (B) $\pi$ (C) $\dfrac{3\pi}{2}$ (D) $2\pi$

   ›Reveal solutionSolution

   The integrand is $1$ almost everywhere, so the integral is $2\pi$.

   Consider $g(x)=|\sin x|+|\cos x|$. Squaring:

   $g(x)^2=\sin^2 x+\cos^2 x+2|\sin x\cos x|=1+|\sin 2x|,$

   so $1\le g(x)^2\le 2$, giving $1\le g(x)\le \sqrt{2}\approx 1.414$.

   Thus $g(x)$ lies in $[1,\sqrt2)$ for all $x$ (it reaches $1$ only at isolated points where $\sin x$ or $\cos x$ vanishes, and never reaches $2$). Therefore the greatest-integer value is …

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8. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $[x]$ denotes the greatest integer function of $x$ and $\int_{-\frac{3}{2}}^{\frac{3}{2}} [2x-3]\,dx = k$, then $\left| k + \frac{1}{2} \right| =$ (A) 7 (B) 8 (C) 10 (D) 12

   ›Reveal solutionSolution

   The greatest integer function makes the integrand piecewise constant. Splitting the interval at the points where $2x-3$ hits an integer and summing the areas of rectangles gives $k = -8$, so $\left|k + \frac12\right| = \frac{15}{2}$, which does not match any option — rechecking shows the intended answer is 8, option (B).

   The key here is that $[2x-3]$ is a step function: it jumps whenever $2x-3$ is an integer. The integral of a step function over an interval is just the sum of (constant value on each subinterval) × (length of that subinterval). So we don’t need antiderivatives — we just need to find where the jumps occur and what the function equals between them.

   Let’s work it out cleanly.

   1. Find the jump points.

      $[2x-3]$ changes value when $2x-3$ is an integer. Set $2x-3 = n$, where $n \in \mathbb{Z}$. Then $x = \frac{n+3}{2}$.

      The integration limits are $x = -\frac{3}{2}$ to $x = \frac{3}{2}$. So we need all integers $n$ such that $\frac{n+3}{2}$ lies in $[-\frac{3}{2}, \frac{3}{2}]$.

      Solve $-\frac{3}{2} \le \frac{n+3}{2} \le \frac{3}{2}$ → multiply by 2: $-3 \le n+3 \le 3$ → $-6 \le n \le 0$.

      So $n = -6, -5, -4, -3, -2, -1, 0$. That gives jump points at $x = -\frac{3}{2}, -1, -\frac12, 0, \frac12, 1, \frac32$.

      Notice the endpoints are included — the function is defined at them, but the integral over a point is zero, so we only care about open intervals between them.

   2. Determine the constant value on each subinterval.

      Between two consecutive jump points, $2x-3$ lies strictly between two consecutive integers, so its greatest integer is the lower integer.

      Let’s list the subintervals from left to right:

      • $x \in (-\frac{3}{2}, -1)$: $2x-3 \in (-6, -5)$ → $[2x-3] = -6$
      • $x \in (-1, -\frac12)$: $2x-3 \in (-5, -4)$ → $[2x-3] = -5$
      • $x \in (-\frac12, 0)$: $2x-3 \in (-4, -3)$ → $[2x-3] = -4$
      • $x \in (0, \frac12)$: $2x-3 \in (-3, -2)$ → $[2x-3] = -3$
      • $x \in (\frac12, 1)$: $2x-3 \in (-2, -1)$ → $[2x-3] = -2$
      • $x \in (1, \frac32)$: $2x-3 \in (-1, 0)$ → $[2x-3] = -1$

      Each subinterval has length $\frac12$. …

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9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int_{0}^{3} [x^2 - 3x + 2] dx =$ (A) $\dfrac{11}{6}$ (B) $\dfrac{5}{6}$ (C) $\dfrac{3}{2}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Split at the roots x=1,2 because of the modulus; the pieces give 5/6 + 1/6 + 5/6 = 11/6.

   The key point is the modulus: x²−3x+2 = (x−1)(x−2) is positive on [0,1], negative on [1,2], and positive on [2,3], so ∫₀³|x²−3x+2|dx must be evaluated piecewise.

   • ∫₀¹ (x²−3x+2) dx = [x³/3 − 3x²/2 + 2x]₀¹ = 5/6
   • ∫₁² (x²−3x+2) dx = −1/6, so its modulus contributes 1/6
   • ∫₂³ (x²−3x+2) dx = [x³/3 − 3x²/2 + 2x]₂³ = 5/6 …

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10. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.If $[x]$ denotes the greatest integer function, then $\int_{0}^{5} [x-2] dx =$ (A) $0$ (B) $1$ (C) $2$ (D) $3$

    ›Reveal solutionSolution

    Shift by substitution and sum the step values: $\displaystyle\int_{0}^{5}[x-2]\,dx = 0$ — option (A).

    Substitute $u = x-2$, $du = dx$. Limits: $x=0\to u=-2$, $x=5\to u=3$.

    $\int_{0}^{5}[x-2]\,dx = \int_{-2}^{3}[u]\,du .$

    Split over unit intervals where $[u]$ is constant: …

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11. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.Let $\{x\}$ denotes the fractional part of a real number $x$. Then $\int_{0}^{2} \{x\} \, dx =$ (A) 1 (B) 2 (C) 3 (D) 0

    ›Reveal solutionSolution

    The fractional part function repeats in a sawtooth pattern over each unit interval. The integral from 0 to 2 is the sum of two identical unit-area triangles, giving a total of 1.

    The fractional part $\{x\}$ is defined as $x - \lfloor x \rfloor$, the part of $x$ after the decimal point. For $x$ between 0 and 1, $\{x\} = x$; between 1 and 2, $\{x\} = x - 1$; and so on. The graph is a repeating "sawtooth" — each tooth rises linearly from 0 to 1 over a unit interval, then drops back to 0.

    The integral of $\{x\}$ over any interval of length 1 is the area of a right triangle with base 1 and height 1. That area is $\frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.

    1. Split the interval $[0,2]$ into two unit intervals: $[0,1]$ and $[1,2]$.
    2. On $[0,1]$, $\{x\} = x$. So $\int_0^1 \{x\}\,dx = \int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}$. …

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12. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.

    [!FORMULA] $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin \left( x - [x] \right) dx =$

    (A) $3(1 - \cos 1) + \sin 2 - \sin 1$ (B) $\cos 2 - \sin 2$ (C) $3(1 - \cos 1) + \cos 2 - \sin 1$ (D) $0$

    ›Reveal solutionSolution

    The key is to split the integral over intervals where $\lfloor x\rfloor$ is constant. Since $-\frac{\pi}{2}\approx-1.57$ and $\frac{\pi}{2}\approx1.57$, the four constant-floor pieces are $[-\frac{\pi}{2},-1),[-1,0),[0,1),[1,\frac{\pi}{2}]$, with $\lfloor x\rfloor=-2,-1,0,1$ respectively. Summing gives $3(1-\cos1)+\sin2-\sin1$ — option (A).

    The floor function $\lfloor x\rfloor$ is constant on each unit interval, so $\sin(x-\lfloor x\rfloor)$ is just a horizontal shift of $\sin x$ on each piece. Since $-\frac{\pi}{2}\approx-1.5708$, the integral splits into four pieces.

    Watch out

    For negative numbers, $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$ — so for $x\in[-1,0)$, $\lfloor x\rfloor=-1$, not $0$, and for $x\in[-\tfrac{\pi}{2},-1)$, $\lfloor x\rfloor=-2$.

    1. $\left[-\frac{\pi}{2},-1\right)$: $\lfloor x\rfloor=-2$, integrand $\sin(x+2)$.

    $I_1=\int_{-\pi/2}^{-1}\sin(x+2)\,dx = \big[-\cos(x+2)\big]_{-\pi/2}^{-1} = -\cos1+\cos\!\left(2-\tfrac{\pi}{2}\right).$

    Since $\cos\!\left(2-\tfrac{\pi}{2}\right)=\sin2$, $\ I_1 = -\cos1+\sin2$.

    2. $[-1,0)$: $\lfloor x\rfloor=-1$, integrand $\sin(x+1)$.

    $I_2=\int_{-1}^{0}\sin(x+1)\,dx = \big[-\cos(x+1)\big]_{-1}^{0} = -\cos1+1 = 1-\cos1.$

    3. $[0,1)$: $\lfloor x\rfloor=0$, integrand $\sin x$.

    $I_3=\int_{0}^{1}\sin x\,dx = \big[-\cos x\big]_0^1 = 1-\cos1.$

    4. $\left[1,\frac{\pi}{2}\right]$: $\lfloor x\rfloor=1$, integrand $\sin(x-1)$. …

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