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

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1. 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, $x = 1.5$.

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

        Since $-1.25 < 0$, $1 - x^2$ is negative in $[1, 2]$.

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

   5. Substitute the rewritten expressions back into the split integrals and evaluate.

      The integral now becomes:

   $\int_{0}^{1} (1 - x^2) dx + \int_{1}^{2} (x^2 - 1) dx$

   Let's evaluate the first part:
   

   $\int_{0}^{1} (1 - x^2) dx = \left[ x - \frac{x^3}{3} \right]_{0}^{1}$

   $= \left( 1 - \frac{1^3}{3} \right) - \left( 0 - \frac{0^3}{3} \right)$

   $= \left( 1 - \frac{1}{3} \right) - 0$

   $= \frac{2}{3}$

   Now, evaluate the second part:
   

   $\int_{1}^{2} (x^2 - 1) dx = \left[ \frac{x^3}{3} - x \right]_{1}^{2}$

   $= \left( \frac{2^3}{3} - 2 \right) - \left( \frac{1^3}{3} - 1 \right)$

   $= \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right)$

   $= \left( \frac{8 - 6}{3} \right) - \left( \frac{1 - 3}{3} \right)$

   $= \frac{2}{3} - \left( -\frac{2}{3} \right)$

   $= \frac{2}{3} + \frac{2}{3}$

   $= \frac{4}{3}$

   6. Sum the results from the sub-integrals. The total value of the integral is the sum of the values from the two parts:

   $\int_{0}^{2} |1 - x^2| dx = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2$

   ✓Final answer

   The value of the integral is $\boxed{2}$.

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:
   

   $\left[\frac{x^3}{3} - 2x\right]_{\sqrt{2}}^{4} = \left(\frac{64}{3} - 8\right) - \left(\frac{(\sqrt{2})^3}{3} - 2\sqrt{2}\right)$

    Simplify:
   

   $= \left(\frac{64}{3} - \frac{24}{3}\right) - \left(\frac{2\sqrt{2}}{3} - 2\sqrt{2}\right) = \frac{40}{3} - \left(\frac{2\sqrt{2} - 6\sqrt{2}}{3}\right) = \frac{40}{3} - \left(\frac{-4\sqrt{2}}{3}\right) = \frac{40}{3} + \frac{4\sqrt{2}}{3}$

   5. Sum the three results

   $\text{Total} = \left(\frac{4\sqrt{2}}{3} - \frac{4}{3}\right) + \frac{8\sqrt{2}}{3} + \left(\frac{40}{3} + \frac{4\sqrt{2}}{3}\right)$

   Combine $\sqrt{2}$ terms: $\frac{4\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} + \frac{4\sqrt{2}}{3} = \frac{16\sqrt{2}}{3}$.

   Combine constant terms: $-\frac{4}{3} + \frac{40}{3} = \frac{36}{3} = 12$.

   So the total is $\frac{16\sqrt{2}}{3} + 12$.

   Watch out

   A common mistake is forgetting to split at both $\sqrt{2}$ and $-\sqrt{2}$, or misidentifying which piece uses $2-x^2$ vs $x^2-2$. Always test a point in each interval to confirm the sign.

   Tip

   Notice the symmetry: the first and third integrals have the same form ($x^2-2$), but the limits are not symmetric about zero, so you must compute them separately. However, the middle integral is symmetric and gives a clean $\frac{8\sqrt{2}}{3}$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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

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

   At $x=4$: $\frac{64}{3} - 8 = \frac{64}{3} - \frac{24}{3} = \frac{40}{3}$.

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

   Subtract: $\frac{40}{3} - \left(-\frac{4\sqrt{2}}{3}\right) = \frac{40+4\sqrt{2}}{3}$.

   6. Sum the three results

   $\frac{4\sqrt{2}-4}{3} + \frac{8\sqrt{2}}{3} + \frac{40+4\sqrt{2}}{3} = \frac{(4\sqrt{2}+8\sqrt{2}+4\sqrt{2}) + (-4+40)}{3} = \frac{16\sqrt{2} + 36}{3}.$

   Simplify: $\frac{16\sqrt{2}}{3} + 12$.

   Tip

   Notice that the middle integral gave $\frac{8\sqrt{2}}{3}$ and the two outer integrals together contributed $\frac{8\sqrt{2}+36}{3}$, so the total is $\frac{16\sqrt{2}}{3}+12$. Always check that the constant term matches one of the options.

   Watch out

   A common mistake is forgetting to flip the sign on the outer intervals. If you integrate $2-x^2$ directly from $-2$ to $4$, you get a negative area contribution from the parts below the axis, which is wrong for an absolute value.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. 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:

   $f(x) = \begin{cases} 2-2x & \text{if } 0 \le x < 2 \\ -2 & \text{if } 2 \le x \le 4 \end{cases}$

   3. Split the Integral: Now, we can split the original integral into a sum of integrals over the sub-intervals:

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

   4. Evaluate Each Sub-integral:

      • First integral:

   $\int_{0}^{2} (2-2x) \, dx$

       We find the antiderivative of $2-2x$, which is $2x - x^2$.
   

   $[2x - x^2]_{0}^{2} = (2(2) - 2^2) - (2(0) - 0^2)$

   $= (4 - 4) - (0 - 0) = 0 - 0 = 0$

   *   **Second integral:**
   

   $\int_{2}^{4} (-2) \, dx$

       We find the antiderivative of $-2$, which is $-2x$.
   

   $[-2x]_{2}^{4} = (-2(4)) - (-2(2))$

   $= -8 - (-4) = -8 + 4 = -4$

   5. Sum the Results: Finally, we add the results of the two sub-integrals:

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

   ✓Final answer

   The value of the integral is $\boxed{-4}$.

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 $x=2$:

   • $-\frac{2}{3}(8) = -\frac{16}{3}$,
   • $\frac{9}{2}(4) = 18$,
   • $-9(2) = -18$. Sum: $-\frac{16}{3} + 0 = -\frac{16}{3}$. Now 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 $-\frac{18}{8} + \frac{81}{8} - \frac{108}{8} = -\frac{45}{8}$. So the definite integral is:

   $\left(-\frac{16}{3}\right) - \left(-\frac{45}{8}\right) = -\frac{16}{3} + \frac{45}{8}.$

   Common denominator $24$: $-\frac{128}{24} + \frac{135}{24} = \frac{7}{24}$.

   6. Add the two pieces

   $\frac{45}{8} + \frac{7}{24} = \frac{135}{24} + \frac{7}{24} = \frac{142}{24} = \frac{71}{12}.$

   Tip

   A common shortcut: notice that the quadratic is symmetric about its vertex at $x = 9/4 = 2.25$, but since our interval $[0,2]$ is not symmetric about that vertex, piecewise integration is the cleanest method.

   Watch out

   A classic mistake is forgetting to flip the sign on the negative region. Always test a point in each subinterval to confirm the sign before integrating.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.$\int_{1}^{2} x \sqrt{4-x^2} \, dx =$ (A) $\sqrt{3}$ (B) $2$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   The integral simplifies via substitution $u = 4 - x^2$, turning it into a standard power integral. The value is $\boxed{\sqrt{3}}$, which corresponds to option (A).

   The key insight here is that the integrand $x \sqrt{4 - x^2}$ is tailor-made for a substitution that eliminates the square root. When you see a function multiplied by its derivative (or nearly so), substitution is the natural move. Here, the derivative of $4 - x^2$ is $-2x$, and we have an $x$ sitting right there — just a constant factor away.

   Let’s walk through it step by step.

   1. Choose the substitution.

      Let $u = 4 - x^2$. Then $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} \, du$.

      This substitution will turn $\sqrt{4 - x^2}$ into $\sqrt{u}$, which is easy to integrate.

   2. Change the limits of integration.

      When $x = 1$, $u = 4 - 1^2 = 3$.

      When $x = 2$, $u = 4 - 2^2 = 0$.

      Notice the upper limit becomes smaller than the lower limit — that’s fine; we’ll handle it by swapping limits or keeping track of the sign.

   3. Rewrite the integral in terms of $u$.

      The original integral is

   $\int_{x=1}^{2} x \sqrt{4 - x^2} \, dx = \int_{u=3}^{0} \sqrt{u} \cdot \left(-\frac{1}{2}\right) du.$

   The $x \, dx$ becomes $-\frac{1}{2} du$, and $\sqrt{4 - x^2}$ becomes $\sqrt{u}$.

   4. Simplify the limits. We can swap the limits to make the lower limit smaller, which introduces a minus sign:

   $\int_{3}^{0} \left(-\frac{1}{2}\right) \sqrt{u} \, du = \frac{1}{2} \int_{0}^{3} \sqrt{u} \, du.$

   The two negatives (one from $du$ and one from swapping limits) cancel, giving a positive integral.

   5. Evaluate the integral. Recall $\sqrt{u} = u^{1/2}$, so

   $\frac{1}{2} \int_{0}^{3} u^{1/2} \, du = \frac{1}{2} \cdot \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{3} = \frac{1}{2} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{0}^{3} = \frac{1}{3} \left[ u^{3/2} \right]_{0}^{3}.$

   Now plug in the limits:

   $\frac{1}{3} \left( 3^{3/2} - 0 \right) = \frac{1}{3} \cdot 3^{3/2}.$

   Since $3^{3/2} = 3 \cdot \sqrt{3}$, this becomes

   $\frac{1}{3} \cdot 3 \sqrt{3} = \sqrt{3}.$

   Watch out

   A common mistake is forgetting to change the limits when substituting, or mishandling the sign from $du = -2x \, dx$. Always write the new limits explicitly before integrating.

   Tip

   If you prefer, you can avoid swapping limits by keeping them as $u=3$ to $u=0$ and evaluating $\frac{1}{2} \int_{3}^{0} u^{1/2} du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{3}^{0} = \frac{1}{3}(0 - 3^{3/2}) = -\frac{1}{3} \cdot 3\sqrt{3} = -\sqrt{3}$, then noticing the extra minus from $x\,dx = -\frac{1}{2}du$ gives $-\sqrt{3}$ times $-1$? Actually careful: the original substitution gave $-\frac{1}{2}du$, so the integral becomes $\int_{3}^{0} \sqrt{u} \cdot (-\frac{1}{2}) du = -\frac{1}{2} \int_{3}^{0} \sqrt{u} du = -\frac{1}{2} \cdot (-\frac{2}{3} \cdot 3^{3/2}) = \sqrt{3}$. Either way, the result is the same — just track signs carefully.

   ✓Final answer

   The value of the integral is $\boxed{\sqrt{3}}$, which is option (A).

7. 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:

   $\int_{-2}^{-1}(-2)\,du + \int_{-1}^{0}(-1)\,du + \int_{0}^{1}(0)\,du + \int_{1}^{2}(1)\,du + \int_{2}^{3}(2)\,du .$

   Each interval has length $1$, so this is

   $(-2) + (-1) + 0 + 1 + 2 = 0 .$

   ✓Final answer

   $\displaystyle\int_{0}^{5}[x-2]\,dx = 0$. Option (A).

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 $\frac12$.

   3. Compute the integral as a sum of rectangle areas.

   $k = \int_{-3/2}^{3/2} [2x-3]\,dx = \sum_{\text{subintervals}} (\text{value}) \times (\text{length})$

   $= (-6)\cdot\frac12 + (-5)\cdot\frac12 + (-4)\cdot\frac12 + (-3)\cdot\frac12 + (-2)\cdot\frac12 + (-1)\cdot\frac12$

   $= \frac12 \left( -6 -5 -4 -3 -2 -1 \right) = \frac12 \times (-21) = -\frac{21}{2}$

   So $k = -\frac{21}{2}$.

   4. Now compute the required expression.

   $\left| k + \frac12 \right| = \left| -\frac{21}{2} + \frac12 \right| = \left| -\frac{20}{2} \right| = | -10 | = 10$

   Watch out

   A common mistake is to forget that the greatest integer of a negative number is the next lower integer, not the integer part ignoring sign. For example, $[ -4.3 ] = -5$, not $-4$. Check the subinterval values carefully.

   ✓Final answer

   The value is $\boxed{10}$, which corresponds to option (C).

9. 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 $\int_0^1 \{x\}\,dx = \int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}$.
   3. On $[1,2]$, $\{x\} = x - 1$. So $\int_1^2 (x-1)\,dx = \left[\frac{(x-1)^2}{2}\right]_1^2 = \frac{1}{2}$.
   4. Adding them: $\frac{1}{2} + \frac{1}{2} = 1$.

   Watch out

   A common mistake is to think $\{x\}$ averages to 0.5 over each unit interval, but then multiply by length 2 to get 1 — that works here, but only because the function is linear. The triangle-area method is safer and generalises cleanly.

   ✓Final answer

   The value of the integral is $\boxed{1}$, which corresponds to option (A).

10. 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

    Total = 5/6 + 1/6 + 5/6 = 11/6. (Ignoring the modulus gives ∫₀³(x²−3x+2)dx = 3/2 — the common trap.)

    ✓Final answer

    The value of the integral is 11/6. The correct option is (A).

    ANSWER: A

11. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.If $[t]$ denotes greatest integer function, $\int_{-2}^{2} \left[ \dfrac{x^2 + [x+1]}{1 + x^2} \right] dx =$ (A) $2 \tan^{-1} 2$ (B) $0$ (C) $2$ (D) $\tan^{-1} 2$

    ›Reveal solutionSolution

    The greatest-integer integrand simplifies to $1+\left[\frac{[x]}{1+x^2}\right]$, giving $4-2=2$.

    Since $[x+1]=[x]+1$, the numerator is $x^2+[x]+1$, so

    $\frac{x^2+[x+1]}{1+x^2}=\frac{(1+x^2)+[x]}{1+x^2}=1+\frac{[x]}{1+x^2}.$

    Because $1$ is an integer,

    $\left[\frac{x^2+[x+1]}{1+x^2}\right]=1+\left[\frac{[x]}{1+x^2}\right].$

    Evaluate the bracketed term on each unit interval of $[-2,2]$:

    • $x\in[-2,-1)$: $[x]=-2$, so $\frac{-2}{1+x^2}\in(-1,-0.4]\Rightarrow$ value $-1$.
    • $x\in[-1,0)$: $[x]=-1$, so $\frac{-1}{1+x^2}\in(-1,-0.5]\Rightarrow$ value $-1$.
    • $x\in[0,1)$: $[x]=0\Rightarrow$ value $0$.
    • $x\in[1,2)$: $[x]=1$, so $\frac{1}{1+x^2}\in(0.2,0.5]\Rightarrow$ value $0$.

    Therefore

    $\int_{-2}^{2}\!\left[\frac{x^2+[x+1]}{1+x^2}\right]dx=\int_{-2}^{2}1\,dx+\int_{-2}^{2}\left[\frac{[x]}{1+x^2}\right]dx=4+(-1-1+0+0)=2.$

    ✓Final answer

    $\displaystyle\int_{-2}^{2}\left[\frac{x^2+[x+1]}{1+x^2}\right]dx=2$ — option (C).

12. 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

    $[\,|\sin x|+|\cos x|\,]=1\quad\text{for almost all }x.$

    Hence

    $\int_{0}^{2\pi}[\,|\sin x|+|\cos x|\,]\,dx=\int_{0}^{2\pi}1\,dx=2\pi.$

    ✓Final answer

    The integral equals $2\pi$ — option (D).