Q.Find the area of the region bounded by the line $y = 3x + 2$, the x-axis and the ordinates $x = -1$ and $x = 1$

Showing the 12 most recent of 22 on this concept.

1. GUJCET 2023Set 091 markMCQ
   Q.Find the area of the region bounded by the line $y = 3 - x$, the X-axis and the ordinates $x = 2$ and $x = 5$. (A) $3$ (B) $\frac{1}{2}$ (C) $\frac{5}{2}$ (D) $\frac{3}{2}$

   ›Reveal solutionSolution

   The line changes sign at x=3, so split the interval and add the absolute areas.

   Concept. Area between a curve and the X-axis is $\int|y|\,dx$; the line $y=3-x$ meets the axis at $x=3$.

   Solution.

   $\int_2^3(3-x)\,dx=\left[3x-\tfrac{x^2}{2}\right]_2^3=4.5-4=0.5,$

   $\int_3^5(x-3)\,dx=\left[\tfrac{x^2}{2}-3x\right]_3^5=(-2.5)-(-4.5)=2.$

   Total area $=0.5+2=\tfrac52$.

   ✓Final answer

   (C) $\frac{5}{2}$

   ANSWER: (C)

2. GUJCET 2025Set 031 markMCQ
   Q.Area of the region bounded by the curve $y = x^3$, $x$-axis and the ordinates $x = -1$ and $x = 2$ is (A) $\frac{17}{4}$ (B) $\frac{19}{4}$ (C) $\frac{15}{4}$ (D) $\frac{9}{4}$

   ›Reveal solutionSolution

   $y=x^3<0$ on $(-1,0)$ and $>0$ on $(0,2)$, so add the magnitudes of the two parts.

   $\int_{-1}^{0}(-x^3)\,dx=\left[-\frac{x^4}{4}\right]_{-1}^{0}=\frac14,\qquad \int_{0}^{2}x^3\,dx=\left[\frac{x^4}{4}\right]_{0}^{2}=4.$

   Total area $=\tfrac14+4=\tfrac{17}{4}$.

   ✓Final answer

   (A) $\dfrac{17}{4}$

   ANSWER: (A)

3. GUJCET 2019Set 171 markMCQ
   Q.The area bounded by curve $y = \sin 2x$ ($x = 0$ to $x = \pi$) and X-axis is $\underline{\hspace{2cm}}$. (A) $2$ (B) $1$ (C) $4$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Take the magnitude of each half-loop area; total is $2$.

   Concept. $\sin 2x \ge 0$ on $[0,\pi/2]$ and $\le 0$ on $[\pi/2,\pi]$; area uses absolute values.

   Steps.

   • $\displaystyle\int_0^{\pi/2}\sin2x\,dx = \Big[-\tfrac{\cos2x}{2}\Big]_0^{\pi/2} = \tfrac12+\tfrac12 = 1.$
   • The second loop has the same magnitude $=1$.
   • Total area $= 1 + 1 = 2.$

   ✓Final answer

   (A) $2$

   ANSWER: (A)

4. GUJCET 2025Set 031 markMCQ
   Q.The area bounded by the curve $y = \sin x$ between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ is _____. (A) 4 (B) 2 (C) 3 (D) 1

   ›Reveal solutionSolution

   [!TLDR]

   Series current is 0.625 A, and the P-Q drop across the 32 $\Omega$ resistor is 20 V.

   Concept

   In a series potential divider the same current flows through both resistors, and the voltage across each is $V = IR$ (Ohm's law).

   Solution

   The 64 $\Omega$ and 32 $\Omega$ resistors are in series between 60 V and 0 V, so the total resistance is

   $R = 64 + 32 = 96\ \Omega,$

   and the current is

   $I = \frac{60}{96} = 0.625\ \text{A}.$

   P and Q are the two ends of the 32 $\Omega$ resistor (Q at 0 V), so

   $V_{PQ} = I \times 32 = 0.625 \times 32 = 20\ \text{V}.$

   [!ANSWER]

   (B) 20 V

5. GUJCET 2026Set x1 markMCQ
   Q.The area bounded by the curve $y = x|x|$, X-axis and the ordinates $x = -1$ and $x = 1$ is ______ (A) 0 (B) $\frac{2}{3}$ (C) $\frac{1}{3}$ (D) $\frac{4}{3}$

   ›Reveal solutionSolution

   By symmetry the area $=2\int_0^1 x^2\,dx=\frac23$.

   The curve is $y=x|x|$: for $x\ge0$, $y=x^2$ (above the axis); for $x<0$, $y=-x^2$ (below the axis). The bounded region between $x=-1$ and $x=1$ is symmetric, so the total (unsigned) area is twice the area from $0$ to $1$:

   $A=2\int_0^1 x^2\,dx=2\left[\frac{x^3}{3}\right]_0^1=2\cdot\frac{1}{3}=\frac{2}{3}$

   ✓Final answer

   Option (B) $\frac{2}{3}$

   ANSWER: (B)

6. GUJCET 2024Set 131 markMCQ
   Q.The area bounded by the curve $y = \cos x$ between $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ is __________. (A) $2$ (B) $1$ (C) $0$ (D) $4$

   ›Reveal solutionSolution

   $\cos x\ge0$ on $[-\pi/2,\pi/2]$, so area $=\int_{-\pi/2}^{\pi/2}\cos x\,dx$.

   Concept. Since $\cos x\ge0$ throughout this interval, the area equals the plain integral.

   $\int_{-\pi/2}^{\pi/2}\cos x\,dx=[\sin x]_{-\pi/2}^{\pi/2}=1-(-1)=2.$

   ✓Final answer

   Option (A) $2$

   ANSWER: (A)

7. GUJCET 2020Set 071 markMCQ
   Q.The smallest area enclosed by circle $x^2 + y^2 = 4$ and line $x + y = 2$ is ________. (A) $\pi + 2$ (B) $\pi - 2$ (C) $\pi$ (D) $2\pi$

   ›Reveal solutionSolution

   Smaller area = quarter-circle area $-$ triangle area $=\pi-2$.

   Concept: The circle has radius $2$. The line $x+y=2$ passes through $(2,0)$ and $(0,2)$, cutting off a region in the first quadrant.

   • Quarter-circle area (bounded by the two axes and the arc) $=\frac{1}{4}\pi r^2=\frac{1}{4}\pi(4)=\pi.$
   • Triangle formed by the line with the axes, vertices $(0,0),(2,0),(0,2)$, area $=\frac{1}{2}(2)(2)=2.$

   The smaller region between the chord and the arc $=\pi-2.$

   ✓Final answer

   (B) $\pi-2$

   ANSWER: (B)

8. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.The area of the region bounded by the curve $y = x|x|$, the $x$-axis, and the ordinates $x=-1$ and $x=1$ is ____.

   (a) $0$

   (b) $1/3$

   (c) $2/3$

   (d) $4/3$

   ›Reveal solutionSolution

   Split the region at $x=0$ since $y=x|x|$ changes sign, then add both areas.

   For $x\ge0$, $y=x^2$; for $x<0$, $y=-x^2$. Both pieces lie below/above the axis symmetrically, so

   Area $=\displaystyle\int_{-1}^0x^2\,dx+\int_0^1x^2\,dx=\frac13+\frac13=\frac23$.

   ✓Final answer

   The correct option is (c) $2/3$.

9. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.The area of the region bounded by the curve $y=\sin x$ between $x=0$ and $x=\pi$ is ____.

   (a) 1

   (b) 2

   (c) 3

   (d) 4

   ›Reveal solutionSolution

   $\sin x\ge0$ on $[0,\pi]$, so directly integrate.

   Area $=\displaystyle\int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=-\cos\pi+\cos0=1+1=2$.

   ✓Final answer

   The correct option is (b) 2.

10. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
    Q.The area of the region, lying in the first quadrant, bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$, $x = 2$ = ____.

    (a) $\pi$

    (b) $\dfrac{\pi}{3}$

    (c) $\dfrac{\pi}{2}$

    (d) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    This region is exactly the quarter-disc of the circle in the first quadrant.

    The circle $x^2+y^2=4$ has radius $2$. Bounded by $x=0$ and $x=2$ in the first quadrant, this describes the full quarter-circle.

    Quarter-circle area $=\dfrac14\pi r^2=\dfrac14\pi(4)=\pi$.

    ✓Final answer

    The correct option is (a) $\pi$.

11. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
    Q.The area of the region bounded by $y = \sin x$, between $x = -\dfrac{\pi}{2}$ and $x = \dfrac{\pi}{2}$ = ____.

    (a) $0$

    (b) $2$

    (c) $1$

    (d) $3$

    ›Reveal solutionSolution

    $\sin x$ is an odd function, so use symmetry and take absolute value for area (sign changes at $x=0$).

    Since $y=\sin x$ is negative on $(-\pi/2,0)$ and positive on $(0,\pi/2)$, the total (unsigned) area is $2\displaystyle\int_0^{\pi/2}\sin x\,dx=2[-\cos x]_0^{\pi/2}=2(0-(-1))=2$.

    ✓Final answer

    The correct option is (b) $2$.

12. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
    Q.The area of the region bounded by the curve $y = x^3$, the X-axis, and the lines $x = -2$ and $x = 1$ = ____.

    (a) $-9$

    (b) $\dfrac{15}{4}$

    (c) $-\dfrac{15}{4}$

    (d) $\dfrac{17}{4}$

    ›Reveal solutionSolution

    $y=x^3$ is negative for $x<0$ and positive for $x>0$, so split at $x=0$ and add the magnitudes.

    $\displaystyle\int x^3dx=\dfrac{x^4}{4}$.

    $\displaystyle\int_{-2}^{0}x^3dx=\left[\dfrac{x^4}{4}\right]_{-2}^{0}=0-4=-4$, magnitude $4$ (curve is below the axis here).

    $\displaystyle\int_{0}^{1}x^3dx=\dfrac14$.

    Total area $=4+\dfrac14=\dfrac{17}{4}$.

    ✓Final answer

    The correct option is (d) $\dfrac{17}{4}$.