Q.Find the area bounded by the lines $y = 4x + 5$, $y = 5 - x$ and $4y = x + 5$.

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.

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

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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. …

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3. 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 …

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4. 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. …

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5. 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. …

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6. 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) $=\frac{1}{4}\pi r^2=\frac{1}{4}\pi(4)=\pi.$ …

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7. 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 $1$: …

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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

   …

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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.

   …

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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.

    …

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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 $x=0$).

    …

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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).

    …

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