Q.The area of the region bounded by the curve $x = 2y + 3$ and the lines $y = 1$ and $y = -1$ is
(A) $4$ sq units
(B) $\frac{3}{2}$ sq units
(C) $6$ sq units
(D) $8$ sq units

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The area (in sq. units) bounded by the curves $x = y^2$ and $x = 3 - 2y^2$ is (A) $8$ (B) $\dfrac{8}{3}$ (C) $4$ (D) $6$

   ›Reveal solutionSolution

   Integrate horizontally (with respect to $y$) since both curves are given as $x=$ function of $y$. Answer: 4 square units.

   Concept and Intuition

   Both curves open sideways (they're expressed as $x$ in terms of $y$), so it's natural to integrate along $y$, treating the region as bounded on the right by $x=3-2y^2$ and on the left by $x=y^2$, between their points of intersection.

   Step-by-Step Solution

   1. Find intersection points: set $y^2 = 3-2y^2 \Rightarrow 3y^2=3 \Rightarrow y^2=1 \Rightarrow y=\pm1$.
   2. For $-1\le y\le 1$, check which curve is to the right: at $y=0$, $x=y^2=0$ vs $x=3-2y^2=3$, so $3-2y^2 \ge y^2$ throughout this range.
   3. Area $= \displaystyle\int_{-1}^{1}\left[(3-2y^2)-y^2\right]dy = \int_{-1}^1 (3-3y^2)\,dy$. …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The area enclosed between the curves $y^2 = x$ and $y = |x|$ is (A) $\frac{1}{6}$ (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   Because $y^2=x$ only exists for $x\ge0$, $y=|x|$ effectively reduces to the single ray $y=x$ there; the enclosed area between the parabola and this line, from $(0,0)$ to $(1,1)$, is $\tfrac16$.

   Concept and Intuition

   $y=|x|$ is a V-shaped pair of rays, but the parabola $y^2=x$ only exists where $x\ge0$ (since $y^2$ can't be negative). So on the left half ($x<0$) there is no parabola to intersect the left ray of $|x|$ — the only relevant intersection is between the parabola and the right ray $y=x$ ($x\ge0$). This reduces the problem to the classic area between $y^2=x$ and $y=x$.

   Step-by-Step Solution

   1. Find intersection points: set $y=x$ into $y^2=x$: $x^2=x \Rightarrow x=0$ or $x=1$. Points: $(0,0)$ and $(1,1)$.
   2. On $[0,1]$, compare $\sqrt x$ (upper parabola branch) with $x$ (the line): at $x=0.25$, $\sqrt x=0.5 > 0.25=x$, so the parabola is above the line throughout $(0,1)$. …

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3. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.The area bounded by $y-1=-|x|$ and $y+1=|x|$ is (A) $\frac{1}{2}$ (B) $1$ (C) $2$ (D) $0$

   ›Reveal solutionSolution

   The two absolute-value "V" graphs cross at $(\pm1,0)$ and enclose a rhombus-shaped region (a square rotated 45°) of area $2$.

   Concept and Intuition

   $y-1=-|x|\Rightarrow y=1-|x|$ is an upside-down V peaking at $(0,1)$ with slopes $\mp1$. $y+1=|x|\Rightarrow y=|x|-1$ is a right-side-up V bottoming at $(0,-1)$ with slopes $\pm1$. Since both have unit slopes, the enclosed figure is actually a square with diagonals along the axes (vertices at $(0,1),(1,0),(0,-1),(-1,0)$), i.e. a rhombus/square of diagonal length 2 each way.

   Step-by-Step Solution

   1. Find intersections: $1-|x|=|x|-1\Rightarrow 2=2|x|\Rightarrow|x|=1\Rightarrow x=\pm1$, giving points $(1,0)$ and $(-1,0)$.
   2. On $(-1,1)$, the top curve is $y=1-|x|$ (value $1$ at $x=0$) and the bottom curve is $y=|x|-1$ (value $-1$ at $x=0$).
   3. Area $=\int_{-1}^{1}\big[(1-|x|)-(|x|-1)\big]dx=\int_{-1}^1(2-2|x|)dx$. …

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4. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The area bounded by the curve $x = \log(|y|)$, the lines $x = -1$ and $x = 0$ is (A) $1 - e^{-1}$ (B) $1 - e$ (C) $2(1-e)$ (D) $2(1 - e^{-1})$

   ›Reveal solutionSolution

   The curve $x=\log|y|$ has two branches $y=\pm e^x$; the area enclosed between $x=-1$ and $x=0$ is $2(1-e^{-1})$.

   Concept and Intuition

   $x=\log|y| \Leftrightarrow |y|=e^x \Leftrightarrow y=\pm e^x$, giving a symmetric pair of curves about the x-axis.

   Step-by-Step Solution

   1. Upper branch: $y=e^x$. Lower branch: $y=-e^x$.
   2. Between $x=-1$ and $x=0$, the vertical gap between the branches is $e^x-(-e^x)=2e^x$. …

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5. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.The area (in sq. units) of the region bounded by the curves $y=4|\cos x|$ and $y=-|\cos x|$ from $x=-\pi/2$ to $\pi/2$ is (A) 6 (B) 8 (C) 12 (D) 10

   ›Reveal solutionSolution

   Since $\cos x\ge0$ throughout $[-\pi/2,\pi/2]$, both curves are simple cosine multiples, and the enclosed area is $5\times\int\cos x\,dx=10$.

   Concept and Intuition

   The area between $y=f(x)$ (top curve) and $y=g(x)$ (bottom curve) over an interval is $\int[f(x)-g(x)]\,dx$. Here the vertical gap between the two curves is constant multiple of $|\cos x|$, so no case-splitting on sign is even needed within this interval.

   Step-by-Step Solution

   1. On $x\in[-\pi/2,\pi/2]$, $\cos x\ge0\Rightarrow|\cos x|=\cos x$.
   2. Top curve: $y=4\cos x$; bottom curve: $y=-\cos x$.
   3. Vertical gap: $4\cos x-(-\cos x)=5\cos x\ge0$ throughout, so this is directly the integrand. …

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6. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.The area (in sq. units) bounded by the curves $y=\dfrac{8}{x}$, $y=2x$ and $x=4$ is (A) $12-8\log 2$ (B) $12+8\log 2$ (C) $12-8\log 4$ (D) $12+8\log 4$

   ›Reveal solutionSolution

   The region enclosed by $y=8/x$, $y=2x$ and $x=4$ runs from their intersection at $x=2$ to $x=4$, with $y=2x$ as the upper boundary; the area works out to $12-8\log 2$.

   Concept and Intuition

   Finding where the two curves meet tells us where the 'wedge'-shaped bounded region starts; the vertical line $x=4$ closes it off on the right. Between the intersection and the line, we need to know which curve is higher to set up $\int(\text{upper}-\text{lower})\,dx$ correctly.

   Step-by-Step Solution

   1. Find the intersection of $y=8/x$ and $y=2x$: $8/x=2x\Rightarrow x^2=4\Rightarrow x=2$ (taking the positive root, matching the given curves), giving $y=4$.
   2. Determine which curve is on top between $x=2$ and $x=4$: at $x=3$, $y=2x=6$ while $y=8/x\approx2.67$. So $2x>8/x$ here — the line is above the hyperbola on this interval.
   3. Set up the area integral from the intersection point to the line $x=4$: $\text{Area}=\displaystyle\int_2^4\left(2x-\dfrac{8}{x}\right)dx$.
   4. Antiderivative: $\displaystyle\int\left(2x-\dfrac8x\right)dx=x^2-8\log x$. …

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7. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The area of the region bounded by the curve $y = x^2 + x$, the lines $y = x$, $x = 1$ and $y = 2$ is (A) $\dfrac{12}{5}$ (B) $\dfrac{7}{2}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The four boundary curves pin down a single closed loop from $x=0$ to $x=1$ between the parabola and the line $y=x$; its area is $\dfrac13$.

   Concept and Intuition

   When several curves are said to "bound a region," first locate every pairwise intersection — the closed loop's vertices are exactly these intersection points, and its area is found by integrating (upper curve minus lower curve) over the right interval.

   Step-by-Step Solution

   1. Intersection of $y=x^2+x$ and $y=x$: $x^2+x=x\Rightarrow x^2=0\Rightarrow x=0$. They only touch at $(0,0)$, and since $x^2+x-x=x^2\ge0$, the parabola is above the line for all other $x$.
   2. Intersection of $y=x$ and $x=1$: point $(1,1)$.
   3. Intersection of $y=x^2+x$ and $x=1$: point $(1,2)$.
   4. Intersection of $y=x^2+x$ and $y=2$: $x^2+x-2=0\Rightarrow(x-1)(x+2)=0\Rightarrow x=1$ (the relevant root, giving $(1,2)$ again). …

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8. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.The area bounded by the curve $y=x^2$ and $y-6=-|x|$ is (A) $37/4$ (B) $22/3$ (C) $44/3$ (D) $38/3$

   ›Reveal solutionSolution

   The parabola and the "tent" line $y=6-|x|$ meet at $x=\pm2$; using the even symmetry of both curves, the enclosed area works out to $44/3$.

   Concept and Intuition

   Both $y=x^2$ and $y=6-|x|$ are even functions, so the region they enclose is symmetric about the Y-axis. It's efficient to compute the area on $[0,2]$ (where $|x|=x$) and double it, rather than integrating across the absolute value directly.

   Step-by-Step Solution

   1. For $x\ge0$: solve $x^2=6-x\Rightarrow x^2+x-6=0\Rightarrow(x-2)(x+3)=0\Rightarrow x=2$ (rejecting $x=-3$). By symmetry the other intersection is $x=-2$.
   2. At $x=2$: $y=4$ for both curves — consistent.
   3. On $[0,2]$, the upper curve is $6-x$ (line) and the lower is $x^2$ (check at $x=1$: line gives $5$, parabola gives $1$ — line is on top). …

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9. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The area of the region enclosed between the curve $y=\log_e(x+e)$ and the coordinate axes is (A) $4$ (B) $3$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   The region bounded by $y=\log(x+e)$ and the two coordinate axes is a simple region between $x=1-e$ (where the curve meets the $x$-axis) and $x=0$ (where it meets the $y$-axis); the area works out to exactly $1$.

   Concept and Intuition

   To find the area enclosed between a curve and the coordinate axes, first locate where the curve crosses each axis — those crossing points bound the finite region. Here $y=\log(x+e)$ is a shifted, increasing logarithm; it crosses the $y$-axis at $x=0$ (giving $y=\log e=1$) and the $x$-axis where $\log(x+e)=0$, i.e. $x+e=1$, so $x=1-e$. Since the curve is positive throughout $(1-e,0)$, the enclosed area is simply the definite integral of $y$ over that interval.

   Step-by-Step Solution

   1. Find the $y$-axis intercept: at $x=0$, $y=\log(0+e)=\log e=1$.
   2. Find the $x$-axis intercept: set $\log(x+e)=0\Rightarrow x+e=1\Rightarrow x=1-e$ (note $1-e\approx-1.718$).
   3. For $x\in(1-e,0)$, the curve is increasing from $0$ up to $1$, staying non-negative, so the enclosed area is $\text{Area}=\int_{1-e}^{0}\log(x+e)\,dx.$ …

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10. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.The area of the region (in sq.units) bounded by the curves $x^2 + y^2 = 16$ and $y^2 = 6x$ is (A) $4\pi + 4\sqrt{3}$ (B) $\dfrac{2}{3}\left(4\pi + \sqrt{3}\right)$ (C) $\dfrac{4}{3}\left(4\pi + \sqrt{3}\right)$ (D) $\dfrac{4\pi+\sqrt{3}}{3}$

    ›Reveal solutionSolution

    The region common to the circle $x^2+y^2=16$ and parabola $y^2=6x$ is bounded by the parabola near the origin and the circle further out; integrating each piece and doubling for symmetry gives $\tfrac43(4\pi+\sqrt3)$.

    Concept and Intuition

    The parabola $y^2=6x$ opens rightward from the origin, and the circle has radius 4. Near the vertex, the parabola is the "narrower" curve (smaller $|y|$ for given $x$), so it bounds the common region; farther out, the circle becomes narrower and takes over as the boundary. The crossover is exactly at their intersection point.

    Step-by-Step Solution

    1. Intersection: substitute $y^2=6x$ into $x^2+y^2=16$: $x^2+6x-16=0\Rightarrow x=2$ or $x=-8$ (rejected, since $y^2=6x\ge0$ needs $x\ge0$). At $x=2$: $y^2=12\Rightarrow y=\pm2\sqrt3$.
    2. For $x\in[0,2]$: parabola gives smaller $|y|$ than the circle (check at $x=1$: parabola $|y|=\sqrt6\approx2.45$, circle $|y|=\sqrt{15}\approx3.87$) — so the parabola bounds the region here.
    3. For $x\in[2,4]$: circle gives smaller $|y|$ (check at $x=3$: parabola $|y|=\sqrt{18}\approx4.24$, circle $|y|=\sqrt7\approx2.65$) — circle bounds here.
    4. Area (using symmetry about the $x$-axis, factor 2):

    $A=2\left[\int_0^2\sqrt{6x}\,dx+\int_2^4\sqrt{16-x^2}\,dx\right].$

    5. $\displaystyle\int_0^2\sqrt{6x}\,dx=\sqrt6\cdot\dfrac23x^{3/2}\Big|_0^2=\sqrt6\cdot\dfrac23\cdot2\sqrt2=\dfrac{8\sqrt3}3$.
    6. Using $\int\sqrt{16-x^2}dx=\dfrac x2\sqrt{16-x^2}+8\sin^{-1}\dfrac x4$: …

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11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.The area of the region under the curve $y = |\sin x - \cos x|$, $0 \le x \le \dfrac{\pi}{2}$ and above x-axis, is (in square units) (A) $2\sqrt{2}$ (B) $2\sqrt{2} - 1$ (C) $2(\sqrt{2}-1)$ (D) $2(\sqrt{2}+1)$

    ›Reveal solutionSolution

    Split the region at $x=\pi/4$ where $\sin x=\cos x$, integrate each branch of the absolute value separately, and add.

    Concept and Intuition

    $|\sin x-\cos x|$ is $\cos x-\sin x$ for $x<\pi/4$ (where cosine dominates) and $\sin x-\cos x$ for $x>\pi/4$ (where sine dominates) — the area under an absolute-value curve must be computed piecewise across the sign change.

    Step-by-Step Solution

    1. $\sin x=\cos x$ at $x=\pi/4$ within $[0,\pi/2]$; for $x<\pi/4$, $\cos x>\sin x$, so $|\sin x-\cos x|=\cos x-\sin x$; for $x>\pi/4$, it's $\sin x-\cos x$.
    2. $\displaystyle\int_0^{\pi/4}(\cos x-\sin x)dx=\Big[\sin x+\cos x\Big]_0^{\pi/4}=\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right)-(0+1)=\sqrt2-1$. …

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12. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
    Q.The area of the region bounded by the curve $xy = -a$ $(a > 1)$ and the lines $x = -a$ and $y = a$ is (A) $a(a - 1 - \log a)$ (B) $a(a + 1 + \log a)$ (C) $a^2 - a + \log a$ (D) $a^2 - a - \log a$

    ›Reveal solutionSolution

    The area enclosed by the rectangular hyperbola $xy=-a$ and the lines $x=-a$, $y=a$ works out, after a direct integration, to $a(a-1-\log a)$.

    Concept and Intuition

    $xy=-a$ ($a>0$) is a hyperbola lying in the second and fourth quadrants (since the product of coordinates must be negative). We only need the branch in the second quadrant here ($x<0,y>0$, i.e. $y=-a/x$). The two given lines pin down a finite region between the curve and the corner point where the lines would meet.

    Step-by-Step Solution

    1. Rewrite the curve as $y=-\dfrac{a}{x}$ (valid for $x<0$ here, giving $y>0$).
    2. Find where the curve meets $x=-a$: $y=-a/(-a)=1$, point $(-a,1)$.
    3. Find where the curve meets $y=a$: $a=-a/x\Rightarrow x=-1$, point $(-1,a)$.
    4. For $x\in[-a,-1]$, the curve $y=-a/x$ lies below the line $y=a$ (check at $x=-1$: curve value $=a$, equal; at $x=-a$: curve value$=1<a$ since $a>1$). So the vertical strip between the curve and the top line $y=a$, from $x=-a$ to $x=-1$, is exactly the bounded region.
    5. Area $=\displaystyle\int_{-a}^{-1}\Big[a-\big(-\tfrac{a}{x}\big)\Big]dx=\int_{-a}^{-1}\Big(a+\tfrac{a}{x}\Big)dx$. …

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