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

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1. 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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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The Area (in sq. units) of the region bounded by $x = 0, x = \dfrac{\pi}{2}$, X-axis, $y = \cos x$ and $y = \tan x$ is (A) $\dfrac{\sqrt{5} - 1}{2} + \dfrac{1}{2}\log\left(\dfrac{\sqrt{5} - 1}{2}\right)$ (B) $\dfrac{3 - \sqrt{5}}{2} + \log\left(\dfrac{\sqrt{5} - 1}{2}\right)$ (C) $\dfrac{\sqrt{5} - 1}{2} - \log\left(\dfrac{\sqrt{5} - 1}{2}\right)$ (D) $\dfrac{3 - \sqrt{5}}{2} + \dfrac{1}{2}\log\left(\dfrac{\sqrt{5} + 1}{2}\right)$

   ›Reveal solutionSolution

   Since $\tan x\to\infty$ at $\pi/2$, the finite region is bounded above by the lower of $\cos x$ and $\tan x$; splitting the integral at their intersection point gives $\dfrac{3-\sqrt5}2+\dfrac12\log\dfrac{\sqrt5+1}2$.

   Concept and Intuition

   $\cos x$ and $\tan x$ cross exactly once in $(0,\pi/2)$. Since $\tan x$ diverges as $x\to\pi/2^-$, a region bounded by the upper envelope of the two curves would have infinite area; the sensible, finite area bounded by the $x$-axis and both curves on $[0,\pi/2]$ is the area under whichever curve is lower at each $x$ — i.e. under $\tan x$ before the crossing and under $\cos x$ after it.

   Step-by-Step Solution

   1. Find the crossing point: $\cos x=\tan x=\dfrac{\sin x}{\cos x}\Rightarrow\cos^2x=\sin x\Rightarrow 1-\sin^2x=\sin x\Rightarrow\sin^2x+\sin x-1=0$.

   $\sin x_0=\frac{-1+\sqrt5}{2}=s\quad(\text{taking the root in }[0,1]).$

   Note also $\cos^2x_0=\sin x_0=s$ (directly from the defining equation), so $\cos x_0=\sqrt s$.

   2. Near $x=0$: $\cos0=1>\tan0=0$, so $\cos x$ is the upper curve, $\tan x$ the lower, for $x\in(0,x_0)$.

   Near $x=\pi/2$: $\tan x\to\infty>\cos x\to0$, so $\tan x$ is upper, $\cos x$ lower, for $x\in(x_0,\pi/2)$.

   3. The finite bounded area is therefore

   $A=\int_0^{x_0}\tan x\,dx+\int_{x_0}^{\pi/2}\cos x\,dx.$

   4. First piece: $\displaystyle\int_0^{x_0}\tan x\,dx=\Big[-\log(\cos x)\Big]_0^{x_0}=-\log(\cos x_0)=-\log\sqrt s=-\tfrac12\log s$.
   5. Second piece: $\displaystyle\int_{x_0}^{\pi/2}\cos x\,dx=\Big[\sin x\Big]_{x_0}^{\pi/2}=1-\sin x_0=1-s$. …

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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 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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5. 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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6. 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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7. 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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8. 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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9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.The area bounded by the curves $y-1=\cos x$, $y=\sin x$ and the X-axis between $x=0$ and $x=\pi$ is (A) $2+\dfrac{\pi}{2}$ (B) $-\dfrac{\pi}{2}$ (C) $2-\dfrac{\pi}{2}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   The area of the region bounded by $y=1+\cos x$, $y=\sin x$ and the X-axis on $[0,\pi]$ is $\dfrac{\pi}{2}$.

   Concept and Intuition

   Three curves fence off one closed region above the X-axis. Its floor is $y=0$; its roof is whichever curve is lower at each $x$ (the lower envelope), because that is what actually caps the region touching the axis. So the area is the integral of $\min(1+\cos x,\ \sin x)$.

   Step-by-Step Solution

   1. Find the crossing: $1+\cos x=\sin x\Rightarrow \sin x-\cos x=1\Rightarrow \sqrt2\sin\!\left(x-\tfrac{\pi}{4}\right)=1$, giving $x=\tfrac{\pi}{2}$ and $x=\pi$.
   2. On $[0,\tfrac{\pi}{2}]$: at $x=0$, $\sin x=0<1+\cos x=2$, so $\sin x$ is the lower (roof) curve.
   3. On $[\tfrac{\pi}{2},\pi]$: at $x=\tfrac{3\pi}{4}$, $1+\cos x\approx0.29<\sin x\approx0.71$, so $1+\cos x$ is the roof.
   4. $\displaystyle\int_0^{\pi/2}\sin x\,dx=[-\cos x]_0^{\pi/2}=1$.
   5. $\displaystyle\int_{\pi/2}^{\pi}(1+\cos x)\,dx=[x+\sin x]_{\pi/2}^{\pi}=\pi-\left(\tfrac{\pi}{2}+1\right)=\tfrac{\pi}{2}-1.$ …

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10. 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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11. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The area of the region lying between the curves $y = \sqrt{4-x^2}$, $y^2 = 3x$ and the Y-axis is (A) $\frac{\pi}{3} - \frac{1}{2\sqrt{3}}$ (B) $\frac{\pi}{6} + \frac{1}{2\sqrt{3}}$ (C) $\frac{\pi}{3} + \frac{1}{2\sqrt{3}}$ (D) $\frac{\pi}{6} - \frac{1}{2\sqrt{3}}$

    ›Reveal solutionSolution

    Integrating with respect to $y$ in two pieces — the parabola from $y=0$ to $\sqrt3$ and the circle from $y=\sqrt3$ to $2$ — gives the enclosed area as $\frac{\pi}{3}-\frac{1}{2\sqrt3}$.

    Concept and Intuition

    When a region's boundary is naturally expressed as $x=x(y)$ for both curves (here $x=\sqrt{4-y^2}$ for the circle and $x=y^2/3$ for the parabola), integrating along $y$ with the y-axis as the common left boundary is far cleaner than integrating along $x$.

    Step-by-Step Solution

    1. Find the intersection of $y=\sqrt{4-x^2}$ (so $y^2=4-x^2$) and $y^2=3x$: $3x=4-x^2\Rightarrow x^2+3x-4=0\Rightarrow(x+4)(x-1)=0$. Since $x\ge0$ on the parabola, $x=1$, giving $y=\sqrt3$.
    2. The circle meets the y-axis at $(0,2)$; the parabola meets it at the origin $(0,0)$.
    3. The enclosed region has: the y-axis as its left edge from $(0,0)$ to $(0,2)$; the parabola $x=y^2/3$ as its right edge for $y\in[0,\sqrt3]$; and the circle $x=\sqrt{4-y^2}$ as its right edge for $y\in[\sqrt3,2]$.
    4. Area $=\displaystyle\int_0^{\sqrt3}\frac{y^2}{3}\,dy+\int_{\sqrt3}^{2}\sqrt{4-y^2}\,dy$.
    5. First integral: $\displaystyle\int_0^{\sqrt3}\frac{y^2}{3}dy=\frac{1}{9}\left[y^3\right]_0^{\sqrt3}=\frac{3\sqrt3}{9}=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}$. …

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12. 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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