Q.Find the area bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$

Showing the 12 most recent of 20 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The area (in sq. units) of the region bounded by the lines $x=0$, $x=\dfrac{\pi}{2}$ and $f(x)=\sin x$, $g(x)=\cos x$ is (A) $2(\sqrt2-1)$ (B) $2(\sqrt3-1)$ (C) $2(\sqrt2+1)$ (D) $3\sqrt2+1$

   ›Reveal solutionSolution

   The two curves $\sin x$ and $\cos x$ cross at $x=\pi/4$ inside $[0,\pi/2]$, so the enclosed area is the sum of two pieces, each evaluating to $\sqrt2-1$, giving total $2(\sqrt2-1)$.

   Concept and Intuition

   Since $\sin x$ and $\cos x$ swap which one is larger at $x=\pi/4$, the area between them over $[0,\pi/2]$ must be split at that crossing point and the absolute difference integrated on each side.

   Step-by-Step Solution

   1. On $[0,\pi/4]$: $\cos x\ge\sin x$, so area contribution is $\int_0^{\pi/4}(\cos x-\sin x)dx=[\sin x+\cos x]_0^{\pi/4}=\left(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}\right)-(0+1)=\sqrt2-1$. …

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8. 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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9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The area of the region (in sq. units) enclosed by the curve $y=x^3-19x+30$ and the X-axis is (A) $\dfrac{167}{2}$ (B) $\dfrac{517}{2}$ (C) 36 (D) 72

   ›Reveal solutionSolution

   The cubic has roots $-5,2,3$; summing the (unsigned) areas of the two lobes between consecutive roots gives $517/2$.

   Concept and Intuition

   The total area enclosed between a cubic and the x-axis over an interval with sign changes is the sum of the absolute areas of each lobe — you cannot simply integrate straight from the leftmost to rightmost root, because the regions above and below the axis would partially cancel.

   Step-by-Step Solution

   1. Find the roots of $x^3-19x+30=0$. Testing $x=2$: $8-38+30=0$ ✓. Dividing out $(x-2)$: $x^3-19x+30=(x-2)(x^2+2x-15)=(x-2)(x+5)(x-3)$.
   2. Roots in order: $x=-5,\,2,\,3$.
   3. Determine sign of $y$ on each interval: at $x=0$ (in $(-5,2)$), $y=30>0$; at $x=2.5$ (in $(2,3)$), $y=15.625-47.5+30=-1.875<0$.
   4. Antiderivative: $F(x)=\dfrac{x^4}{4}-\dfrac{19x^2}{2}+30x$.
   5. $F(-5)=\dfrac{625}{4}-\dfrac{19\cdot25}{2}-150=156.25-237.5-150=-231.25$.
   6. $F(2)=4-38+60=26$.
   7. $F(3)=20.25-85.5+90=24.75$. …

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10. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The area (in sq. units) of the smaller region lying above the X-axis and bounded between the circle $x^2+y^2=2ax$ and the parabola $y^2=ax$ is (A) $2a^2\left(\dfrac{\pi}{4}-\dfrac{2}{3}\right)$ (B) $a^2\left(\dfrac{\pi}{4}-\dfrac{2}{3}\right)$ (C) $a^2\left(\dfrac{\pi}{4}+\dfrac{2}{3}\right)$ (D) $a^2\left(\dfrac{\pi^2}{4}-\dfrac{1}{3}\right)$

    ›Reveal solutionSolution

    The circle and parabola meet at $x=0$ and $x=a$; integrating the gap between the (higher) circle and the (lower) parabola over $[0,a]$ gives the smaller enclosed area $a^2\left(\dfrac\pi4-\dfrac23\right)$.

    Concept and Intuition

    The circle $x^2+y^2=2ax$, i.e. $(x-a)^2+y^2=a^2$, is centered at $(a,0)$ with radius $a$, so it passes through the origin and through $(2a,0)$. The parabola $y^2=ax$ opens rightward from the origin. Near $x=0$ the circle bulges upward faster than the parabola (its upper-half slope near the origin is steeper), and they cross again at $x=a$ — the "smaller" lens-shaped region above the x-axis is exactly the strip between the two curves over $x\in[0,a]$.

    Step-by-Step Solution

    1. Find intersections: substitute $y^2=ax$ into $x^2+y^2=2ax$: $x^2+ax=2ax \Rightarrow x^2-ax=0 \Rightarrow x(x-a)=0 \Rightarrow x=0$ or $x=a$.
    2. At $x=a$: $y^2=a\cdot a=a^2 \Rightarrow y=\pm a$; take $y=a$ for the region above the x-axis. So the curves cross at $(0,0)$ and $(a,a)$.
    3. For $0<x<a$, compare the upper branches: circle gives $y=\sqrt{2ax-x^2}$, parabola gives $y=\sqrt{ax}$. Near $x=0^+$, $2ax-x^2\approx2ax > ax$, so the circle lies above the parabola throughout $(0,a)$ — this strip is the smaller enclosed region.
    4. Area $=\displaystyle\int_0^a\left[\sqrt{2ax-x^2}-\sqrt{ax}\right]dx$.
    5. For $\int_0^a\sqrt{ax}\,dx = \sqrt a\cdot\dfrac23x^{3/2}\Big|_0^a = \sqrt a\cdot\dfrac23a^{3/2}=\dfrac23a^2$. …

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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 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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