Q.Find the area of the region bounded by the triangle whose vertices are $(-1, 1)$, $(0, 5)$ and $(3, 2)$, using integration.

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-M1 markMCQ
   Q.The area of the region bounded by the line $y=x+2$ and the curve $x=-y^2$ is (A) 13.5 sq units (B) $\frac{7}{6}$ sq units (C) 4.5 sq units (D) 2.5 sq units

   ›Reveal solutionSolution

   The region is bounded by a line and a left‑opening parabola; we integrate with respect to $y$ to avoid splitting the region, and the area is $4.5$ square units.

   Concept & Intuition

   When a region is bounded by a curve that is not a function of $x$ (here $x = -y^2$ gives two $y$-values for most $x$), it is often easier to integrate with respect to $y$. The line $y = x+2$ can be rewritten as $x = y-2$. The area between two curves expressed as functions of $y$ is $\int_{y_{\text{low}}}^{y_{\text{high}}} \bigl(x_{\text{right}}(y) - x_{\text{left}}(y)\bigr)\,dy$. Here the parabola $x = -y^2$ is the left boundary and the line $x = y-2$ is the right boundary. We find the intersection points to set the limits, then integrate.

   Step‑by‑Step Solution

   1. Find the intersection points of $y = x+2$ and $x = -y^2$. Substitute $x = -y^2$ into $y = x+2$:

   $y = -y^2 + 2 \quad\Rightarrow\quad y^2 + y - 2 = 0.$

   Factor: $(y+2)(y-1)=0$, so $y = -2$ and $y = 1$.

   The corresponding $x$-values: for $y=-2$, $x = -4$; for $y=1$, $x = -1$.

   Intersection points: $(-4,-2)$ and $(-1,1)$.

   2. Decide the variable of integration.

      The parabola $x = -y^2$ opens left; for a fixed $y$, it gives a single $x$. The line $x = y-2$ is also single‑valued in $y$. Over the interval $y \in [-2, 1]$, the line lies to the right of the parabola. So we integrate with respect to $y$.

   3. Set up the area integral.

      Right boundary: $x_{\text{right}} = y-2$.

      Left boundary: $x_{\text{left}} = -y^2$.

      Area = $\displaystyle \int_{y=-2}^{1} \bigl[(y-2) - (-y^2)\bigr]\,dy = \int_{-2}^{1} (y - 2 + y^2)\,dy$.

   4. Evaluate the integral.

   $\int_{-2}^{1} (y^2 + y - 2)\,dy = \left[ \frac{y^3}{3} + \frac{y^2}{2} - 2y \right]_{-2}^{1}.$

   At $y=1$: $\frac{1}{3} + \frac{1}{2} - 2 = \frac{2}{6} + \frac{3}{6} - \frac{12}{6} = -\frac{7}{6}$. …

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2. COMEDK 2025Set 2025-A1 markMCQ
   Q.The area of the region enclosed by the lines $2 x+y=10, y=1, y=5$ and the $y$-axis is (A) 28 sq units (B) 9.5 sq units (C) 14 sq units (D) 37.5 sq units

   ›Reveal solutionSolution

   Integrating $x=\frac{10-y}{2}$ from $y=1$ to $y=5$ gives area $=14$ sq units.

   The line $2x+y=10$ gives $x=\dfrac{10-y}{2}$. The region is bounded on the left by the $y$-axis ($x=0$), on the right by this line, and horizontally between $y=1$ and $y=5$. Integrating with respect to $y$:

   $A=\int_{1}^{5} x\,dy=\int_{1}^{5}\frac{10-y}{2}\,dy=\frac12\left[10y-\frac{y^2}{2}\right]_{1}^{5}.$

   Evaluate: …

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3. COMEDK 2026Set 2026-A1 markMCQ
   Q.Find the area bounded by the curve $y=|2-x|$, the $x$-axis, and the lines $x=0$ and $x=5$ (A) 4.5 sq units (B) 6.5 sq units (C) 12 .5 sq units (D) 8.5 sq units

   ›Reveal solutionSolution

   The area is the sum of two triangular regions formed by the V‑shaped absolute‑value function, giving a total of 6.5 square units.

   We are asked for the area bounded by $y = |2 - x|$, the $x$-axis, and the vertical lines $x = 0$ and $x = 5$.

   The key idea: the absolute value creates a V‑shaped graph with a vertex at $x = 2$. The area under this curve (above the $x$-axis) from $x = 0$ to $x = 5$ is simply the sum of two triangles — one on each side of the vertex. No integration is strictly needed, but we can also integrate piecewise.

   ---

   1. Understand the shape of $y = |2 - x|$

   The expression $|2 - x|$ equals:

   • $2 - x$ when $2 - x \ge 0$, i.e. $x \le 2$
   • $x - 2$ when $2 - x < 0$, i.e. $x > 2$

   So the graph is a straight line of slope $-1$ from $x = 0$ to $x = 2$, then slope $+1$ from $x = 2$ onward. At $x = 2$, $y = 0$. This is a V‑shape with the vertex on the $x$-axis.

   2. Identify the region

   We are bounded by:

   • The curve $y = |2 - x|$
   • The $x$-axis ($y = 0$)
   • The left vertical line $x = 0$
   • The right vertical line $x = 5$

   The curve is always above or on the $x$-axis, so the region is simply the area under the V from $x = 0$ to $x = 5$.

   3. Break into two parts at the vertex $x = 2$

   Left part ($0 \le x \le 2$):

   Here $y = 2 - x$. At $x = 0$, $y = 2$; at $x = 2$, $y = 0$.

   This is a right triangle with base $2$ (from $x=0$ to $x=2$) and height $2$.

   Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$.

   Right part ($2 \le x \le 5$):

   Here $y = x - 2$. At $x = 2$, $y = 0$; at $x = 5$, $y = 3$. …

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4. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { The area of the region (in square units) bounded by the line } y+3=x ; x=1 \text { and } x=5 \text { is }$

   (A) 2 (B) 32 (C) 24 (D) 4

   ›Reveal solutionSolution

   The line $y=x-3$ crosses the x-axis at $x=3$, splitting the region between $x=1$ and $x=5$ into two equal triangular pieces; the total area is 4 square units — option (D).

   Step-by-step solution

   1. Rewrite the line.

      $y+3=x \Rightarrow y=x-3$, which crosses the x-axis at $x=3$: below the axis for $1\le x\le3$, above it for $3\le x\le5$.

   2. Set up the area as the sum of two pieces (taking the region between the line and the x-axis, the standard convention for this type of "area bounded by a line and two verticals" problem):

   $\text{Area} = \int_1^3(3-x)\,dx + \int_3^5(x-3)\,dx$

   3. First integral.

   $\left[3x-\frac{x^2}{2}\right]_1^3 = (9-4.5)-(3-0.5) = 4.5-2.5 = 2$

   4. Second integral. $\left[\frac{x^2}{2}-3x\right]_3^5 = (12.5-15)-(4.5-9) = -2.5-(-4.5) = 2$ …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.Area of the region bounded by the curve $y=\cos x$ between $x=-\frac{\pi}{2}$ and $x=\pi$ is ------------------ (A) 3 sq units (B) 4 sq units (C) 1 sq units (D) 2 sq units

   ›Reveal solutionSolution

   The area is the integral of the absolute value of $\cos x$ from $-\pi/2$ to $\pi$, because area counts all region regardless of sign. The total area is $3$ square units, so the correct option is (A).

   The key idea: When we ask for the "area bounded by the curve," we mean the total geometric area, not the signed area (which would be the net integral). Since $\cos x$ is positive on some intervals and negative on others, we must split the interval at the zeros and take the absolute value of each piece.

   ---

   1. Identify where $\cos x$ changes sign

   Between $x = -\pi/2$ and $x = \pi$, the function $\cos x$ is zero at $x = -\pi/2$, $x = \pi/2$, and $x = 3\pi/2$ (but $3\pi/2$ is outside our range). So the only zero inside $(-\pi/2, \pi)$ is at $x = \pi/2$.

   • On $[-\pi/2, \pi/2]$, $\cos x \ge 0$
   • On $[\pi/2, \pi]$, $\cos x \le 0$

   2. Set up the area as a sum of absolute integrals

   $\text{Area} = \int_{-\pi/2}^{\pi/2} \cos x \, dx \;+\; \int_{\pi/2}^{\pi} (-\cos x) \, dx$

   3. Compute the first integral

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

   4. Compute the second integral

   $$$$

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6. KCET 2023Set A-21 markMCQ
   Q.In the interval $(0, \pi/2)$, area lying between the curves $y = \tan x$ and $y = \cot x$ and the X-axis is (A) $2 \log 2$ sq. units (B) $4 \log 2$ sq. units (C) $\log 2$ sq. units (D) $3 \log 2$ sq. units

   ›Reveal solutionSolution

   The curves cross at $x=\pi/4$; integrate $\tan x$ from $0$ to $\pi/4$ and $\cot x$ from $\pi/4$ to $\pi/2$, giving $\tfrac12\log2+\tfrac12\log2=\log 2$.

   1. Point of intersection. $\tan x = \cot x \Rightarrow \tan^2 x = 1 \Rightarrow \tan x = 1$ (positive on $(0,\pi/2)$), so $x = \dfrac{\pi}{4}$, where both equal $1$.

   2. Set up the region. Between the X-axis and the two curves, the lower of the two curves forms the boundary: for $0<x<\pi/4$, $\tan x < \cot x$; for $\pi/4<x<\pi/2$, $\cot x < \tan x$. So

   $A=\int_{0}^{\pi/4}\tan x\,dx+\int_{\pi/4}^{\pi/2}\cot x\,dx$

   (This also keeps both integrals finite — $\tan x$ blows up at $\pi/2$ and $\cot x$ at $0$, and neither divergence is included.)

   3. Evaluate the first integral. …

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7. COMEDK 2023Set 2023-E1 markMCQ
   Q.The area of the upper half of the circle whose equation is $(x-1)^2+y^2=1$ is given by (A) $\frac{\pi}{4} \text { sq units }$ (B) $\int_0^2 \sqrt{2-x^2} d x \text { sq units }$ (C) $\int_0^2 \sqrt{2x-x^2} d x \text { sq units }$ (D) $\int_0^1 \sqrt{2x-x^2} d x \text { sq units }$

   ›Reveal solutionSolution

   The upper half of the circle $(x-1)^2+y^2=1$ is $y=\sqrt{2x-x^2}$ with $x$ ranging from $0$ to $2$, so its area is $\int_0^2 \sqrt{2x-x^2}\,dx$. The correct option is (C).

   ---

   Concept

   The area bounded by a curve $y=f(x)\ge 0$ and the $x$-axis between $x=a$ and $x=b$ is the definite integral $\int_a^b f(x)\,dx$. For the upper half of a circle we take the positive square root for $y$ and integrate across the $x$-extent of the circle.

   Solution

   1. Identify the circle. $(x-1)^2+y^2=1$ has centre $(1,0)$ and radius $1$. It crosses the $x$-axis where $y=0$, i.e. $(x-1)^2=1 \Rightarrow x=0$ or $x=2$.

   2. Solve for the upper half. $y^2 = 1-(x-1)^2 = 1-(x^2-2x+1) = 2x-x^2$. The upper half takes the positive root: $y=\sqrt{2x-x^2}$.

   3. Set up the integral. As $x$ runs over the full width of the circle, from $0$ to $2$, the area is

      $\text{Area}=\int_0^2 \sqrt{2x-x^2}\,dx.$ …

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8. COMEDK 2025Set 2025-M1 markMCQ
   Q.Area of the region bounded by the curve $y=\sin \left(\frac{x}{2}\right)$ between $-4 \pi$ and 0 is (A) 4 sq units (B) 8 sq units (C) 6 sq units (D) 1 sq units

   ›Reveal solutionSolution

   The area is the integral of the absolute value of $\sin(x/2)$ from $-4\pi$ to $0$. Because the sine function is odd and symmetric, the total area equals $4$ times the area of one positive lobe, giving $8$ square units.

   The key idea is that "area bounded by a curve" means the total geometric area, not the signed area. The function $y = \sin(x/2)$ crosses the x-axis at multiples of $2\pi$, so between $-4\pi$ and $0$ it has two full positive lobes and two full negative lobes. Since area is always positive, we must integrate the absolute value.

   1. Find the zeros of $\sin(x/2)$ in $[-4\pi, 0]$

      $\sin(x/2) = 0$ when $x/2 = n\pi$, i.e., $x = 2n\pi$.

      For $n = 0, -1, -2$, we get $x = 0, -2\pi, -4\pi$. These split the interval into two equal subintervals: $[-4\pi, -2\pi]$ and $[-2\pi, 0]$.

   2. Determine the sign of $\sin(x/2)$ on each subinterval

      • On $(-4\pi, -2\pi)$: $x/2$ is between $-2\pi$ and $-\pi$. Sine is negative there (since sine is negative in the third quadrant).
      • On $(-2\pi, 0)$: $x/2$ is between $-\pi$ and $0$. Sine is positive there (since sine is positive in the fourth quadrant? Actually careful: $\sin(-\theta) = -\sin\theta$, so from $-\pi$ to $0$, sine goes from $0$ to $0$ through negative values? Let's check: $\sin(-\pi/2) = -1$, so it's negative. Wait — let's re-evaluate.)
      Watch out

      A common mistake: thinking $\sin(x/2)$ is positive on $(-2\pi, 0)$. Actually, for $x$ in $(-2\pi, 0)$, $x/2$ is in $(-\pi, 0)$. Sine is negative on $(-\pi, 0)$ because $\sin(-\theta) = -\sin\theta$ and $\sin\theta > 0$ for $\theta$ in $(0, \pi)$. So $\sin(x/2) < 0$ on $(-2\pi, 0)$ as well.

      Let's check a test point: $x = -\pi$, then $x/2 = -\pi/2$, $\sin(-\pi/2) = -1$. So indeed both subintervals give negative values. That means the curve lies entirely below the x-axis from $-4\pi$ to $0$.

   3. Set up the area integral

      Since the function is negative everywhere on $[-4\pi, 0]$, the absolute value is $-\sin(x/2)$. The area is:

   $\text{Area} = \int_{-4\pi}^{0} \left| \sin\left(\frac{x}{2}\right) \right| \, dx = \int_{-4\pi}^{0} -\sin\left(\frac{x}{2}\right) dx$

   4. Compute the integral Let $u = x/2$, so $dx = 2\,du$. When $x = -4\pi$, $u = -2\pi$; when $x = 0$, $u = 0$. Then:

   $\text{Area} = \int_{u=-2\pi}^{0} -\sin(u) \cdot 2\,du = -2 \int_{-2\pi}^{0} \sin u \, du$

   The antiderivative of $\sin u$ is $-\cos u$, so:

   $-2 \left[ -\cos u \right]_{-2\pi}^{0} = -2 \left( -\cos(0) + \cos(-2\pi) \right) = -2 \left( -1 + 1 \right) = -2(0) = 0$

   That can't be right — we got zero? That's because we forgot that the integral of $\sin u$ over a full period is zero, but we are integrating the absolute value. The mistake: we assumed the function is negative everywhere, but actually it alternates sign.

   Tip

   Always check the sign by testing a point in each subinterval. For $x = -3\pi$ (in $(-4\pi, -2\pi)$), $x/2 = -3\pi/2$, $\sin(-3\pi/2) = 1$ (positive!). So the sign flips: on $(-4\pi, -2\pi)$ it's positive, on $(-2\pi, 0)$ it's negative. …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.The area (in sq units) of the minor segment bounded by the circle $x^2+y^2=a^2$ and the line $x=\frac{a}{\sqrt{2}}$ is (A) $\frac{a^2}{4}(\pi-2)$ (B) $\frac{a^2}{4}(\pi+2)$ (C) $\frac{\pi a^2}{4}$ (D) $\frac{a^2}{4}(3 \pi-2)$

   ›Reveal solutionSolution

   The minor segment area is found by subtracting the area of the triangle formed by the chord and the center from the area of the circular sector. The result is $\frac{a^2}{4}(\pi - 2)$, which corresponds to option (A).

   ---

   The key idea is that the area of a segment of a circle can be computed as the difference between the area of a sector and the area of the triangle formed by the two radii and the chord. Here, the chord is vertical, given by $x = \frac{a}{\sqrt{2}}$, and the circle is centered at the origin with radius $a$. The "minor segment" is the smaller part cut off by this chord.

   Why this works: Instead of integrating directly (which is also possible), the sector-triangle method is cleaner because the geometry is symmetric and the angle subtended by the chord at the center is easy to find. The sector area is $\frac{1}{2} a^2 \theta$, and the triangle area is $\frac{1}{2} a^2 \sin\theta$. Their difference gives the segment area.

   ---

   1. Find the angle subtended by the chord at the center. The line $x = \frac{a}{\sqrt{2}}$ meets the circle $x^2 + y^2 = a^2$ at points where

   $\left(\frac{a}{\sqrt{2}}\right)^2 + y^2 = a^2 \quad\Rightarrow\quad \frac{a^2}{2} + y^2 = a^2 \quad\Rightarrow\quad y^2 = \frac{a^2}{2}.$

   So $y = \pm \frac{a}{\sqrt{2}}$. The chord is vertical, and the radii to these intersection points make angles with the positive x-axis:

   $\cos\theta = \frac{x}{a} = \frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta = \frac{\pi}{4}.$

   Thus the central angle between the two radii is $2\theta = \frac{\pi}{2}$ (90°). This is the angle of the sector that contains the segment.

   2. Area of the sector. For a circle of radius $a$, the area of a sector with central angle $\frac{\pi}{2}$ is

   $\text{Sector area} = \frac{1}{2} a^2 \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}.$

   3. Area of the triangle. The triangle formed by the two radii and the chord is isosceles with sides $a, a$ and included angle $\frac{\pi}{2}$. Its area is $$$$

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10. COMEDK 2024Set 2024-E1 markMCQ
    Q.The area bounded by the curve $y=\cos x, x=0$ and $x=\pi$ is (A) 2 sq units (B) 1 sq units (C) 4 sq units (D) 3 sq units

    ›Reveal solutionSolution

    The area bounded by $y = \cos x$, $x = 0$, and $x = \pi$ is found by integrating the absolute value of $\cos x$ because the curve dips below the x‑axis. The total area is $2$ square units, so the correct option is (A).

    The key idea here is that "area bounded by a curve and the x‑axis" means the geometric area — always positive — not the signed area (which would be the net integral). Since $\cos x$ is positive on $[0, \pi/2]$ and negative on $[\pi/2, \pi]$, we must split the integral and take the absolute value of the function.

    Why this works:

    If we simply integrated $\cos x$ from $0$ to $\pi$, the positive part (from $0$ to $\pi/2$) and the negative part (from $\pi/2$ to $\pi$) would cancel, giving zero — which is clearly not the actual area. The geometric area is the sum of the magnitudes of the regions above and below the axis.

    Step‑by‑step reasoning:

    1. Identify where the curve crosses the x‑axis.

       $y = \cos x = 0$ when $x = \pi/2$ within $[0, \pi]$. So the curve is above the axis for $0 \le x < \pi/2$ and below for $\pi/2 < x \le \pi$.

    2. Set up the area as the sum of two absolute integrals.

    $\text{Area} = \int_{0}^{\pi/2} \cos x \, dx \;+\; \int_{\pi/2}^{\pi} (-\cos x) \, dx$

    The second integral uses $-\cos x$ because $\cos x$ is negative there, and we want the positive height.

    3. Evaluate the first integral.

    $\int_{0}^{\pi/2} \cos x \, dx = \left[ \sin x \right]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1.$

    4. Evaluate the second integral. $$$$

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11. COMEDK 2026Set 2026-M1 markMCQ
    Q.The area of the region in the first quadrant enclosed by the $x$-axis, the line $x=\sqrt{3} y$ and the circle $x^2+y^2=4$ is (A) $\frac{\pi}{6}$ sq units (B) $\frac{2 \pi}{3}$ sq units (C) $\pi$ sq units (D) $\frac{\pi}{3}$ sq units

    ›Reveal solutionSolution

    The region is a circular sector minus a triangle; its area is $\frac{\pi}{3}$ square units, so the correct option is (D).

    ---

    We want the area in the first quadrant bounded by:

    • the $x$-axis,
    • the line $x = \sqrt{3}\, y$,
    • and the circle $x^2 + y^2 = 4$.

    Concept & Intuition

    The circle has radius $2$. The line $x = \sqrt{3}\, y$ passes through the origin and makes an angle with the $x$-axis. In the first quadrant, the $x$-axis is the ray at angle $0$, and the line is another ray. The region is essentially a sector of the circle (from angle $0$ to the angle of the line) minus a right triangle that lies under the line but outside the circle? Actually, careful: the region is bounded by the $x$-axis, the line, and the circle. That means we take the part of the circle in the first quadrant that lies above the $x$-axis and below the line? Let’s check: The line $x = \sqrt{3}\, y$ can be rewritten as $y = \frac{x}{\sqrt{3}}$. For a given $x$, this line gives a $y$ value. The circle gives $y = \sqrt{4 - x^2}$. The region is enclosed: the $x$-axis is the bottom, the line is one side, and the circular arc is the outer boundary. So the region is like a circular sector from angle $0$ to the angle where the line meets the circle, minus the triangle formed by the origin, the intersection point, and the foot on the $x$-axis? Actually, no — the sector itself is bounded by two radii and the arc. Here one radius is the $x$-axis, the other is the line. So the region is exactly that sector. But wait: the $x$-axis and the line meet at the origin, and the arc of the circle from the $x$-axis intersection to the line intersection gives the curved boundary. So the region is just a sector of the circle. That makes the problem simple: find the angle between the $x$-axis and the line, then compute sector area.

    ---

    Step-by-step solution

    1. Find the angle of the line

       The line is $x = \sqrt{3}\, y$, i.e. $y = \frac{1}{\sqrt{3}} x$. Its slope is $\frac{1}{\sqrt{3}}$, so the angle $\theta$ it makes with the positive $x$-axis satisfies $\tan\theta = \frac{1}{\sqrt{3}}$. Hence $\theta = \frac{\pi}{6}$.

    2. Sector area formula …

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12. COMEDK 2024Set 2024-A1 markMCQ
    Q.The area of the region (in sq units) bounded by the curve $y=\sqrt{16-x^2} \text { and } x \text {-axis is }$ (A) 20$\pi$ (B) 16$\pi$ (C) 256$\pi$ (D) 8$\pi$

    ›Reveal solutionSolution

    The curve $y=\sqrt{16-x^2}$ is the upper half of a circle of radius 4 centered at the origin. The area bounded by it and the x‑axis is half the area of that circle, which is $8\pi$. The correct option is (D).

    ---

    The key idea is recognizing the geometric shape hiding in the equation. When you see $y = \sqrt{16 - x^2}$, think: if you square both sides, you get $y^2 = 16 - x^2$, or $x^2 + y^2 = 16$. That’s the equation of a circle centered at $(0,0)$ with radius $4$. But because the original equation takes only the non‑negative square root, $y \ge 0$. So the graph is just the top half of that circle — a semicircle.

    The x‑axis ($y=0$) is the diameter. So the region bounded by the curve and the x‑axis is exactly that semicircle. The area is therefore half the area of the full circle.

    ---

    1. Identify the shape

       The equation $y = \sqrt{16 - x^2}$ implies $x^2 + y^2 = 16$ with $y \ge 0$. This is a semicircle of radius $r = 4$.

    2. Recall the area of a full circle

       The area of a circle is $A_{\text{circle}} = \pi r^2$. Here $r = 4$, so

    $A_{\text{circle}} = \pi \cdot 4^2 = 16\pi.$

    3. Take half for the semicircle The region is only the upper half, so

    $A_{\text{semicircle}} = \frac{1}{2} \times 16\pi = 8\pi.$

    4. Match with the options …

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