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

Showing the 12 most recent of 16 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.A curve with equation $y = x^3 - 8x^2 + 16x$ meets the $x$-axis at the origin $O$ and at a point $A$. Then the area of the region, bounded by the curve and the straight-line segment $OA$, is (A) $\dfrac{61}{3}$ (B) $\dfrac{62}{3}$ (C) $\dfrac{64}{3}$ (D) $\dfrac{65}{3}$ (E) $\dfrac{68}{3}$

   ›Reveal solutionSolution

   Factor to find the roots, then integrate from $0$ to $4$.

   $y=x^3-8x^2+16x=x(x-4)^2$, so the curve meets the $x$-axis at $O(0,0)$ and $A(4,0)$, and $y\ge 0$ on $[0,4]$. The area between the curve and $OA$ (the $x$-axis) is …

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2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The area of the region bounded by $y = \sqrt{x},\ y = -x$, and $x = 4$ (in square units) is (A) $\dfrac{8}{3}$ (B) $\dfrac{20}{3}$ (C) $\dfrac{40}{3}$ (D) $\dfrac{26}{3}$ (E) $\dfrac{32}{3}$

   ›Reveal solutionSolution

   Between $x=0$ and $x=4$ the upper curve is $y=\sqrt x$ and the lower is $y=-x$; integrating their difference gives $\dfrac{40}{3}$.

   For $0\le x\le 4$, $y=\sqrt x$ lies above $y=-x$, so

   $A=\int_0^4\bigl(\sqrt x-(-x)\bigr)dx=\int_0^4(\sqrt x+x)\,dx=\left[\tfrac23 x^{3/2}+\tfrac{x^2}{2}\right]_0^4.$ …

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3. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The area of the region bounded by the lines, $y = x + 2,\ x = 0,\ x = 1$ and $y = 0$ is (A) $2$ sq.units (B) $\dfrac{5}{2}$ sq.units (C) $\dfrac{9}{2}$ sq.units (D) $9$ sq.units (E) $12$ sq.units

   ›Reveal solutionSolution

   The region is under $y=x+2$ from $x=0$ to $1$; its area is $\int_0^1(x+2)dx=\tfrac52$.

   On $[0,1]$ the line $y=x+2$ lies above $y=0$. The bounded area is

   $\int_0^1 (x+2)\,dx=\left[\dfrac{x^2}{2}+2x\right]_0^1=\dfrac{1}{2}+2=\dfrac{5}{2}$ sq. units. …

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4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.The area of the region bounded by $y = x^{5/2}$ and $y = x$ (in square units) is (A) $\dfrac{3}{7}$ (B) $\dfrac{2}{7}$ (C) $\dfrac{3}{14}$ (D) $\dfrac{5}{14}$ (E) $\dfrac{4}{7}$

   ›Reveal solutionSolution

   The curves meet at x=0 and x=1, and the enclosed area is 1/2 - 2/7 = 3/14.

   Concept and Intuition

   Between the intersection points, the line y=x lies above y=x^{5/2} on (0,1), so the area is the integral of the difference.

   Step-by-Step Solution

   1. Intersection: x^{5/2} = x gives x(x^{3/2}-1)=0, so x = 0 and x = 1.
   2. On (0,1), x > x^{5/2}, so area = integral from 0 to 1 of (x - x^{5/2}) dx. …

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5. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.The area bounded by $y = x - 1, 1 \le x \le 2, y = 0$ (in sq.units) is (A) $2$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $4$ (E) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Integrate $x-1$ from $1$ to $2$.

   On $[1,2]$ the line $y=x-1$ is nonnegative, so the area between it and $y=0$ is …

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6. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.Area of the region bounded by the function $f(x) = \begin{cases} x & x \le 3 \\ -x + 6 & x > 3 \end{cases}$ with the $x$-axis (in square units) in the first quadrant is (A) $18$ (B) $9$ (C) $6$ (D) $3$ (E) $4.5$

   ›Reveal solutionSolution

   The graph forms a triangle with base $6$ (from $0$ to $6$) and peak height $3$ at $x=3$; area $=\tfrac12(6)(3)=9$.

   For $x\le3$, $y=x$ rises from $(0,0)$ to $(3,3)$. For $x>3$, $y=-x+6$ falls from $(3,3)$ to $(6,0)$. Together with the $x$-axis this bounds a triangle with base $6$ (from $x=0$ to $x=6$) and height $3$. Computing directly: …

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7. KEAM 2025Set eng-2025-04294 marksMCQ
   Q.Area of the region bounded by $y = |x|$ and $x = 4$ is (A) $4$ sq.units (B) $6$ sq.units (C) $8$ sq.units (D) $12$ sq.units (E) $13$ sq.units

   ›Reveal solutionSolution

   The region under $y=|x|$ up to $x=4$ (first-quadrant triangle) has area $\tfrac{1}{2}\times 4\times 4 = 8$ sq. units.

   Setting up the region. The curve $y=|x|$ for $x\geq 0$ is the straight line $y=x$. Bounded by the $x$-axis and the vertical line $x=4$, this traces a right triangle with vertices $(0,0)$, $(4,0)$ and $(4,4)$.

   Computing the area. …

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8. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.The area bounded by the curves $y = x^2$ and $y = 2x$ in the first quadrant, is equal to (A) $\frac{2}{3}$ (B) $\frac{4}{3}$ (C) $\frac{1}{3}$ (D) $\frac{8}{3}$ (E) $\frac{7}{3}$

   ›Reveal solutionSolution

   Intersection at $x=0,2$; integrate the difference of the upper and lower curves.

   Set $x^2=2x\Rightarrow x=0$ or $x=2$. On $(0,2)$ the line $y=2x$ lies above $y=x^2$. So …

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9. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.The area bounded by the parabola $y=x^{2}+2$ and the lines $y=x$, $x=1$ and $x=2$ is (in square units) (A) $\frac{31}{6}$ (B) $\frac{29}{6}$ (C) $\frac{25}{6}$ (D) $\frac{17}{6}$ (E) $\frac{13}{6}$

   ›Reveal solutionSolution

   Integrate (upper curve $-$ lower line) over $[1,2]$.

   On $[1,2]$, $x^2+2>x$, so

   $A=\int_1^2\big(x^2+2-x\big)dx=\left[\frac{x^3}{3}+2x-\frac{x^2}{2}\right]_1^2.$ …

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10. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.The area bounded by the parabola $y=x^2+4$ and the straight line passing through the points $(-1,2)$ and $(1,6)$ is (in square units) (A) $\dfrac{20}{3}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{8}{3}$ (D) $\dfrac{16}{3}$ (E) $\dfrac{14}{3}$

    ›Reveal solutionSolution

    Find the line, its intersections with the parabola, and integrate the gap.

    Slope of the line $=\frac{6-2}{1-(-1)}=2$, so $y-6=2(x-1)\Rightarrow y=2x+4$.

    Intersections with $y=x^2+4$:

    $x^2+4=2x+4\Rightarrow x^2-2x=0\Rightarrow x=0,\,2.$

    On $(0,2)$ the line lies above the parabola (at $x=1$: line $6$, parabola $5$): …

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11. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.The area bounded by the curves $y = 2x$ and $y = x^2$ (in square units) is (A) $\dfrac{2}{3}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{4}{3}$ (D) $\dfrac{3}{2}$ (E) $0$

    ›Reveal solutionSolution

    The line and parabola meet at $x=0,2$; between them the line is above, so the area is $\int_0^2(2x-x^2)dx=\frac43$.

    Setting $2x=x^2$ gives $x=0$ and $x=2$. On $(0,2)$, $2x>x^2$, so …

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12. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.The area of the smaller segment cut-off from the circle $x^2 + y^2 = 25$ by $x = 3$ is (in Sq. units) (A) $75\cos^{-1}\left(\dfrac{3}{5}\right) - 12$ (B) $25\cos^{-1}\left(\dfrac{3}{5}\right) - 24$ (C) $25\cos^{-1}\left(\dfrac{3}{5}\right) - 12$ (D) $25\cos^{-1}\left(\dfrac{3}{5}\right) - 6$ (E) $50\cos^{-1}\left(\dfrac{3}{5}\right) - 12$

    ›Reveal solutionSolution

    Integrating $2\sqrt{25-x^2}$ from $x=3$ to $5$ and converting $\sin^{-1}$ to $\cos^{-1}$ gives $25\cos^{-1}\frac35-12$.

    The smaller segment lies to the right of $x=3$:

    $A=\int_3^5 2\sqrt{25-x^2}\,dx=2\left[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}\sin^{-1}\frac{x}{5}\right]_3^5.$

    At $x=5$: $0+\frac{25}{2}\cdot\frac{\pi}{2}$; at $x=3$: $\frac{3}{2}\cdot4+\frac{25}{2}\sin^{-1}\frac35=6+\frac{25}{2}\sin^{-1}\frac35$. …

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