Q.Find the area of the region bounded by the line $y = 3x + 2$, the x-axis and the ordinates $x = -1$ and $x = 1$

Showing the 12 most recent of 16 on this concept.

1. 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.

   (Equivalently, a trapezium with parallel sides $2$ and $3$, width $1$: area $=\tfrac12(2+3)(1)=\tfrac52$.)

   ✓Final answer

   The correct option is (B).

2. 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

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

   At $x=2$: $\dfrac{8}{3}+4-2=\dfrac{14}{3}$. At $x=1$: $\dfrac{1}{3}+2-\dfrac{1}{2}=\dfrac{11}{6}$.

   $A=\frac{14}{3}-\frac{11}{6}=\frac{28-11}{6}=\frac{17}{6}.$

   ✓Final answer

   The correct option is (D).

3. 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

   $\int_1^2 (x-1)\,dx=\left[\frac{(x-1)^2}{2}\right]_1^2=\frac{1}{2}-0=\frac12\text{ sq. units}.$

   ✓Final answer

   The correct option is (C).

4. 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

   $\text{Area}=\int_0^2(2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac83=\frac43.$

   ✓Final answer

   The correct option is (C).

5. 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

   $\int_{0}^{4}(x^3-8x^2+16x)\,dx=\left[\frac{x^4}{4}-\frac{8x^3}{3}+8x^2\right]_0^4=64-\frac{512}{3}+128=192-\frac{512}{3}=\frac{64}{3}.$

   ✓Final answer

   The correct option is (C).

6. 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 $5$):

   $A=\int_0^2\big[(2x+4)-(x^2+4)\big]dx=\int_0^2(2x-x^2)dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}.$

   ✓Final answer

   The correct option is (B).

7. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.The area bounded by the curve $y = x(2-x)$ and the line $y = x$ is (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{5}{6}$ (E) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   The enclosed area is $\frac16$.

   Concept and Intuition

   Find the intersection points, determine which curve is on top, then integrate the difference between them.

   Step-by-Step Solution

   1. Set $x(2-x)=x\Rightarrow 2x-x^2=x\Rightarrow x-x^2=0\Rightarrow x=0,1$.
   2. At $x=0.5$: parabola $=0.75$, line $=0.5$, so parabola is above.
   3. Area $=\int_0^1\big[(2x-x^2)-x\big]dx=\int_0^1(x-x^2)\,dx$.
   4. $=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac12-\frac13=\frac16$.

   Common Mistakes

   • Choosing the wrong upper curve.
   • Integrating $x(2-x)$ without subtracting the line.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{6}$.

   ANSWER: A

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

   $A=\int_0^2 (2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac83=\frac43.$

   ✓Final answer

   The correct option is (B).

9. 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

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

   At $x=4$: $\tfrac23\cdot 8+\tfrac{16}{2}=\tfrac{16}{3}+8=\tfrac{16}{3}+\tfrac{24}{3}=\tfrac{40}{3}$ square units.

   ✓Final answer

   The correct option is (C).

10. 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:

    $\int_0^3 x\,dx+\int_3^6(-x+6)\,dx=\left[\frac{x^2}{2}\right]_0^3+\left[-\frac{x^2}{2}+6x\right]_3^6=\frac{9}{2}+\frac{9}{2}=9.$

    ✓Final answer

    The correct option is (B).

11. 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.
    3. = [x^2/2 - x^{7/2}/(7/2)] from 0 to 1 = 1/2 - 2/7.
    4. = 7/14 - 4/14 = 3/14.

    Common Mistakes

    • Integrating x^{7/2} without dividing by 7/2 (i.e. forgetting the factor 2/7).

    ✓Final answer

    The correct option is (C) — 3/14.

    ANSWER: C

12. KEAM 2021Set eng-2021-P2-B14 marksMCQ
    Q.The area of the region bounded by the curves $y = x^2$ and $y = \sqrt{x}$ is (in square units) (A) $\dfrac{2}{3}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{6}$ (D) $\dfrac{5}{6}$ (E) $1$

    ›Reveal solutionSolution

    The enclosed area between $y=x^2$ and $y=\sqrt{x}$ is $\dfrac13$.

    Concept and Intuition

    Between their intersection points, the upper curve minus the lower curve integrated gives the area. On $[0,1]$, $\sqrt{x}\ge x^2$.

    Step-by-Step Solution

    1. Intersections: $x^2=\sqrt{x}\Rightarrow x^4=x\Rightarrow x=0,1$.
    2. Area $=\int_0^1(\sqrt{x}-x^2)\,dx$.
    3. $=\big[\tfrac23 x^{3/2}-\tfrac{x^3}{3}\big]_0^1=\tfrac23-\tfrac13=\tfrac13$.

    Common Mistakes

    • Taking $x^2$ as the upper curve on $[0,1]$.

    ✓Final answer

    The correct option is (B) — $\dfrac{1}{3}$.

    ANSWER: B