Q.Sketch the region $\{(x, 0) : y = \sqrt{4 - x^2}\}$ and x-axis. Find the area of the region using integration.

Showing the 12 most recent of 20 on this concept.

1. 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.
   4. Area $=\displaystyle\int_{-\pi/2}^{\pi/2}5\cos x\,dx=5[\sin x]_{-\pi/2}^{\pi/2}=5(1-(-1))=10$.

   Common Mistakes

   • Forgetting to drop the absolute value correctly (it's valid to drop it directly here since $\cos x\ge0$ on the whole given interval).
   • Sign error when subtracting the lower curve from the upper curve.

   ✓Final answer

   The correct option is (D) — 10.

   ANSWER: D

2. 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: $\displaystyle\int_0^{\sqrt3}\frac{y^2}{3}dy=\frac{1}{9}\left[y^3\right]_0^{\sqrt3}=\frac{3\sqrt3}{9}=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}$.
   6. Second integral, using $\int\sqrt{4-y^2}\,dy=\frac{y}{2}\sqrt{4-y^2}+2\arcsin\frac{y}{2}+C$: at $y=2$, value $=0+2\cdot\frac\pi2=\pi$; at $y=\sqrt3$, value $=\frac{\sqrt3}{2}\cdot1+2\cdot\frac\pi3=\frac{\sqrt3}{2}+\frac{2\pi}{3}$. So the integral $=\pi-\left(\frac{\sqrt3}{2}+\frac{2\pi}{3}\right)=\frac{\pi}{3}-\frac{\sqrt3}{2}$.
   7. Total area $=\dfrac1{\sqrt3}+\dfrac\pi3-\dfrac{\sqrt3}{2}=\dfrac\pi3+\sqrt3\left(\dfrac13-\dfrac12\right)=\dfrac\pi3-\dfrac{\sqrt3}{6}=\dfrac\pi3-\dfrac{1}{2\sqrt3}$ (since $\sqrt3/6=1/(2\sqrt3)$).

   Common Mistakes

   • Integrating over $x$ instead of $y$, which requires splitting the circle piece awkwardly since it isn't a single-valued function of $x$ over the needed range in a convenient way.
   • Sign/simplification slips converting $\sqrt3/6$ to $1/(2\sqrt3)$ when matching the answer to the option's form.

   ✓Final answer

   The correct option is (A) — $\frac{\pi}{3} - \frac{1}{2\sqrt{3}}$.

   ANSWER: A

3. 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. $\displaystyle\int_0^{\pi/4}(\cos x-\sin x)dx=\Big[\sin x+\cos x\Big]_0^{\pi/4}=\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\right)-(0+1)=\sqrt2-1$.
   3. $\displaystyle\int_{\pi/4}^{\pi/2}(\sin x-\cos x)dx=\Big[-\cos x-\sin x\Big]_{\pi/4}^{\pi/2}=(0-1)-\left(-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\right)=-1+\sqrt2=\sqrt2-1$.
   4. Total area $=(\sqrt2-1)+(\sqrt2-1)=2\sqrt2-2=2(\sqrt2-1)$.

   Common Mistakes

   • Integrating $\sin x-\cos x$ across the whole interval without splitting at the sign change, which would give the wrong (partially cancelled) value.
   • Sign slips evaluating the boundary terms.

   ✓Final answer

   The correct option is (C) — $2(\sqrt{2}-1)$.

   ANSWER: C

4. 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 $\int_0^a\sqrt{ax}\,dx = \sqrt a\cdot\dfrac23x^{3/2}\Big|_0^a = \sqrt a\cdot\dfrac23a^{3/2}=\dfrac23a^2$.
   6. For $\int_0^a\sqrt{2ax-x^2}\,dx = \int_0^a\sqrt{a^2-(x-a)^2}\,dx$: substituting $u=x-a$ (ranging from $-a$ to $0$), this is $\int_{-a}^0\sqrt{a^2-u^2}\,du$, which is exactly one quarter of the full circle's area, $\dfrac{\pi a^2}{4}$.
   7. Area $= \dfrac{\pi a^2}{4} - \dfrac23a^2 = a^2\left(\dfrac\pi4-\dfrac23\right)$.

   Common Mistakes

   • Mixing up which curve is on top over $(0,a)$ — a quick check near $x\to0^+$ (circle grows like $\sqrt{2ax}$, parabola like $\sqrt{ax}$, so circle is bigger) resolves this.
   • Forgetting that $\int_0^a\sqrt{2ax-x^2}\,dx$ is a quarter-circle (not a semicircle) — the shifted range $u\in[-a,0]$ covers only one quadrant of the circle.

   ✓Final answer

   The correct option is (B) — $a^2\left(\dfrac\pi4-\dfrac23\right)$.

   ANSWER: B

5. 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 $\int\sqrt{16-x^2}dx=\dfrac x2\sqrt{16-x^2}+8\sin^{-1}\dfrac x4$: At $x=4$: $0+8\cdot\dfrac\pi2=4\pi$. At $x=2$: $\sqrt{12}+8\cdot\dfrac\pi6=2\sqrt3+\dfrac{4\pi}3$. So $\displaystyle\int_2^4\sqrt{16-x^2}dx=4\pi-2\sqrt3-\dfrac{4\pi}3=\dfrac{8\pi}3-2\sqrt3$.
   7. Sum: $\dfrac{8\sqrt3}3+\dfrac{8\pi}3-2\sqrt3=\dfrac{8\pi}3+\dfrac{2\sqrt3}3$.
   8. Area $=2\left(\dfrac{8\pi}3+\dfrac{2\sqrt3}3\right)=\dfrac{16\pi+4\sqrt3}3=\dfrac43(4\pi+\sqrt3)$.

   Common Mistakes

   • Using only the circle or only the parabola over the whole $[0,4]$ range instead of switching boundary at $x=2$.
   • Sign/evaluation slips in the standard $\int\sqrt{a^2-x^2}dx$ formula.

   ✓Final answer

   The correct option is (C) — $\dfrac{4}{3}\left(4\pi + \sqrt{3}\right)$.

   ANSWER: C

6. 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 $= \displaystyle\int_{-1}^{1}\left[(3-2y^2)-y^2\right]dy = \int_{-1}^1 (3-3y^2)\,dy$.
   4. By symmetry (even integrand): $=2\displaystyle\int_0^1(3-3y^2)\,dy = 2\left[3y-y^3\right]_0^1 = 2(3-1)=4$.

   Common Mistakes

   • Trying to integrate with respect to $x$ directly, which requires splitting into two branches ($y=\pm\sqrt{x}$) and is more error-prone.
   • Sign error in determining which curve is "outer" over the interval.

   ✓Final answer

   The correct option is (C) — $4$.

   ANSWER: C

7. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If the area of the region enclosed by the curve $x^2 + y^2 = 16$ and the lines $x = 2$ and $x = 3$ is $\left(3\sqrt{7} - 4\sqrt{3} - \dfrac{8\pi}{3} + k\right)$ sq.units, then 'k' equals ______ (A) $16\sin^{-1}\left(\dfrac{3}{4}\right)$ (B) $8\sin^{-1}\left(\dfrac{3}{4}\right)$ (C) $4\sin^{-1}\left(\dfrac{3}{4}\right)$ (D) $2\sin^{-1}\left(\dfrac{3}{4}\right)$

   ›Reveal solutionSolution

   Compute the area between $x=2$ and $x=3$ under the full circle (upper + lower) using the standard $\int\sqrt{r^2-x^2}\,dx$ formula, then match the constant term to identify $k$.

   Concept and Intuition

   The region bounded by the circle between two vertical lines $x=2$ and $x=3$ (with both halves counted) is twice the area under the upper semicircle over that range — a direct application of the standard circle-area antiderivative.

   Step-by-Step Solution

   1. Area $=2\displaystyle\int_2^3\sqrt{16-x^2}\,dx$ (factor 2 for upper + lower half).
   2. Antiderivative: $\displaystyle\int\sqrt{16-x^2}\,dx=\dfrac{x}2\sqrt{16-x^2}+8\sin^{-1}\left(\dfrac x4\right)+c$.
   3. At $x=3$: $\dfrac32\sqrt7+8\sin^{-1}\left(\dfrac34\right)$.
   4. At $x=2$: $1\cdot\sqrt{12}+8\sin^{-1}\left(\dfrac12\right)=2\sqrt3+8\cdot\dfrac{\pi}6=2\sqrt3+\dfrac{4\pi}3$.
   5. Definite integral $=\dfrac32\sqrt7+8\sin^{-1}\left(\dfrac34\right)-2\sqrt3-\dfrac{4\pi}3$.
   6. Multiply by 2 (both halves): Area $=3\sqrt7-4\sqrt3-\dfrac{8\pi}3+16\sin^{-1}\left(\dfrac34\right)$.
   7. Comparing to the given form $\left(3\sqrt7-4\sqrt3-\dfrac{8\pi}3+k\right)$, we read off $k=16\sin^{-1}\left(\dfrac34\right)$.

   Common Mistakes

   • Forgetting the factor of 2 for counting both the upper and lower halves of the circle in the enclosed region.
   • Arithmetic slip evaluating $\sqrt{16-4}=\sqrt{12}=2\sqrt3$ or $\sin^{-1}(1/2)=\pi/6$.

   ✓Final answer

   The correct option is (A) — $16\sin^{-1}\left(\dfrac34\right)$.

   ANSWER: A

8. 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. $\displaystyle\int_{\pi/2}^{\pi}(1+\cos x)\,dx=[x+\sin x]_{\pi/2}^{\pi}=\pi-\left(\tfrac{\pi}{2}+1\right)=\tfrac{\pi}{2}-1.$
   6. Total area $=1+\left(\tfrac{\pi}{2}-1\right)=\tfrac{\pi}{2}.$

   Common Mistakes

   • Integrating the difference of the two curves ($\int(\sin x-(1+\cos x))$ on $[\tfrac{\pi}{2},\pi]$ gives $2-\tfrac{\pi}{2}$) — that is the lens between the curves, which does NOT use the X-axis, so it ignores a stated boundary.
   • Forgetting that the roof switches curves at $x=\tfrac{\pi}{2}$.

   ✓Final answer

   The correct option is (D) — $\dfrac{\pi}{2}$.

   ANSWER: D

9. 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 $=\int_{-1}^{1}\big[(1-|x|)-(|x|-1)\big]dx=\int_{-1}^1(2-2|x|)dx$.
   4. By symmetry, $=2\int_0^1(2-2x)dx=2\big[2x-x^2\big]_0^1=2(2-1)=2$.
   5. (Geometric check: vertices $(0,1),(1,0),(0,-1),(-1,0)$ form a square with diagonals of length 2 each; area $=\frac12 d_1 d_2=\frac12(2)(2)=2$.)

   Common Mistakes

   • Forgetting the factor from symmetry and only integrating over $[0,1]$, halving the true area.
   • Mixing up which piecewise line is "on top" for $x<0$ vs $x>0$.

   ✓Final answer

   The correct option is (C) — $2$.

   ANSWER: C

10. 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. $F(3)=20.25-85.5+90=24.75$.
    8. Area on $(-5,2)$ (curve above axis) $=F(2)-F(-5)=26-(-231.25)=257.25$.
    9. Area on $(2,3)$ (curve below axis) $=|F(3)-F(2)|=|24.75-26|=1.25$.
    10. Total enclosed area $=257.25+1.25=258.5=\dfrac{517}{2}$ sq. units.

    Common Mistakes

    • Integrating $y$ directly from $-5$ to $3$ in one go, which lets the negative lobe cancel part of the positive lobe, giving a wrong (too small) answer.
    • Sign/arithmetic slips evaluating $F$ at each root.

    ✓Final answer

    The correct option is (B) — $\dfrac{517}{2}$.

    ANSWER: B

11. 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: $\displaystyle\int\left(2x-\dfrac8x\right)dx=x^2-8\log x$.
    5. Evaluate: $\left[x^2-8\log x\right]_2^4=(16-8\ln4)-(4-8\ln2)=12-8\ln4+8\ln2=12-8(\ln4-\ln2)$.
    6. Since $\ln4-\ln2=\log(4/2)=\ln2$, the area $=12-8\ln2$.

    Common Mistakes

    • Integrating from $x=0$ instead of the actual curve intersection at $x=2$ (the region is bounded by the two curves' crossing, not the y-axis).
    • Mixing up which curve is on top, which would flip the sign of the log term.

    ✓Final answer

    The correct option is (A) — $12-8\log 2$.

    ANSWER: A

12. 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 $(0,1)$.
    3. The lower parabola branch $y=-\sqrt x$ is always negative for $x>0$, while the line $y=x$ (for $x\ge0$, the relevant part of $|x|$) is always non-negative, so they meet only at the origin — they don't bound any extra region.
    4. Area $=\displaystyle\int_0^1(\sqrt x - x)\,dx = \left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_0^1 = \frac23-\frac12 = \frac{4-3}{6}=\frac16$.

    Common Mistakes

    • Trying to include a symmetric mirror region for $x<0$, forgetting that the parabola simply doesn't exist there.
    • Mixing up which curve is on top when setting up the integrand (must be upper-curve minus lower-curve).

    ✓Final answer

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

    ANSWER: A