Q.Calculate the area under the curve $y = 2\sqrt{x}$ included between the lines $x = 0$ and $x = 1$.

Showing the 12 most recent of 20 on this concept.

1. 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

2. 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

3. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.The area bounded by the curve $x = \log(|y|)$, the lines $x = -1$ and $x = 0$ is (A) $1 - e^{-1}$ (B) $1 - e$ (C) $2(1-e)$ (D) $2(1 - e^{-1})$

   ›Reveal solutionSolution

   The curve $x=\log|y|$ has two branches $y=\pm e^x$; the area enclosed between $x=-1$ and $x=0$ is $2(1-e^{-1})$.

   Concept and Intuition

   $x=\log|y| \Leftrightarrow |y|=e^x \Leftrightarrow y=\pm e^x$, giving a symmetric pair of curves about the x-axis.

   Step-by-Step Solution

   1. Upper branch: $y=e^x$. Lower branch: $y=-e^x$.
   2. Between $x=-1$ and $x=0$, the vertical gap between the branches is $e^x-(-e^x)=2e^x$.
   3. Area $=\int_{-1}^0 2e^x\,dx = 2\left[e^x\right]_{-1}^0 = 2(e^0-e^{-1}) = 2(1-e^{-1})$.

   Common Mistakes

   • Only considering one branch ($y=e^x$) instead of the full region between both branches.

   ✓Final answer

   The correct option is (D) — $2(1 - e^{-1})$.

   ANSWER: D

4. 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).
   5. So all four curves pass through the triangle with vertices $(0,0),(1,1),(1,2)$ — the "$x=1$" side and the "$y=2$" boundary coincide at the single corner $(1,2)$, so the closed region is bounded below by $y=x$, on the right by $x=1$, and above/left by the parabola, for $x\in[0,1]$.
   6. Area $=\displaystyle\int_0^1\Big[(x^2+x)-x\Big]dx=\int_0^1 x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13$.

   Common Mistakes

   • Assuming $y=2$ cuts off a separate strip, when in fact it passes exactly through the same corner point as the other two boundaries.
   • Integrating in the wrong order (line minus parabola) and getting a negative or wrong magnitude.

   ✓Final answer

   The correct option is (D) — $\dfrac13$.

   ANSWER: D

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

   Common Mistakes

   • Integrating $\cos x-\sin x$ across the whole interval without splitting at the crossing point, which gives a wrong (too small or signed) result.
   • Sign errors in the antiderivative of $\sin x-\cos x$.

   ✓Final answer

   The correct option is (A) — $2(\sqrt2-1)$.

   ANSWER: A

6. 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

   $\text{Area}=\int_{1-e}^{0}\log(x+e)\,dx.$

   4. Substitute $u=x+e$, $du=dx$: when $x=1-e$, $u=1$; when $x=0$, $u=e$. So Area $=\displaystyle\int_{1}^{e}\log u\,du$.
   5. Use $\int\log u\,du = u\log u - u+c$: Area $= \big[u\log u-u\big]_1^{e} = (e\cdot1-e)-(1\cdot0-1) = 0-(-1)=1$.

   Common Mistakes

   • Forgetting to shift the limits of integration when substituting $u=x+e$.
   • Mixing up which axis intercept bounds the region (using $x=-e$, the vertical asymptote, instead of $x=1-e$, the actual zero of the curve).

   ✓Final answer

   The correct option is (D) — $1$.

   ANSWER: D

7. 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

8. 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 $=\displaystyle\int_{-a}^{-1}\Big[a-\big(-\tfrac{a}{x}\big)\Big]dx=\int_{-a}^{-1}\Big(a+\tfrac{a}{x}\Big)dx$.
   6. $\displaystyle\int_{-a}^{-1} a\,dx=a[(-1)-(-a)]=a(a-1)$.
   7. $\displaystyle\int_{-a}^{-1}\dfrac{a}{x}dx=a\big[\log|x|\big]_{-a}^{-1}=a[\ln1-\log a]=-a\log a$.
   8. Total area $=a(a-1)-a\log a=a(a-1-\log a)$.

   Common Mistakes

   • Sign confusion working with negative $x$ values inside $\log|x|$.
   • Forgetting $a>1$ is what guarantees the curve stays below $y=a$ throughout the strip (otherwise the region description would need revisiting).

   ✓Final answer

   The correct option is (A) — $a(a-1-\log a)$.

   ANSWER: A

9. 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

10. 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

11. 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

12. 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: $\displaystyle\int_{x_0}^{\pi/2}\cos x\,dx=\Big[\sin x\Big]_{x_0}^{\pi/2}=1-\sin x_0=1-s$.
    6. So $A=(1-s)-\tfrac12\log s$. Since $s=\dfrac{\sqrt5-1}2$, we have $\dfrac1s=\dfrac{\sqrt5+1}2$ (rationalize: $\frac{2}{\sqrt5-1}=\frac{2(\sqrt5+1)}{4}=\frac{\sqrt5+1}2$), so $-\tfrac12\log s=\tfrac12\log\dfrac1s=\tfrac12\log\dfrac{\sqrt5+1}2$.
    7. Also $1-s=1-\dfrac{\sqrt5-1}2=\dfrac{3-\sqrt5}2$.
    8. So

    $A=\frac{3-\sqrt5}{2}+\frac12\log\left(\frac{\sqrt5+1}{2}\right).$

    Numerically, $A\approx0.382+0.241=0.623$, consistent with the shape of the region.

    Common Mistakes

    • Taking the area under the upper envelope over the full interval, which diverges because $\tan x\to\infty$ at $\pi/2$.
    • Sign errors converting $-\tfrac12\log s$ to $\tfrac12\log(1/s)$, or not rationalizing $1/s$ back into the $(\sqrt5+1)/2$ form that matches the answer choices.

    ✓Final answer

    The correct option is (D) — $\dfrac{3 - \sqrt{5}}{2} + \dfrac{1}{2}\log\left(\dfrac{\sqrt{5} + 1}{2}\right)$.

    ANSWER: D