Q.A student observes an open-air Honeybee nest on the branch of a tree, whose plane figure is parabolic shape given by $x^{2}=4y$. Then the area (in sq units) of the region bounded by parabola $x^{2}=4y$ and the line $y=4$ is
(A) $\frac{32}{3}$
(B) $\frac{64}{3}$
(C) $\frac{128}{3}$
(D) $\frac{256}{3}$

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The area of a triangle (in sq. units) formed by the latus rectum of the parabola $x^2 = 16y$ and the lines joining the vertex of the parabola to the ends of the latus rectum is (A) 24 (B) 28 (C) 32 (D) 64

   ›Reveal solutionSolution

   The problem asks for the area of a triangle formed by the vertex of the parabola $x^2 = 16y$ and the endpoints of its latus rectum. By identifying the parabola's properties, finding the coordinates of the vertex and the latus rectum's endpoints, and then using the formula for the area of a triangle, we find the area to be $\boxed{32}$ square units.

   The core idea here is to first understand the geometry of a parabola and its key features: the vertex and the latus rectum. Once these geometric elements are identified and their coordinates determined, calculating the area of the triangle becomes a straightforward application of the area formula.

   The given parabola is $x^2 = 16y$. This equation is in the standard form $x^2 = 4ay$, which describes a parabola opening upwards with its vertex at the origin $(0,0)$.

   The latus rectum of a parabola is a line segment that passes through the focus, is perpendicular to the axis of symmetry, and has its endpoints on the parabola. For a parabola of the form $x^2 = 4ay$, the focus is at $(0, a)$, and the latus rectum is the horizontal line segment $y=a$ with length $4a$.

   The triangle in question has three vertices:

   1. The vertex of the parabola.
   2. One end of the latus rectum.
   3. The other end of the latus rectum.

   We can find the coordinates of these three points and then use the standard formula for the area of a triangle given its vertices. Alternatively, recognizing that the base of this triangle (the latus rectum) is horizontal and the vertex of the parabola is on the y-axis, we can use the simple formula $\frac{1}{2} \times \text{base} \times \text{height}$.

   1. Identify the properties of the parabola:

      The given equation of the parabola is $x^2 = 16y$.

      Comparing this with the standard form $x^2 = 4ay$, we can determine the value of $a$.

      $4a = 16$

      $a = \frac{16}{4}$

      $a = 4$

      For a parabola of the form $x^2 = 4ay$:

      • The vertex is at the origin, $V = (0, 0)$.
      • The focus is at $(0, a)$, which is $F = (0, 4)$.
      • The axis of the parabola is the y-axis.

   2. Determine the endpoints of the latus rectum:

      The latus rectum is a line segment passing through the focus $(0, 4)$ and perpendicular to the axis of the parabola (the y-axis). Therefore, the equation of the line containing the latus rectum is $y = 4$.

      To find the endpoints of the latus rectum, substitute $y=4$ into the parabola's equation:

      $x^2 = 16(4)$

      $x^2 = 64$

      Taking the square root of both sides:

      $x = \pm\sqrt{64}$

      $x = \pm 8$

      So, the endpoints of the latus rectum are $L_1 = (-8, 4)$ and $L_2 = (8, 4)$.

      Tip

      The length of the latus rectum for $x^2=4ay$ is $4a$. In this case, $4a = 16$. The endpoints are $(\pm 2a, a)$. Here, $2a = 2(4) = 8$, so the endpoints are $(\pm 8, 4)$, which matches our calculation.

   3. Identify the vertices of the triangle:

      The triangle is formed by the vertex of the parabola and the ends of the latus rectum.

      The vertices are:

      • $V = (0, 0)$
      • $L_1 = (-8, 4)$
      • $L_2 = (8, 4)$

   4. Calculate the area of the triangle: …

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2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The area of the region bounded by $y = x^3$, x-axis, $x = -2$ and $x = 4$ is (A) 64 (B) $\dfrac{81}{4}$ (C) $\dfrac{66}{5}$ (D) 68

   ›Reveal solutionSolution

   The area is the sum of the absolute values of the definite integrals from $x=-2$ to $x=0$ and from $x=0$ to $x=4$, because the curve lies below the x-axis on the left interval and above it on the right. The total area is $\boxed{68}$, so the correct option is (D).

   The key concept here is that area between a curve and the x-axis is always taken as a positive quantity. When a function dips below the x-axis, the definite integral gives a negative number, but area is the absolute value of that signed area. So we must split the region at the x-intercept(s) where the curve crosses the axis, integrate each piece separately, take the absolute value, and then add them.

   For $y = x^3$, the only x-intercept is at $x = 0$. Between $x = -2$ and $x = 0$, the curve is below the axis (since $x^3$ is negative for negative $x$). Between $x = 0$ and $x = 4$, the curve is above the axis. Therefore, the total area is:

   $\text{Area} = \left| \int_{-2}^{0} x^3 \, dx \right| + \int_{0}^{4} x^3 \, dx$

   Now we compute step by step.

   1. Integrate $x^3$

      The antiderivative is $\frac{x^4}{4}$.

   2. Evaluate from $x = -2$ to $x = 0$

   $\int_{-2}^{0} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{-2}^{0} = \frac{0^4}{4} - \frac{(-2)^4}{4} = 0 - \frac{16}{4} = -4$

   The signed area is $-4$, so the actual (positive) area is $| -4 | = 4$.

   3. Evaluate from $x = 0$ to $x = 4$ $$\int_{0}^{4} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{4} = \frac{4^4}{4} - \frac{0^4}{4} = \frac{256}{4} - 0 = 64 …$$

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3. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The area of the region enclosed by the curves $y^2 = 4(x+1)$ and $y^2 = 5(x-4)$ is (A) $\dfrac{280}{3}$ (B) $150$ (C) $140$ (D) $\dfrac{200}{3}$

   ›Reveal solutionSolution

   The parabolas meet at $y=\pm10$; integrating the horizontal width $5-\dfrac{y^2}{20}$ gives area $\dfrac{200}{3}$.

   Set up horizontal strips. Solving each curve for $x$:

   $x_1=\frac{y^2}{4}-1\quad(\text{from }y^2=4(x+1)),\qquad x_2=\frac{y^2}{5}+4\quad(\text{from }y^2=5(x-4)).$

   Intersection points. Setting $x_1=x_2$:

   $\frac{y^2}{4}-\frac{y^2}{5}=5\;\Longrightarrow\;\frac{y^2}{20}=5\;\Longrightarrow\;y^2=100\;\Longrightarrow\;y=\pm10,$

   with common $x=24$.

   Width of a strip. For $|y|<10$ the curve $x_2$ lies to the right of $x_1$:

   $x_2-x_1=\left(\frac{y^2}{5}+4\right)-\left(\frac{y^2}{4}-1\right)=5-\frac{y^2}{20}.$ …

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The area of the region bounded by the curves $y = x^2 + 1$, $x = y^2 + 1$, X-axis, Y-axis and $x = 2$ (in Sq. units) is (A) $\dfrac{13}{3}$ (B) $4$ (C) $6$ (D) $\dfrac{15}{2}$

   ›Reveal solutionSolution

   Area $=\displaystyle\int_0^2 (x^2+1)\,dx-\int_1^2\sqrt{x-1}\,dx=\frac{14}{3}-\frac{2}{3}=4$ sq. units.

   The two curves are:

   • $y=x^2+1$ — an upward parabola, vertex $(0,1)$; at $x=2,\ y=5$.
   • $x=y^2+1$, i.e. $y=\sqrt{x-1}$ — a rightward parabola through $(1,0)$ and $(2,1)$.

   The region is bounded above by $y=x^2+1$, on the left by the $Y$-axis, on the right by $x=2$, below by the $X$-axis for $0\le x\le 1$ and by $y=\sqrt{x-1}$ for $1\le x\le 2$.

   Area under the top curve from $x=0$ to $2$:

   $\int_0^2 (x^2+1)\,dx=\left[\frac{x^3}{3}+x\right]_0^2=\frac{8}{3}+2=\frac{14}{3}.$ …

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5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If the points of intersection of the parabolas $y^2 = 5x$ and $x^2 = 5y$ lie on the line L, then the area of the triangle formed by the directrix of one parabola, latus rectum of another parabola and the line L is (A) $\dfrac{15}{32}$ (B) $\dfrac{12}{25}$ (C) $\dfrac{25}{8}$ (D) $\dfrac{25}{32}$

   ›Reveal solutionSolution

   The two parabolas meet on the line $L:y=x$. The directrix of one, the latus rectum of the other, and $L$ form a right triangle with both legs $\tfrac{5}{2}$, so its area is $\tfrac{25}{8}$.

   Step 1 — line $L$. Solving $y^2=5x$ and $x^2=5y$: substituting $x=\tfrac{y^2}{5}$ gives $y^4=125y$, so $y=0$ or $y=5$, i.e. intersections $(0,0)$ and $(5,5)$. Thus $L:y=x$.

   Step 2 — directrix and latus rectum. Each parabola has $4a=5$, so $a=\tfrac{5}{4}$.

   • For $y^2=5x$: directrix $x=-\tfrac{5}{4}$; latus rectum $x=\tfrac{5}{4}$.
   • For $x^2=5y$: directrix $y=-\tfrac{5}{4}$; latus rectum $y=\tfrac{5}{4}$.

   Take the directrix of the first, $x=-\tfrac{5}{4}$, and the latus rectum of the second, $y=\tfrac{5}{4}$ (any one-from-each choice gives the same area by symmetry).

   Step 3 — vertices.

   • $x=-\tfrac{5}{4}\cap y=x:\;\left(-\tfrac{5}{4},-\tfrac{5}{4}\right)$ …

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6. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.The area of the triangle formed by the tangent and the normal drawn to the curve $y^2 = 4x$ at $(1, 2)$ with Y-axis is (in square units) (A) 4 (B) 3 (C) 2 (D) 1

   ›Reveal solutionSolution

   The area is found by computing the intercepts of the tangent and normal lines with the Y‑axis, then using the triangle’s base (distance between intercepts) and height (x‑coordinate of the point). The result is 3 square units.

   We are given the parabola $y^2 = 4x$ and the point $(1,2)$. The tangent and normal at this point, together with the Y‑axis, form a triangle. The Y‑axis is the line $x = 0$, so the triangle’s vertices are: the intersection of the tangent with the Y‑axis, the intersection of the normal with the Y‑axis, and the point of tangency $(1,2)$. The base of the triangle lies along the Y‑axis, and the height is the horizontal distance from the point to the Y‑axis (which is simply the x‑coordinate of the point).

   Why this approach works: Instead of finding all three vertices and using a determinant, we can note that the Y‑axis is vertical, so the triangle’s base is the segment between the two Y‑intercepts. The height is the perpendicular distance from the point to the Y‑axis, which is just $|x| = 1$. Then area = $\frac12 \times \text{base} \times \text{height}$.

   1. Find the equation of the tangent. For the parabola $y^2 = 4x$, the standard tangent at $(x_1, y_1)$ is $y y_1 = 2(x + x_1)$. Here $(x_1, y_1) = (1, 2)$, so:

   $y \cdot 2 = 2(x + 1) \quad\Rightarrow\quad 2y = 2x + 2 \quad\Rightarrow\quad y = x + 1.$

   The Y‑intercept (set $x = 0$) is $y = 1$. So the tangent meets the Y‑axis at $A(0, 1)$.

   2. Find the equation of the normal. The slope of the tangent is $1$ (from $y = x + 1$). The slope of the normal is the negative reciprocal: $-1$. Using point-slope form at $(1,2)$:

   $y - 2 = -1(x - 1) \quad\Rightarrow\quad y - 2 = -x + 1 \quad\Rightarrow\quad y = -x + 3.$

   The Y‑intercept (set $x = 0$) is $y = 3$. So the normal meets the Y‑axis at $B(0, 3)$.

   3. Determine the base of the triangle. The base is the segment $AB$ on the Y‑axis from $y = 1$ to $y = 3$. Its length is:

   $\text{base} = |3 - 1| = 2.$

   4. Determine the height. …

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7. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The area of the region bounded by the curve $y = |x-2| + |x-8|$, X-axis and the lines $x = 0$, $x = 10$ is (A) 50 (B) 68 (C) 100 (D) 98

   ›Reveal solutionSolution

   The area is found by splitting the absolute-value expression into linear pieces over intervals where the signs of the terms are constant, then integrating the resulting piecewise linear function from 0 to 10. The total area is 68, so the correct option is (B).

   Concept & Intuition

   The function $y = |x-2| + |x-8|$ is a sum of two absolute values. Each absolute value changes slope at its “corner” (where the inside is zero). So the whole function is piecewise linear, with possible corners at $x=2$ and $x=8$. Instead of integrating an absolute value directly, we break the domain $[0,10]$ into subintervals where the expressions inside the absolute values have constant sign. On each subinterval, the absolute values become simple linear expressions, and the area under the curve (above the X‑axis) is just the integral of that linear function. Since the curve is always non‑negative, the area is exactly the definite integral from 0 to 10.

   ---

   Step‑by‑step solution

   1. Identify the critical points The expressions inside the absolute values are zero at $x=2$ and $x=8$. These split the real line into three intervals:

   $(-\infty, 2),\quad [2,8),\quad [8,\infty).$

   Our integration limits are $x=0$ to $x=10$, so we consider the subintervals $[0,2]$, $[2,8]$, and $[8,10]$.

   2. Write the function without absolute values on each subinterval

      • On $[0,2]$: $x-2 \le 0$ so $|x-2| = -(x-2) = 2-x$. $x-8 < 0$ so $|x-8| = -(x-8) = 8-x$. Hence

   $y = (2-x) + (8-x) = 10 - 2x.$

   • On $[2,8]$: $x-2 \ge 0$ so $|x-2| = x-2$. $x-8 \le 0$ so $|x-8| = 8-x$. Hence

   $y = (x-2) + (8-x) = 6.$

   • On $[8,10]$: $x-2 > 0$ so $|x-2| = x-2$. $x-8 \ge 0$ so $|x-8| = x-8$. Hence

   $y = (x-2) + (x-8) = 2x - 10.$

   3. Integrate piecewise from 0 to 10

   $\text{Area} = \int_{0}^{2} (10 - 2x)\,dx \;+\; \int_{2}^{8} 6\,dx \;+\; \int_{8}^{10} (2x - 10)\,dx.$

   • First integral: $$$$

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8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The area of the region bounded by the curves $y = x^3$, $y = x^2$ and the lines $x = 0$ and $x = 2$ is (A) $\dfrac{4}{3}$ (B) $\dfrac{3}{2}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{5}{3}$

   ›Reveal solutionSolution

   Area $=\dfrac{3}{2}$ sq. units — option (B).

   Compare the curves: on $[0,1]$, $x^{2}\ge x^{3}$; on $[1,2]$, $x^{3}\ge x^{2}$. So

   $\text{Area}=\int_{0}^{1}\!\left(x^{2}-x^{3}\right)dx+\int_{1}^{2}\!\left(x^{3}-x^{2}\right)dx.$

   $\int_{0}^{1}\!\left(x^{2}-x^{3}\right)dx=\frac{1}{3}-\frac{1}{4}=\frac{1}{12},$ …

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9. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The area (in sq. units) bounded by the curve $y = 2x - x^2$ and the line $y = -x$ is (A) $\dfrac{9}{2}$ (B) $\dfrac{11}{2}$ (C) $\dfrac{16}{3}$ (D) $\dfrac{22}{5}$

   ›Reveal solutionSolution

   The area between the parabola $y = 2x - x^2$ and the line $y = -x$ is found by integrating the difference of the functions between their intersection points. The result is $\frac{9}{2}$, so the correct option is (A).

   ---

   We want the area bounded by a downward-opening parabola and a straight line. The key idea: area between two curves is the integral of (top function minus bottom function) over the interval where they cross. So first, find where they intersect; then decide which curve is above the other; then integrate.

   1. Find intersection points

      Set $2x - x^2 = -x$.

      Bring all terms to one side:

      $2x - x^2 + x = 0 \quad\Rightarrow\quad 3x - x^2 = 0$

      Factor: $x(3 - x) = 0$

      So $x = 0$ and $x = 3$.

      These are the boundaries of the region.

   2. Determine which curve is on top

      Pick a test point between $0$ and $3$, say $x = 1$:

      • Parabola: $y = 2(1) - 1^2 = 1$
      • Line: $y = -1$ Since $1 > -1$, the parabola is above the line on $(0, 3)$. So the vertical height between them is $(2x - x^2) - (-x) = 2x - x^2 + x = 3x - x^2$.

   3. Set up the definite integral

      Area $= \int_{0}^{3} (3x - x^2) \, dx$ …

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10. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
    Q.Let $P\left(\dfrac{\pi}{4}\right), Q\left(\dfrac{5\pi}{4}\right), R\left(\dfrac{3\pi}{4}\right), T\left(\dfrac{7\pi}{4}\right)$ be the points on the hyperbola $x^2 - 4y^2 - 4 = 0$ in the parametric form. Then the area of the quadrilateral $PQRT$ is (in square units) (A) $4\sqrt{2}$ (B) $16\sqrt{2}$ (C) $32\sqrt{2}$ (D) $8\sqrt{2}$

    ›Reveal solutionSolution

    The quadrilateral is a rectangle whose sides are aligned with the asymptotes of the hyperbola. Its area is $8\sqrt{2}$ square units, option (D).

    The hyperbola is $x^2 - 4y^2 = 4$, which we rewrite in standard form:

    $\frac{x^2}{4} - \frac{y^2}{1} = 1.$

    Its standard parametric form is $x = 2\sec\theta$, $y = \tan\theta$. The given points use parameters $\frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}$ — these are spaced by $\frac{\pi}{2}$ in the parameter. That spacing is the key: on a rectangular hyperbola, equal parametric spacing often gives a rectangle, but here the hyperbola is not rectangular ($a \neq b$). Still, the geometry of the secant/tangent parametrization creates a special shape.

    Let’s find the coordinates.

    1. Compute the four points.

       For $\theta = \frac{\pi}{4}$:

       $x = 2\sec\frac{\pi}{4} = 2\cdot\sqrt{2} = 2\sqrt{2}$,

       $y = \tan\frac{\pi}{4} = 1$.

       So $P = (2\sqrt{2}, 1)$.

       For $\theta = \frac{5\pi}{4}$:

       $\sec\frac{5\pi}{4} = \sec(\pi + \frac{\pi}{4}) = -\sec\frac{\pi}{4} = -\sqrt{2}$,

       $\tan\frac{5\pi}{4} = \tan(\pi + \frac{\pi}{4}) = \tan\frac{\pi}{4} = 1$.

       So $Q = (-2\sqrt{2}, 1)$.

       For $\theta = \frac{3\pi}{4}$:

       $\sec\frac{3\pi}{4} = \sec(\pi - \frac{\pi}{4}) = -\sec\frac{\pi}{4} = -\sqrt{2}$,

       $\tan\frac{3\pi}{4} = \tan(\pi - \frac{\pi}{4}) = -\tan\frac{\pi}{4} = -1$.

       So $R = (-2\sqrt{2}, -1)$.

       For $\theta = \frac{7\pi}{4}$:

       $\sec\frac{7\pi}{4} = \sec(2\pi - \frac{\pi}{4}) = \sec\frac{\pi}{4} = \sqrt{2}$,

       $\tan\frac{7\pi}{4} = \tan(2\pi - \frac{\pi}{4}) = -\tan\frac{\pi}{4} = -1$.

       So $T = (2\sqrt{2}, -1)$.

    2. Identify the shape.

       The four points are:

    $P(2\sqrt{2}, 1),\quad Q(-2\sqrt{2}, 1),\quad R(-2\sqrt{2}, -1),\quad T(2\sqrt{2}, -1).$

    These are the vertices of a rectangle centered at the origin, with sides parallel to the axes. …

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11. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.The area of the region bounded by the curves $y = x^2 - 3x + 3$ and $y = 2x^2 - 1$ is (A) $\dfrac{403}{6}$ (B) $27$ (C) $19$ (D) $\dfrac{125}{6}$

    ›Reveal solutionSolution

    The region is bounded by two parabolas; find their intersection points, determine which curve is on top, then integrate the difference of the functions. The area is $\frac{125}{6}$, which corresponds to option (D).

    The problem asks for the area between two curves. The core idea is always the same: area between curves is the integral of (top function minus bottom function) over the interval where they cross. But the real work is in finding where they intersect and which one lies above the other.

    We have $y = x^2 - 3x + 3$ (call it $f(x)$) and $y = 2x^2 - 1$ (call it $g(x)$). Both are upward-opening parabolas. The first has vertex at $x = 1.5$, the second at $x = 0$. They will cross at two points, enclosing a finite region.

    1. Find the intersection points. Set $f(x) = g(x)$:

    $x^2 - 3x + 3 = 2x^2 - 1$

    Bring all terms to one side:

    $0 = 2x^2 - 1 - x^2 + 3x - 3 = x^2 + 3x - 4$

    So $x^2 + 3x - 4 = 0$. Factor:

    $(x + 4)(x - 1) = 0$

    Hence $x = -4$ and $x = 1$. These are the limits of integration.

    2. Determine which curve is on top in $[-4, 1]$. Pick a test point between the intersections, say $x = 0$:

    $f(0) = 0^2 - 3(0) + 3 = 3$

    $g(0) = 2(0)^2 - 1 = -1$

    Since $3 > -1$, $f(x)$ is above $g(x)$ throughout the interval. (Both are continuous parabolas crossing only at the endpoints, so the ordering cannot flip inside.)

    3. Set up the area integral. Area $A = \int_{x=-4}^{1} [f(x) - g(x)] \, dx$:

    $f(x) - g(x) = (x^2 - 3x + 3) - (2x^2 - 1) = -x^2 - 3x + 4$

    So

    $A = \int_{-4}^{1} (-x^2 - 3x + 4) \, dx$

    4. Integrate term by term.

    $\int (-x^2) \, dx = -\frac{x^3}{3}, \quad \int (-3x) \, dx = -\frac{3x^2}{2}, \quad \int 4 \, dx = 4x$

    Thus

    $A = \left[ -\frac{x^3}{3} - \frac{3x^2}{2} + 4x \right]_{-4}^{1}$

    5. Evaluate at the upper limit $x = 1$:

    $-\frac{1^3}{3} - \frac{3(1)^2}{2} + 4(1) = -\frac{1}{3} - \frac{3}{2} + 4$

    Convert to sixths: $-\frac{2}{6} - \frac{9}{6} + \frac{24}{6} = \frac{13}{6}$

    6. Evaluate at the lower limit $x = -4$: …

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12. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.Area of the region bounded by the curves $y=2^x$, $y^2=4x$ and the lines $x=\dfrac{1}{2}$, $x=1$ is (A) $\dfrac{2-\sqrt{2}}{\log 2} - \dfrac{4-\sqrt{2}}{3}$ (B) $\dfrac{4-\sqrt{2}}{3} - \dfrac{2-\sqrt{2}}{\log 2}$ (C) $\dfrac{4+\sqrt{2}}{3} - \dfrac{2-\sqrt{2}}{\log 2}$ (D) $\dfrac{2-\sqrt{2}}{\log 2} - \dfrac{4+\sqrt{2}}{3}$

    ›Reveal solutionSolution

    The area is the difference between the integral of $y=2^x$ and the integral of $y=2\sqrt{x}$ (the upper branch of $y^2=4x$) from $x=1/2$ to $x=1$, giving option (B).

    The region is bounded by two curves and two vertical lines. The key is to see which curve lies above the other in the given interval. The curve $y^2 = 4x$ is a right-opening parabola; solving for $y$ gives $y = \pm 2\sqrt{x}$. Since the region is bounded also by $y = 2^x$ (which is always positive), we only consider the upper branch $y = 2\sqrt{x}$.

    Between $x = 1/2$ and $x = 1$, we need to compare $2^x$ and $2\sqrt{x}$. At $x = 1$, $2^1 = 2$ and $2\sqrt{1} = 2$, so they meet. At $x = 1/2$, $2^{1/2} = \sqrt{2} \approx 1.414$ and $2\sqrt{1/2} = 2/\sqrt{2} = \sqrt{2} \approx 1.414$ — they also meet! So the two curves actually intersect at both endpoints. But between them, which is larger? Test a point, say $x = 3/4$: $2^{3/4} = \sqrt[4]{8} \approx 1.68$, while $2\sqrt{3/4} = \sqrt{3} \approx 1.73$. So $2\sqrt{x} > 2^x$ in the interior. That means the parabola lies above the exponential curve in this interval.

    Thus the area is the integral of (upper curve minus lower curve) from $x = 1/2$ to $x = 1$:

    $\text{Area} = \int_{1/2}^{1} \left( 2\sqrt{x} - 2^x \right) dx$

    Now we compute step by step.

    1. Integrate $2\sqrt{x}$.

       Write $2\sqrt{x} = 2x^{1/2}$. Its antiderivative is $2 \cdot \frac{x^{3/2}}{3/2} = \frac{4}{3} x^{3/2}$.

    2. Integrate $2^x$.

       The antiderivative of $a^x$ is $\frac{a^x}{\log a}$. So $\int 2^x dx = \frac{2^x}{\log 2}$.

    3. Evaluate the definite integral.

    $\left[ \frac{4}{3} x^{3/2} - \frac{2^x}{\log 2} \right]_{1/2}^{1}$

    At $x = 1$: $\frac{4}{3}(1)^{3/2} - \frac{2^1}{\log 2} = \frac{4}{3} - \frac{2}{\log 2}$.

    At $x = 1/2$: $\frac{4}{3}\left(\frac{1}{2}\right)^{3/2} - \frac{2^{1/2}}{\log 2} = \frac{4}{3} \cdot \frac{1}{2\sqrt{2}} - \frac{\sqrt{2}}{\log 2} = \frac{4}{6\sqrt{2}} - \frac{\sqrt{2}}{\log 2} = \frac{2}{3\sqrt{2}} - \frac{\sqrt{2}}{\log 2}$. …

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