Q.Using integration, find the area lying above x-axis and included between the circle $x^2 + y^2 = 8x$ and inside the parabola $y^2 = 4x$.

Showing the 12 most recent of 17 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.The area bounded by the curve $y = x|x|$, $x$-axis and the ordinates $x = -1$ and $x = 1$ is given by (A) $0$ (B) $\frac{1}{3}$ (C) $\frac{2}{3}$ (D) $3$

   ›Reveal solutionSolution

   The curve $y = x|x|$ is an odd function, so the signed area cancels to zero, but the bounded area (absolute area) is the sum of two equal positive lobes, giving $\frac{2}{3}$.

   The key here is to understand what $y = x|x|$ actually looks like. The absolute value on $x$ splits the definition into two cases:

   • When $x \ge 0$, $|x| = x$, so $y = x \cdot x = x^2$.
   • When $x < 0$, $|x| = -x$, so $y = x \cdot (-x) = -x^2$.

   So the curve is a parabola opening upward on the right side, and a parabola opening downward on the left side. It is an odd function: $f(-x) = -f(x)$. This symmetry is the heart of the problem.

   The question asks for the area bounded by the curve, the $x$-axis, and the vertical lines $x = -1$ and $x = 1$. "Area bounded" means geometric area — always positive — not signed area (the integral). This is a classic trap.

   Let’s work through it.

   1. Set up the absolute area integral. The geometric area between a curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is $\int_a^b |f(x)| \, dx$. Here:

   $\text{Area} = \int_{-1}^{1} |x|x|| \, dx.$

   2. Simplify $|x|x||$. Since $|x|x|| = |x| \cdot |x| = |x|^2 = x^2$ (because squaring removes the sign), we have:

   $|x|x|| = x^2 \quad \text{for all real } x.$

   That’s a neat simplification: the absolute value of the function is just $x^2$, a simple upward parabola.

   3. Compute the integral.

   $\text{Area} = \int_{-1}^{1} x^2 \, dx.$

   The antiderivative of $x^2$ is $\frac{x^3}{3}$. So: …

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2. CBSE 2026Set 65/2/11 markMCQ
   Q.Which of the following expressions will give the area of region bounded by the curve $y = x^2$ and line $y = 16$? (A) $\int_{0}^{4} x^2\, dx$ (B) $2 \int_{0}^{4} x^2\, dx$ (C) $\int_{0}^{16} \sqrt{y}\, dy$ (D) $2 \int_{0}^{16} \sqrt{y}\, dy$

   ›Reveal solutionSolution

   The region between $y = x^2$ and $y = 16$ is symmetric about the $y$-axis; integrating horizontally from $y = 0$ to $y = 16$ with $x = \sqrt{y}$ and doubling for both sides gives $2 \int_{0}^{16} \sqrt{y}\, dy$.

   The parabola $y = x^2$ opens upward with vertex at the origin, and the horizontal line $y = 16$ cuts it at two points. Finding those intersection points: $x^2 = 16$ gives $x = \pm 4$. So the bounded region sits between $x = -4$ and $x = 4$, below the line and above the parabola.

   The key decision is whether to integrate with respect to $x$ (vertical slices) or $y$ (horizontal slices). Both are valid, but the setup differs.

   Vertical slices (integrating with respect to $x$):

   At any $x$ between $-4$ and $4$, a vertical strip runs from the parabola $y = x^2$ up to the line $y = 16$. The height of that strip is $16 - x^2$. The area is

   $A = \int_{-4}^{4} (16 - x^2)\, dx.$

   Because the integrand $16 - x^2$ is even (symmetric about $x = 0$), this equals

   $A = 2 \int_{0}^{4} (16 - x^2)\, dx = 2 \int_{0}^{4} 16\, dx - 2 \int_{0}^{4} x^2\, dx.$

   Notice that $\int_{0}^{4} x^2\, dx$ alone is not the area; it gives the area under the parabola from $0$ to $4$, not the region between the parabola and the line. So option (A) is incorrect, and option (B) is also incorrect (it's twice the area under the parabola, not the region we want).

   Horizontal slices (integrating with respect to $y$):

   At any height $y$ between $0$ and $16$, a horizontal strip extends from the left branch of the parabola to the right branch. Solving $y = x^2$ for $x$ gives $x = \pm \sqrt{y}$. The width of the strip is

   $\sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}.$

   The area is then

   $A = \int_{0}^{16} 2\sqrt{y}\, dy = 2 \int_{0}^{16} \sqrt{y}\, dy.$

   This matches option (D). …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is

   (a) $2$

   (b) $4/9$

   (c) $9/4$

   (d) $9/2$

   ›Reveal solutionSolution

   Integrate $x$ as a function of $y$ (since the boundary is the $y$-axis and the line $y=3$) using $x=y^2/4$ from the parabola.

   From $y^2=4x$, $x=\dfrac{y^2}{4}$.

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is:

   (a) 2

   (b) $\dfrac{9}{4}$

   (c) $\dfrac{9}{3}$

   (d) $\dfrac{9}{2}$

   ›Reveal solutionSolution

   Integrate $x=\dfrac{y^2}{4}$ with respect to $y$ from $0$ to $3$, since the region is bounded by the $y$-axis.

   For $y^2=4x$, we have $x=\dfrac{y^2}{4}$.

   The area bounded by the curve, the $y$-axis, and $y=3$ (from $y=0$ to $y=3$) is

   …

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5. CBSE 2025Set 65/1/11 markMCQ
   Q.The area of the shaded region bounded by the curves $y^2 = x, x = 4$ and the $x$-axis is given by (A) $\int_{0}^{4} x \, dx$ (B) $\int_{0}^{2} y^2 \, dy$ (C) $2 \int_{0}^{4} \sqrt{x} \, dx$ (D) $\int_{0}^{4} \sqrt{x} \, dx$

   ›Reveal solutionSolution

   The shaded region is the area under the parabola $y^2 = x$ from $x=0$ to $x=4$, above the $x$-axis. This area equals $\int_{0}^{4} \sqrt{x} \, dx$, which matches option (D).

   The problem asks for the area of the region bounded by $y^2 = x$, the vertical line $x = 4$, and the $x$-axis. Let’s first picture what’s happening.

   The curve $y^2 = x$ is a right-opening parabola with its vertex at the origin. For a given $x$, $y = \pm \sqrt{x}$. The $x$-axis is $y=0$, and the line $x=4$ cuts off the region on the right. The “shaded region” is typically the part above the $x$-axis — that is, the area under the upper half of the parabola from $x=0$ to $x=4$.

   So we want the area between $y = \sqrt{x}$ (the upper branch), the $x$-axis, and the vertical line $x=4$.

   ---

   1. Set up the integral with respect to $x$

   The upper boundary is $y = \sqrt{x}$, the lower boundary is $y = 0$, and $x$ runs from $0$ to $4$. The area is:

   $\text{Area} = \int_{x=0}^{4} (\sqrt{x} - 0) \, dx = \int_{0}^{4} \sqrt{x} \, dx$

   That’s exactly option (D).

   ---

   2. Check the other options

   • (A) $\int_{0}^{4} x \, dx$ would give the area under the line $y = x$, not under $\sqrt{x}$. That’s a different shape entirely.
   • (B) $\int_{0}^{2} y^2 \, dy$ comes from rewriting $x = y^2$ and integrating with respect to $y$ from $y=0$ to $y=2$ (since at $x=4$, $y=2$). That actually gives the same numerical area — but the question asks for the area of the region bounded by the given curves and the $x$-axis, and the standard representation in $x$ is $\int \sqrt{x} \, dx$. Option (B) is a valid alternative form, but it’s not the one listed that matches the direct $x$-integral.
   • (C) $2 \int_{0}^{4} \sqrt{x} \, dx$ would give the area of the full parabola (both upper and lower halves), which is twice the shaded region. …

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6. CBSE 2025Set 65/2/11 markMCQ
   Q.The area of the shaded region (figure) represented by the curves $y = x^2,\ 0 \le x \le 2$, and the $y$-axis is given by: (A) $\displaystyle\int_0^2 x^2\,dx$ (B) $\displaystyle\int_0^2 \sqrt{y}\,dy$ (C) $\displaystyle\int_0^4 x^2\,dx$ (D) $\displaystyle\int_0^4 \sqrt{y}\,dy$

   ›Reveal solutionSolution

   When integrating along the $y$-axis for a region bounded by $y = x^2$ from $x = 0$ to $x = 2$, we express $x$ in terms of $y$ and integrate with respect to $y$ over the corresponding range $0 \le y \le 4$. The answer is (D) $\displaystyle\int_0^4 \sqrt{y}\,dy$.

   The question asks for the area of a region bounded by the parabola $y = x^2$ (from $x = 0$ to $x = 2$) and the $y$-axis. The key is recognizing that we can compute area by integrating either horizontally or vertically, and the setup depends entirely on which variable we choose as our integration variable.

   When we integrate with respect to $x$, we sum vertical strips of width $dx$ and height $y = x^2$. When we integrate with respect to $y$, we sum horizontal strips of width $dy$ and length equal to the horizontal distance from the $y$-axis to the curve.

   Let me identify what happens at the boundaries. At $x = 0$, we have $y = 0^2 = 0$. At $x = 2$, we have $y = 2^2 = 4$. So as $x$ ranges from $0$ to $2$, the variable $y$ ranges from $0$ to $4$.

   Now let's examine each option:

   1. Option (A): $\displaystyle\int_0^2 x^2\,dx$

      This integrates with respect to $x$ from $0$ to $2$, summing vertical strips of height $x^2$. This gives the area under the curve $y = x^2$ from $x = 0$ to $x = 2$, which is indeed the region described. This is a valid representation.

   2. Option (B): $\displaystyle\int_0^2 \sqrt{y}\,dy$

      This integrates with respect to $y$, but only from $0$ to $2$. Since the curve reaches $y = 4$ when $x = 2$, stopping at $y = 2$ would only capture part of the region. This is incorrect.

   3. Option (C): $\displaystyle\int_0^4 x^2\,dx$

      This integrates with respect to $x$ from $0$ to $4$, which extends beyond the given domain $0 \le x \le 2$. This would compute the area under the parabola all the way to $x = 4$, which is not our region. This is incorrect.

   4. Option (D): $\displaystyle\int_0^4 \sqrt{y}\,dy$ …

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7. CBSE 2025Set ANNUAL1 mark
   Q.If area of triangle is 35 sq. units with vertices $(2, -6)$, $(5, 4)$ and $(k, 4)$, then $k$ is ______ .

   ›Reveal solutionSolution

   Using the determinant formula for the area of a triangle with the given vertices and setting it to 35 gives two valid values of $k$.

   For vertices $(x_1,y_1)=(2,-6)$, $(x_2,y_2)=(5,4)$, $(x_3,y_3)=(k,4)$, the area is

   $\text{Area} = \frac12\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big|$

   $= \frac12\big|2(4-4)+5(4-(-6))+k(-6-4)\big| = \frac12|0+50-10k| = \frac12|50-10k|$ …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is:

   (a) 2

   (b) $\dfrac{9}{4}$

   (c) $\dfrac{9}{8}$

   (d) $\dfrac{9}{2}$

   ›Reveal solutionSolution

   Integrate $x$ as a function of $y$ (since $x=y^2/4$) from $y=0$ to $y=3$.

   The parabola is $y^2 = 4x$, so $x = \dfrac{y^2}{4}$.

   Area bounded by the curve, the $y$-axis, and the line $y=3$:

   …

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9. CBSE 2023Set 65/2/11 markMCQ
   Q.If $(a, b)$, $(c, d)$ and $(e, f)$ are the vertices of $\triangle ABC$ and $\Delta$ denotes the area of $\triangle ABC$, then $\begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$ is equal to:

   (a) $2\Delta^2$

   (b) $4\Delta^2$

   (c) $2\Delta$

   (d) $4\Delta$

   ›Reveal solutionSolution

   The area of a triangle can be expressed using a determinant of its vertices. The given expression is the square of a determinant which is the transpose of the one used in the area formula, leading to a result of $\mathbf{4\Delta^2}$.

   Concept and Intuition

   The area of a triangle whose vertices are given by coordinates is a fundamental concept in coordinate geometry. While you might be familiar with the base-height formula or Heron's formula, when coordinates are involved, a powerful tool is the determinant.

   The determinant method for calculating the area of a triangle arises from vector geometry. If we consider two vectors forming two sides of a triangle, say $\vec{AB}$ and $\vec{AC}$, then the area of the triangle is half the magnitude of their cross product, i.e., $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. When these vectors are expressed in coordinates, this cross product magnitude simplifies to a determinant.

   Alternatively, you can think of it as a generalization of the "shoelace formula" for polygon areas. The determinant essentially calculates a signed area, where the sign depends on the order of vertices (clockwise or counter-clockwise). Since area is always positive, we take the absolute value of the determinant.

   The area $\Delta$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

   $\Delta = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

   The absolute value bars are crucial because the determinant itself can be negative, but area must be positive.

   Step-by-Step Solution

   1. Identify the vertices and the standard area formula: The vertices of $\triangle ABC$ are given as $(a, b)$, $(c, d)$, and $(e, f)$. Using the determinant formula for the area of a triangle, we can write:

   $\Delta = \frac{1}{2} \left| \begin{vmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{vmatrix} \right|$

   2. Isolate the determinant from the area formula: From the formula above, we can multiply both sides by 2:

   $2\Delta = \left| \begin{vmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{vmatrix} \right|$

   Let's denote the determinant inside the absolute value as $D$:
   

   $D = \begin{vmatrix} a & b & 1 \\ c & d & 1 \\ e & f & 1 \end{vmatrix}$

   So, we have $2\Delta = |D|$.
   

   3. Consider the given expression:

   We need to evaluate $\begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$.

   Let's call the determinant in this expression $D'$.

   $D' = \begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}$

   4. Relate $D'$ to $D$ using determinant properties: A fundamental property of determinants states that the determinant of a matrix is equal to the determinant of its transpose. That is, $\det(A) = \det(A^T)$. If we compare $D$ and $D'$, we can see that $D'$ is the transpose of $D$. …

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10. CBSE 2023Set 65/3/11 markMCQ
    Q.Let $A$ be the area of a triangle having vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$. Which of the following is correct ?

    (a) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm A$

    (b) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A$

    (c) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm \frac{A}{2}$

    (d) $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = A^2$

    ›Reveal solutionSolution

    The area of a triangle $A$ with given vertices is half the absolute value of a specific $3 \times 3$ determinant. This means the determinant itself is equal to $\pm 2A$.

    The area of a triangle in coordinate geometry is a fundamental concept. While you might be familiar with the base-height formula, when the vertices are given as coordinates, a more direct formula exists. This formula can be elegantly expressed using a determinant, which is what this question explores.

    The core idea is that a determinant involving the coordinates of the vertices provides a value that is directly proportional to the area of the triangle. The sign of this determinant tells us about the orientation of the vertices (whether they are listed in a clockwise or counter-clockwise order), while its absolute value gives twice the area. Since area is always a positive quantity, we take the absolute value of the determinant expression.

    1. Recall the Area Formula for a Triangle with Given Vertices The area $A$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

    $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

    The absolute value is crucial here because area must be non-negative. The expression inside the absolute value can be positive or negative depending on the order in which the vertices are taken.
    
    > [!FORMULA]
    > The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:
    > $$A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$
    

    2. Define the Determinant in Question

    Let's consider the determinant given in the options:

    $D = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

    3. Expand the Determinant We expand this $3 \times 3$ determinant along the first row:

    $D = x_1 \begin{vmatrix} y_2 & 1 \\ y_3 & 1 \end{vmatrix} - y_1 \begin{vmatrix} x_2 & 1 \\ x_3 & 1 \end{vmatrix} + 1 \begin{vmatrix} x_2 & y_2 \\ x_3 & y_3 \end{vmatrix}$

    Now, evaluate the $2 \times 2$ determinants:
    

    $D = x_1(y_2 \cdot 1 - 1 \cdot y_3) - y_1(x_2 \cdot 1 - 1 \cdot x_3) + 1(x_2 y_3 - y_2 x_3)$

    $D = x_1(y_2 - y_3) - y_1(x_2 - x_3) + (x_2 y_3 - x_3 y_2)$

    Rearranging the terms to match the area formula's structure:
    

    $D = x_1(y_2 - y_3) + x_2 y_3 - x_2 y_1 + x_3 y_1 - x_3 y_2$

    This can be rewritten as:
    

    $D = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$

    Notice that this is exactly the expression inside the absolute value in the area formula from Step 1.
    

    4. Relate the Determinant to the Area

    From Step 1, we have $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.

    From Step 3, we found that $D = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$.

    Therefore, we can write:

    $A = \frac{1}{2} |D|$

    Multiplying both sides by 2, we get: …
    

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11. CBSE 2023Set ANNUAL1 markMCQ
    Q.The area of the region bounded by the curve $y=x^2$ and the line $y=4$ is:

    (a) $\frac{33}{2}$

    (b) $\frac{8}{3}$

    (c) $\frac{32}{3}$

    (d) $\frac{4}{3}$

    ›Reveal solutionSolution

    Find where $y=x^2$ meets $y=4$, then integrate the vertical strip $(4-x^2)$ between those limits.

    The curve $y=x^2$ meets $y=4$ where $x^2=4$, i.e. $x=\pm2$.

    By symmetry about the $y$-axis, the required area is

    …

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12. CBSE 2023Set ANNUAL1 mark
    Q.Find the area of the region bounded by $y^2=9x; x=2, x=4$ and the $x$-axis in the first quadrant.

    ›Reveal solutionSolution

    Express $y$ in terms of $x$ from $y^2=9x$ (first quadrant, so $y\ge0$) and integrate between $x=2$ and $x=4$.

    $y^2=9x\Rightarrow y=3\sqrt{x}$ (taking the positive root, first quadrant)

    …

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