Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The area enclosed by the curve $y = \min \{ \sin x, \cos x \}$ and the $x$-axis between $x = -\pi$ and $x = \pi$ is determined by summing the integrals across the intervals defined by the intersection points $x = -3\pi/4$ and $x = \pi/4$. Evaluating the components yields $\int_{0}^{\frac{\pi}{4}} \sin x \, dx = 1 - \frac{1}{\sqrt{2}}$, $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx = 1 - \frac{1}{\sqrt{2}}$, and $\int_{-\pi}^{-\frac{3\pi}{4}} (\sin x - \cos x) \, dx = -1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$. Summing these parts leads to a total area $A = 4$, which results in $A^2 = 16$.

The area of a region enclosed by two curves is determined by integrating the difference between the upper and lower boundaries over the interval defined by their intersection points. Equating $y=4x-x^{2}$ and $y=\frac{(x-4)^{2}}{3}$ reveals that the curves intersect at $x=1$ and $x=4$. Within this interval, the parabola $y=4x-x^{2}$ lies above the other curve, meaning the area is found by evaluating \int_{1}^{4} \left[ (4x-x^{2}) - \frac{(x-4)^{2}}{3} \right]\ , dx$.

Calculating the antiderivative of this expression provides $\left[ 2x^{2} - \frac{x^{3}}{3} - \frac{(x-4)^{3}}{9} \right]_{1}^{4}$. Evaluating at the upper boundary $x=4$ yields a value of $\frac{32}{3}$, while substituting the lower boundary $x=1$ results in $\frac{14}{3}$. Subtracting these two values, $\frac{32}{3} - \frac{14}{3}$, gives the final area of 6.

The area trapped between two parabolas is determined by integrating the vertical distance between them, which is the difference of their equations, from the smaller intersection point to the larger one. Equating $x^{2-}5x$ and $7x-x^2$ leads to $2x^{2-}12x=0$, which factors as $2x(x-6)=0$, establishing the integration limits as $x=0$ and $x=6$.

In this interval, $7x-x^2$ acts as the upper function, making the integral setup
$\int_{0}^{6} ((7x-x^2) - (x^{2-}5x)) \, dx = \int_{0}^{6} (12x - 2x^2) \, dx$
Finding the antiderivative yields
$\left[ 6x^2 - \frac{2}{3}x^3 \right]_0^6$
Substituting the upper limit $x=6$ results in $6(36) - \frac{2}{3}(216) = 216 - 144$, confirming the enclosed area is $72$.

The region bounded by the ellipse $x^{2+}3y^2=18$ and the line $y=x$ in the first quadrant consists of two distinct areas: a triangle and an elliptical sector. Substituting $y=x$ into the ellipse equation $x^{2+}3y^2=18$ gives $4x^2=18$, establishing the intersection point at $x=\frac{3}{\sqrt{2}}$. The area of the triangle formed by the origin, the point $(\frac{3}{\sqrt{2}}, 0)$, and the intersection point $(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})$ is calculated as $\frac{1}{2} \times \frac{3}{\sqrt{2}} \times \frac{3}{\sqrt{2}} = \frac{9}{4}$. Adding this triangular area to the integral of the ellipse from $x=\frac{3}{\sqrt{2}}$ to the x-intercept $x=3\sqrt{2}$, which is $\int_{\frac{3}{\sqrt{2}}}^{3\sqrt{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} \, dx = \sqrt{3}\pi - \frac{9}{4}$, results in a total area of $\sqrt{3}\pi$.

The area is bounded by the lower envelope of the functions $y = x^2 + 2$ and $y = 2x + 2$ over the interval $[0, 3]$. These curves intersect where $x^2 + 2 = 2x + 2$, which simplifies to $x^2 - 2x = 0$, giving intersection points at $x = 0$ and $x = 2$. Within the interval $[0, 2]$, the quadratic function $x^2 + 2$ lies below the linear function, while from $2$ to $3$, the linear function $2x + 2$ represents the lower boundary. The total area $A$ is defined by the sum of these two definite integrals: $\int_0^2 (x^2 + 2) dx + \int_2^3 (2x + 2) dx$. Evaluating the first integral yields $[\frac{x^3}{3} + 2x]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}$, and calculating the second integral results in $[x^2 + 2x]_2^3 = (9 + 6) - (4 + 4) = 7$. Combining these gives $A = \frac{20}{3} + \frac{21}{3} = \frac{41}{3}$. Thus, the value of $12A$ is $12 \times \frac{41}{3} = 164$.

To determine the region bounded by the parabolas $2y^2 = kx$ and $ky^2 = 2(y - x)$, we first find their intersection points by expressing $x$ in terms of $y$ for both equations. From the first equation, $x = \frac{2y^2}{k}$, and from the second, $x = y - \frac{ky^2}{2}$. Equating these expressions gives $\frac{2y^2}{k} = y - \frac{ky^2}{2}$, which simplifies to $y^2(\frac{2}{k} + \frac{k}{2}) = y$. Solving for $y$ yields the intersection points at $y = 0$ and $y = \frac{2k}{k^{2+}4}$.

The area of the bounded region is determined by the integral $\int_0^{\frac{2k}{k^{2+}4}} (y - (\frac{k}{2} + \frac{2}{k})y^2) \, dy$. Evaluating this integral results in the area formula $A = \frac{2}{3(k + \frac{4}{k})^2}$. To maximize the area, the denominator $(k + \frac{4}{k})^2$ must be minimized. By the arithmetic mean-geometric mean inequality, $|k + \frac{4}{k}| \geq 2\sqrt{|k \cdot \frac{4}{k}|} = 4$, which implies that the denominator reaches its minimum value of $16$ when $k = \frac{4}{k}$. Solving $k^2 = 4$ gives $k = 2$ and $k = -2$. The sum of the squares of these possible values for $k$ is $2^2 + (-2)^2 = 8$.

Calculating the area enclosed by a parabola and a line is most efficiently done by expressing the curves as functions of $y$, which treats the horizontal distance between the boundaries as the integrand. Given the parabola $y^2 = 4(x-2)$ and the line $y = 2x - 8$, these equations rearrange to $x = \frac{y^2}{4} + 2$ and $x = \frac{y}{2} + 4$, respectively.

Equating these expressions to identify the intersection points leads to $\frac{y^2}{4} + 2 = \frac{y}{2} + 4$, which simplifies to $y^2 - 2y - 8 = 0$. Factoring this quadratic gives $(y-4)(y+2) = 0$, establishing the integration limits at $y=-2$ and $y=4$. The area is found by integrating the horizontal difference $(\frac{y}{2} + 4) - (\frac{y^2}{4} + 2)$ with respect to $y$ over this interval, resulting in $\int_{-2}^{4} (-\frac{y^2}{4} + \frac{y}{2} + 2) dy$. Evaluating this definite integral yields $[-\frac{y^3}{12} + \frac{y^2}{4} + 2y]_{-2}^{4}$, which simplifies to $9$.

Calculating the area of a region bounded by curves involves integrating the difference between the upper curve and the lower curve within the relevant interval. For the circle $x^{2+}y^2=8$ and the parabola $y^2=2x$ in the first quadrant, finding the intersection point by substituting $y^2=2x$ into the circle's equation gives $x^{2+}2x-8=0$, which simplifies to $(x+4)(x-2)=0$, identifying $x=2$ as the boundary limit for the region.

The area is determined by evaluating the definite integral $\int_0^2 (\sqrt{8-x^2} - \sqrt{2x}) dx$. Evaluating the circular component $\int_0^2 \sqrt{8-x^2} dx$ using standard trigonometric substitution results in $[\frac{x}{2}\sqrt{8-x^2} + 4 \sin^{-1}(\frac{x}{2\sqrt{2}})]_0^2$, which simplifies to $2+\pi$. For the parabolic component, computing $\int_0^2 \sqrt{2}x^{1/2} dx$ yields \sqrt{2} \[\frac{2}{3}x^{3/2}]_0^2$, resulting \in a value of$\frac{8}{3}$. Subtracting the parabolic area from the circular area gives$2 + \pi - \frac{8}{3}$, which simplifies to$\pi - \frac{2}{3}$.

Finding the area enclosed by two curves relies on identifying their intersection points and then computing the definite integral of the difference between the upper and lower functions over the resulting interval. The two curves are represented as $y = \frac{16}{x+4}$ and $y = 6-x$. Equating these expressions to find the intersection points leads to $6-x = \frac{16}{x+4}$, which transforms into the quadratic equation $(6-x)(x+4) = 16$. Expanding the terms gives $x^2 - 2x - 8 = 0$, which defines the limits of integration at $x = -2$ and $x = 4$.

The area is determined by the integral:
$\int_{-2}^{4} \left( (6-x) - \frac{16}{x+4} \right) dx$
Integrating these terms gives the expression $6x - \frac{x^2}{2} - 16 \log_e|x+4|$. Evaluating this at the upper limit $x=4$ gives $6(4) - \frac{4^2}{2} - 16 \log_e(8)$, which simplifies to $16 - 48 \log_e 2$. Evaluating the same expression at the lower limit $x=-2$ yields $6(-2) - \frac{(-2)^2}{2} - 16 \log_e(2)$, or $-14 - 16 \log_e 2$. Subtracting the value at the lower limit from the value at the upper limit, $(16 - 48 \log_e 2) - (-14 - 16 \log_e 2)$, results in $30 - 32 \log_e 2$.

The region is enclosed by three distinct curves, which means finding the intersection points is essential for splitting the total area into manageable parts. Comparing $y=3x$ and $y=3x-x^{3/2}$ identifies the starting point at $x=0$, while equating $y=3x$ and $y=\frac{27-3x}{2}$ marks the transition point at $x=3$. The final boundary for the region occurs where the curves $y=3x-x^{3/2}$ and $y=\frac{27-3x}{2}$ meet at $x=9$, defining our limits of integration.

The total area $A$ is calculated by integrating the difference between the upper and lower functions over these two distinct intervals. For the interval $0 \le x \le 3$, the upper function is $y=3x$ and the lower is $y=3x-x^{3/2}$, which simplifies the integrand to $x^{3/2}$. This results in the integral $\int_{0}^{3} x^{3/2} \, dx$, evaluating to $\frac{2}{5}3^{5/2}$. For the second interval $3 \le x \le 9$, the upper function is $y=\frac{27-3x}{2}$ and the lower remains $y=3x-x^{3/2}$, creating the integral $\int_{3}^{9} (\frac{27}{2} - \frac{9x}{2} + x^{3/2}) \, dx$.

Evaluating this second integral yields $[\frac{27x}{2} - \frac{9x^2}{4} + \frac{2x^{5/2}}{5}]_{3}^{9}$, which calculates to $81 - 162 + \frac{2}{5}3^5 - \frac{2}{5}3^{5/2}$. When the areas from both intervals are combined, the terms involving $3^{5/2}$ cancel each other out. This leaves $A = \frac{486}{5} - 81$, which simplifies to $A = \frac{81}{5}$. Applying the final calculation, $10A$ is equal to 162.

The ratio of the area of a parabolic segment to the area of the triangle formed by its vertices is inherently dictated by the geometry of the curve $y=x^2$. The triangle $OPQ$ with vertices $O(0,0)$, $P(a, a^2)$, and $Q(-b, b^2)$ has an area $S_2$ calculated via the determinant formula for coordinates, resulting in
$S_2 = \frac{1}{2} \left| 0(a^2 - b^2) + a(b^2 - 0) + (-b)(0 - a^2) \right| = \frac{1}{2} ab(a+b)$
The line segment $PQ$ connecting $(a, a^2)$ and $(-b, b^2)$ possesses a slope of $\frac{a^{2-}b^2}{a+b} = a-b$. Consequently, the equation for the line $PQ$ is $y - a^2 = (a-b)(x-a)$, which simplifies to $y = (a-b)x + ab$. The area $S_1$ of the region bounded by this chord and the parabola is the definite integral of the difference between the linear equation and the parabola $y=x^2$ over the interval $[-b, a]$:
$S_1 = \int_{-b}^{a} \left( (a-b)x + ab - x^2 \right) dx = \left[ \frac{(a-b)x^2}{2} + abx - \frac{x^3}{3} \right]_{-b}^{a}$
Evaluating this integral yields $\frac{(a+b)^3}{6}$. The ratio of these two areas is then expressed as
$\frac{S_1}{S_2} = \frac{\frac{(a+b)^3}{6}}{\frac{ab(a+b)}{2}} = \frac{(a+b)^2}{3ab} = \frac{a^2 + b^2 + 2ab}{3ab} = \frac{1}{3} \left( \frac{a}{b} + \frac{b}{a} + 2 \right)$
Since $a, b > 0$, the arithmetic mean-geometric mean inequality implies $\frac{a}{b} + \frac{b}{a} \geq 2$, with equality occurring when $a=b$. Therefore, the minimum value of the expression is $\frac{1}{3}(2 + 2) = \frac{4}{3}$. With $\frac{m}{n} = \frac{4}{3}$, where $\gcd(4,3)=1$, the sum $m+n$ equals $4+3=7$.

Finding the area of a region defined by curves is most efficient when integrating with respect to the variable along which the boundaries are continuous, which in this case is the y-axis. The region is bounded by the parabola $x = (y-2)^2 + 1$ and the line $x = 2y - 4$, while being constrained by the positive coordinate axes. By calculating the total area trapped between the parabola and the y-axis from $y=0$ to $y=3$, we encompass the total area under the curve; however, we must subtract the portion beneath the line segment to isolate the exact region bounded by these three specific constraints.

The total area between the parabola and the y-axis is determined by the definite integral $\int_{0}^{3} ((y-2)^2 + 1) \, dy$, which expands to the integral $\int_{0}^{3} (y^2 - 4y + 5) \, dy$. Evaluating this expression gives $\left[ \frac{y^3}{3} - 2y^2 + 5y \right]_{0}^{3}$, resulting in a value of $9 - 18 + 15$, which equals $6$. From this sum, we subtract the area of the triangular region formed by the line $x - 2y + 4 = 0$ as it intersects the y-axis and the parabola, which is calculated as $\frac{1}{2} \times 1 \times 2 = 1$. Subtracting this triangular area from the integral result leaves a final value of $5$.

The absolute value components in these curves necessitate splitting the analysis into piecewise intervals based on the sign of $x$. For non-negative $x$, the equations simplify to $y = x^2$ and $y = 0$, which meet only at the origin, whereas for negative $x$, the equations become $y = -x^2$ and $y = 2x$. Setting $-x^2 = 2x$ reveals an intersection at $x = -2$, establishing the integration bounds from $-2$ to $0$. Over this range, the curve $y = -x^2$ remains above $y = 2x$, so the area is defined by the definite integral $\int_{-2}^{0} (-x^2 - 2x) \, dx$. Evaluating this expression results in $[-\frac{x^3}{3} - x^2]_{-2}^{0}$, which simplifies to $0 - (\frac{8}{3} - 4)$, ultimately producing a final value of $\frac{4}{3}$.

The area of the region is bounded by the parabola $y^2 = 4x$, restricting the vertical height to $2\sqrt{x}$ in magnitude, while the rational inequality $\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ determines the valid $x$-intervals. Analyzing the sign of $y$ reveals that when $y > 0$, the inequality holds for $x \in (0,1) \cup (2,3)$, and when $y < 0$, it holds for $x \in (1,2) \cup (3,4)$. Because these intervals partition the $x$-axis continuously from $0$ to $4$, the total area is found by integrating $2\sqrt{x}$ over the combined span of these ranges. Evaluating $\int_0^4 2\sqrt{x} \, dx$ as 2 \cdot \[\frac{2}{3} x^{3/2}]_0^4$provides the final result of$32/3$.

Determining the area of a region bounded by multiple inequalities requires identifying the intersection points of the boundary curves to establish appropriate limits for integration. By expressing the bounding curves with $x$ as the subject, we can compute the total area by integrating the difference between the right and left boundaries with respect to $y$, splitting the integral where the boundary definitions shift at their point of intersection.

From $y=0$ to $y=1$, the region is bounded by $x = -2y^2$ on the left and $x = 8-4y^2$ on the right, providing an area component of $\int_0^1 (8-4y^2 - (-2y^2)) dy = \int_0^1 (8-2y^2) dy$. Evaluating this expression results in $\left[8y - \frac{2y^3}{3}\right]_0^1 = \frac{22}{3}$, which is equivalent to $\frac{88}{12}$.

For the second segment from $y=1$ to $y=\frac{3}{2}$, the left boundary transitions to the line $x = 2y-4$ while the right boundary remains $x = 8-4y^2$. The corresponding integral is $\int_1^{3/2} (8-4y^2 - (2y-4)) dy = \int_1^{3/2} (12 - 2y - 4y^2) dy$, which yields $\left[12y - y^2 - \frac{4y^3}{3}\right]_1^{3/2} = \frac{19}{12}$. Summing these parts results in a total area of $\frac{107}{12}$, and because $107$ and $12$ are coprime, the sum $m+n$ equals $119$.

To find the area of the region bounded by the curves $\frac{1}{x}$ and $\frac{a}{x^2}$ over the interval from $x=1$ to $x=2$, we evaluate the definite integral $\int_{1}^{2} (\frac{1}{x} - \frac{a}{x^2}) \, dx$. Performing the integration yields the expression $[\ln x + \frac{a}{x}]_{1}^{2}$, and substituting the upper and lower limits results in $(\ln 2 + \frac{a}{2}) - (\ln 1 + a)$, which simplifies to $\ln 2 - \frac{a}{2}$. Given that this calculated area equals $\ln 2 - \frac{1}{7}$, we equate $\ln 2 - \frac{a}{2} = \ln 2 - \frac{1}{7}$ to find that $\frac{a}{2} = \frac{1}{7}$, leading to $a = \frac{2}{7}$. Finally, substituting this value into the expression $7a - 3$ gives $7(\frac{2}{7}) - 3$, which results in $-1$.

Calculating the area of the region bounded by the circle $x^2 + y^2 = 169$ and the line $5x - y = 13$ involves integrating the circular function $\sqrt{169 - y^2}$ with respect to $y$ over the appropriate bounds determined by the intersection points of the two curves. By evaluating this integral and subtracting the portion of the area defined by the line, the result is expressed as $\frac{\pi}{2} \times \frac{169}{2} - \frac{65}{2} + \frac{169}{2} \sin^{-1}\left(\frac{12}{13}\right)$.

Matching this result term-by-term with the given expression $\frac{\pi \alpha}{2 \beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1}\left(\frac{12}{13}\right)$ allows for the identification of the unknown variables. The coefficient of the inverse sine function, $\frac{\alpha}{\beta}$, is clearly $\frac{169}{2}$. Given that $\alpha$ and $\beta$ must be coprime, we assign $\alpha = 169$ and $\beta = 2$. Consequently, summing these two values yields $\alpha + \beta = 169 + 2 = 171$.

The area is bounded by the minimum of two curves, so we first determine the intersection point by solving $2x = 6x - x^2$, which occurs at $x = 0$ and $x = 4$. Within the interval $0 \le x \le 4$, the function $y = 2x$ is smaller, forming a triangle with a base of 4 and a height of 8 that contributes $\frac{1}{2} \times 4 \times 8 = 16$ to the total area. For the interval $4 \le x \le 6$, the parabola $y = 6x - x^2$ lies below the line, so we compute the area using the integral $\int_{4}^{6} (6x - x^2) \, dx$, which evaluates to $\frac{28}{3}$. Adding these two components results in a total area of $A = 16 + \frac{28}{3} = \frac{76}{3}$, and multiplying this final value by 12 gives 304.

The region defined by $y \ge x^2$, $y \ge (1-x)^2$, and $y \le 2x(1-x)$ is bounded above by the parabola $y = 2x(1-x)$ and below by the curves $y = x^2$ and $y = (1-x)^2$. Because these curves exhibit symmetry about the vertical line $x = \frac{1}{2}$, the total area can be determined by integrating the difference between the upper and lower boundaries over the interval $[\frac{1}{3}, \frac{1}{2}]$ and doubling the result. Within this specific interval, the upper boundary is $y = 2x - 2x^2$ and the lower boundary is $y = (1-x)^2$.

The total area $A$ is expressed by the definite integral:
$A = 2 \int_{\frac{1}{3}}^{\frac{1}{2}} (2x - 2x^2 - (1-x)^2) dx$
Simplifying the integrand yields $2x - 2x^2 - (1 - 2x + x^2) = -3x^2 + 4x - 1$. Evaluating the integral provides:
$A = 2 \left[ -x^3 + 2x^2 - x \right]_{\frac{1}{3}}^{\frac{1}{2}} = 2 \left( \left( -\frac{1}{8} + 2\left(\frac{1}{4}\right) - \frac{1}{2} \right) - \left( -\frac{1}{27} + 2\left(\frac{1}{9}\right) - \frac{1}{3} \right) \right)$
$A = 2 \left( -\frac{1}{8} - \left( -\frac{4}{27} \right) \right) = 2 \left( \frac{-27 + 32}{216} \right) = 2 \left( \frac{5}{216} \right) = \frac{5}{108}$
Multiplying this result by $540$ confirms the final value:
$540 A = 540 \times \frac{5}{108} = 25$

The roots of a quadratic equation with rational coefficients are rational if and only if its discriminant is a perfect square. For the equation $x^{2}-p x+\frac{5}{4} p=0$, the discriminant is $D=(-p)^{2}-4(1)(\frac{5}{4} p)=p^{2}-5p$. We require $D$ to be a perfect square for integer values of $p$ within the range $[0, 10]$. Checking the values in this interval, $p=9$ yields $D=81-45=36$, which is $6^{2}$, while no larger integer $p \le 10$ results in a perfect square. Thus, we identify $q=9$.

The area of the region is defined by the definite integral $\int_{0}^{9}(x-9)^{2} dx$, which represents the area under the parabola $(x-9)^{2}$ from $x=0$ to $x=9$. Evaluating this integral, we obtain $\left[\frac{(x-9)^{3}}{3}\right]_{0}^{9}=\frac{0^{3}}{3}-\frac{(-9)^{3}}{3} = \frac{729}{3}=243$.