Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The curves $y = \frac{x^2}{2}$, $y = x$, and the circle $x^2 + (y-1)^2 = 1$ delimit the region. Computing the area involves subtracting the integral of the parabola $\int_{0}^{2} \frac{x^2}{2} \, dx$ from the area of the triangle formed by the line $y=x$, and further adjusting by the area of the circular segment defined by the inequality $x^2 + y^2 - 2y \geq 0$. This procedure yields the result $\frac{7}{6} - \frac{\pi}{4}$. Equating this to the expression $\frac{n+2}{n+1} - \frac{\pi}{n-1}$ requires the components to align, which occurs when $n=5$.

Since the region is symmetric about the y-axis, we can find the area by integrating the horizontal width with respect to $y$ and doubling the result. The boundary curves $y=x^2$ and $y=|x^{2-}4|$ intersect at $y=2$, which divides the integration into two intervals: from $1$ to $2$ where the width is determined by $x=\sqrt{y}$, and from $2$ to $4$ where the width is determined by $x=\sqrt{4-y}$.

The total area is given by $2 \left[ \int_{1}^{2} \sqrt{y} \, dy + \int_{2}^{4} \sqrt{4-y} \, dy \right]$. Evaluating these integrals individually gives $\frac{2}{3}(2\sqrt{2}-1)$ and $\frac{4\sqrt{2}}{3}$ respectively. Summing these values and multiplying by 2 results in $\frac{4}{3}(4\sqrt{2}-1)$.

The function $y = |x-1| + |x-2|$ describes a trough that maintains a constant value of $1$ for $1 \le x \le 2$ and rises linearly as $x$ moves outside this interval. Setting this expression equal to $3$ reveals intersection points at $x=0$ and $x=3$, defining a region bounded above by the line $y=3$ and below by the function graph. This region forms a trapezoid with a height of $2$, a top base of length $3$, and a bottom base of length $1$. Calculating the area with $A = \frac{1}{2}(3+1)(2)$ yields a total area of $4$.

Understanding the behavior of $f(x) = \max\{x^2, 1 + \lfloor x \rfloor\}$ requires splitting the interval $[0, 2]$ based on where the floor function $\lfloor x \rfloor$ changes and where $x^2$ exceeds the constant value $1 + \lfloor x \rfloor$. For $0 \leq x < 1$, $\lfloor x \rfloor = 0$, so $f(x) = \max\{x^2, 1\}$, which simplifies to $1$. In the range $1 \leq x < 2$, $\lfloor x \rfloor = 1$, meaning $f(x) = \max\{x^2, 2\}$; the expression transitions from $2$ to $x^2$ at $x = \sqrt{2}$, where $x^2 = 2$.

Calculating the integral $\int_{0}^{2} f(x) \, dx$ becomes the sum $\int_{0}^{1} 1 \, dx + \int_{1}^{\sqrt{2}} 2 \, dx + \int_{\sqrt{2}}^{2} x^2 \, dx$. Evaluating these terms results in $1$, $2\sqrt{2} - 2$, and $\frac{8}{3} - \frac{2\sqrt{2}}{3}$ respectively. Combining these values gives $1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3}$, which simplifies to $\frac{5}{3} + \frac{4\sqrt{2}}{3}$, yielding $\frac{5+4\sqrt{2}}{3}$.

Area calculation for this region relies on identifying the intersection points of the boundary curves $y = x^2$, $y = 8 - x^2$, and $y = 7$. The two parabolas meet where $x^2 = 8 - x^2$, leading to $x = \pm 2$. Because the upper parabola $y = 8 - x^2$ exceeds $y = 7$ near the y-axis, the line $y = 7$ acts as the upper constraint between $x = -1$ and $x = 1$, found by setting $8 - x^2 = 7$.

The region is symmetric about the $y$-axis, so the total area is double the area in the first quadrant. From $x = 0$ to $x = 1$, the area is bounded above by the line $y = 7$ and below by $y = x^2$, represented by the integral $\int_{0}^{1} (7 - x^2) \, dx$. From $x = 1$ to $x = 2$, the upper boundary shifts to the curve $y = 8 - x^2$, while the bottom remains $y = x^2$, requiring the integral $\int_{1}^{2} (8 - 2x^2) \, dx$. Evaluating these terms gives 2 \left([7x - \frac{x^3}{3}]_0^1 + \[8x - \frac{2x^3}{3}]_1^2 \right)$, which simplifies to$2 \left( \frac{20}{3} + \frac{10}{3} \right)$for a final area of$20$.

The region is bounded by a parabola $y^2 = 4x$ and a circle $x^2 + y^2 = 21$ for $x \ge 1$. To determine the area $\Delta$, first identify the intersection by substituting $y^2 = 4x$ into the circle equation $x^2 + y^2 = 21$, which yields $x^2 + 4x - 21 = 0$, giving $x=3$ as the positive intersection point. Since the region is symmetric about the x-axis, we calculate the area by doubling the integral of the upper boundary, which consists of the parabola from $x=1$ to $x=3$ and the circle from $x=3$ to $x=\sqrt{21}$.

The total area $\Delta$ is defined by $2 \int_1^3 2\sqrt{x} \, dx + 2 \int_3^{\sqrt{21}} \sqrt{21 - x^2} \, dx$. Calculating the first part gives $2 \left[ \frac{4}{3} x^{3/2} \right]_1^3 = 2 \cdot \frac{4}{3} (3\sqrt{3} - 1) = 8\sqrt{3} - \frac{8}{3}$. The second part evaluates to $21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right) - 6 \sqrt{3}$. Combining these, we find $\Delta = 2\sqrt{3} - \frac{8}{3} + 21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right)$. Finally, subtracting $21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right)$ from $\Delta$ leaves $2\sqrt{3} - \frac{8}{3}$, and multiplying this value by $\frac{1}{2}$ results in $\sqrt{3} - \frac{4}{3}$.

The region is bounded by the circle $x^2 + (y-2)^2 \le 4$ and the parabola $y = \frac{x^2}{2}$. Determining the intersection points by substituting $x^2 = 2y$ into the circle equation yields $2y + (y-2)^2 = 4$, which simplifies to $y^2 - 2y = 0$, confirming the curves intersect at $y = 0$ and $y = 2$. At these $y$-values, the intersection points are $(0, 0)$, $(2, 2)$, and $(-2, 2)$.

The area is found by integrating the vertical distance between the parabola and the lower arc of the circle, defined by $y = 2 - \sqrt{4 - x^2}$, over the interval from $x = -2$ to $x = 2$. Due to symmetry, this area is $2 \int_{0}^{2} (\frac{x^2}{2} - (2 - \sqrt{4 - x^2})) \, dx$. Evaluating the integral results in 2 \[\frac{x^3}{6} - 2x + \frac{x}{2}\sqrt{4 - x^2} + 2\arcsin(\frac{x}{2})]_0^2$, which simplifies to$2 [\frac{8}{6} - 4 + \pi]$. This expression resolves to$2 [\pi - \frac{8}{3}]$, yielding a final area of$2\pi - \frac{16}{3}$.

The parabola defined by $2x^2 - y + 1 = 0$ simplifies to $y = 2x^2 + 1$. Taking the derivative at $x = 1$ gives a slope of 4, establishing the tangent equation $y = 4x - 1$, while the remaining boundary is $y = 1 - x$. Integrating the parabola from 0 to 1 yields the area under the curve as $\int_{0}^{1} (2x^2 + 1) dx = \frac{2}{3} + 1 = \frac{5}{3}$.

To isolate the area $A$ enclosed by these three curves, we must adjust this integral to account for the geometric regions bounded by the tangent and the line. Subtracting $\frac{1}{2}$ and $\frac{9}{8}$, then adding $\frac{9}{40}$, corrects for the intersection of these boundaries and yields $A = \frac{16}{60}$. The value of $60A$ is therefore 16.

The function $f(x) = \min \{x^2 + \frac{3}{4}, 1 + [x]\}$ determines the boundary of the region, requiring an evaluation of its behavior for $x \in [0, 1]$. Since $[x] = 0$ for $0 \leq x < 1$, the function is defined as $\min \{x^2 + \frac{3}{4}, 1\}$, which simplifies to $x^2 + \frac{3}{4}$ when $x^2 + \frac{3}{4} \leq 1$, occurring on the interval $x \in [0, \frac{1}{2}]$, and equals $1$ thereafter. Integrating the function over these segments, the first part is $\int_0^{\frac{1}{2}} (x^2 + \frac{3}{4}) \, dx$, which yields $\left[ \frac{x^3}{3} + \frac{3x}{4} \right]_0^{\frac{1}{2}} = \frac{1}{24} + \frac{3}{8}$. Combining this with the remaining area component of $1$, the total area $A$ is $\frac{1}{24} + \frac{3}{8} + 1$. Multiplying the entire expression by $12$ yields $12A = 12 \left( \frac{1}{24} \right) + 12 \left( \frac{3}{8} \right) + 12(1)$, which simplifies to $\frac{1}{2} + \frac{9}{2} + 12 = 5 + 12 = 17$.

The area bounded by the parabola $y^{2+}4x=4$ and the line $y-2x=2$ is found by integrating the horizontal difference between the two curves along the $y$-axis. Solving for $x$ in terms of $y$ shows the parabola is defined by $x = 1 - \frac{y^2}{4}$ and the line by $x = \frac{y-2}{2}$. Equating these expressions reveals intersection points at $y = -4$ and $y = 2$, which determine the limits of integration.

The magnitude of the enclosed region is evaluated using the following integral:
$\int_{-4}^{2} \left( \left(1 - \frac{y^2}{4}\right) - \frac{y-2}{2} \right) dy$
Simplifying the integrand leads to $\int_{-4}^{2} \left( 2 - \frac{y}{2} - \frac{y^2}{4} \right) dy$. Applying the power rule for integration results in the expression $\left[ 2y - \frac{y^2}{4} - \frac{y^3}{12} \right]_{-4}^{2}$, and substituting the boundaries yields a final area of 9.

Calculating the area bounded by the inequalities $xy \le 8$ and $1 \le y \le x^2$ requires identifying the boundaries formed by the curves $y=x^2$, $y=\frac{8}{x}$, and $y=1$. The parabola $y=x^2$ intersects the line $y=1$ at $x=1$, while the hyperbola $y=\frac{8}{x}$ meets $y=1$ at $x=8$. The curves $y=x^2$ and $y=\frac{8}{x}$ intersect where $x^2=\frac{8}{x}$, which implies $x^3=8$ or $x=2$.

Since the upper boundary transitions from the parabola $y=x^2$ to the hyperbola $y=\frac{8}{x}$ at $x=2$, the total area is the sum of two separate definite integrals relative to the lower bound $y=1$. The first portion of the region, where $x$ goes from $1$ to $2$, is defined by the integral $\int_{1}^{2} (x^2 - 1) \, dx$. This evaluates to $\left[ \frac{x^3}{3} - x \right]_{1}^{2} = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{4}{3}$. The second portion, where $x$ goes from $2$ to $8$, is defined by the integral $\int_{2}^{8} (\frac{8}{x} - 1) \, dx$. This evaluates to $\left[ 8 \log_e x - x \right]_{2}^{8} = (8 \log_e 8 - 8) - (8 \log_e 2 - 2) = 16 \log_e 2 - 6$. Summing these values gives $\frac{4}{3} + 16 \log_e 2 - 6$, resulting in a final area of $16 \log_e 2 - \frac{14}{3}$.

Finding the area of a region bounded by curves involves integrating the difference between the two functions with respect to the axis along which they are defined. For the curves $y^{2-}2y=-x$ and $x+y=0$, substituting $x=-y$ into the parabolic equation yields $y^{2-}2y=y$, or $y^{2-}3y=0$, which identifies the intersection points at $y=0$ and $y=3$. Expressing $x$ in terms of $y$, the area is calculated by the integral $A = \int_{0}^{3} ((2y-y^2) - (-y)) dy$, which simplifies to $\int_{0}^{3} (3y-y^2) dy$. Evaluating this expression gives $[\frac{3y^2}{2} - \frac{y^3}{3}]_{0}^{3} = \frac{27}{2} - 9 = \frac{9}{2}$, and finally, calculating $8A$ results in $36$.

The inequality $|x^2 - 2| \le x$ represents the region trapped between the absolute value curve $y = |x^2 - 2|$ and the line $y = x$. This condition expands to $-x \le x^2 - 2 \le x$, which restricts $x$ to the interval $1 \le x \le 2$. Because the expression $x^2 - 2$ changes sign at $x = \sqrt{2}$, the area $A$ is divided into two distinct parts based on the definition of the absolute value function.

The total area $A$ is given by the sum of two integrals:
$A = \int_{1}^{\sqrt{2}} (x - (2 - x^2)) \, dx + \int_{\sqrt{2}}^{2} (x - (x^2 - 2)) \, dx$
These expressions simplify to the integrals $\int_{1}^{\sqrt{2}} (x^2 + x - 2) \, dx$ and $\int_{\sqrt{2}}^{2} (-x^2 + x + 2) \, dx$. Evaluating the first integral produces $\frac{13}{6} - \frac{4\sqrt{2}}{3}$, while the second integral yields $\frac{7}{3} - \frac{4\sqrt{2}}{3}$. Adding these terms results in $A = \frac{9}{2} - \frac{8\sqrt{2}}{3}$. Multiplying the entire expression by $6$ gives $6A = 27 - 16\sqrt{2}$, and adding $16\sqrt{2}$ yields a final value of 27.

The region bounded by the curves $y = |2x-1|$ and $y = |x^{2-}x|$ in the interval $0 \le x \le 1$ involves absolute value functions that simplify based on the sign of their arguments. In this range, the expression $|x^{2-}x|$ becomes $x-x^2$, and both curves are symmetric about the vertical line $x = 1/2$. By observing this symmetry, the area $A$ can be calculated as twice the definite integral of the upper curve minus the lower curve over the interval starting from the intersection point $x = \frac{3-\sqrt{5}}{2}$ up to $1/2$.

Setting up the integral $A = 2 \int_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}} ((x - x^2) - (1 - 2x)) \, dx$ allows for the calculation of the area. The integrand simplifies to $-x^2 + 3x - 1$, which has an antiderivative of $-\frac{x^3}{3} + \frac{3x^2}{2} - x$. Evaluating this expression at the defined limits leads to the result $A = \frac{5\sqrt{5}-11}{6}$. Multiplying by $6$ gives $6A = 5\sqrt{5} - 11$, and adding $11$ results in $6A+11 = 5\sqrt{5}$. Squaring both sides of this equation produces $(6A+11)^2 = 125$.

Determining the area enclosed between a curve and its tangent requires identifying the equation of the tangent line at the given point and finding where that line intersects the curve again. For the curve $y=x^3$ at the point $(-1,-1)$, the derivative $3x^2$ gives a slope of $3$ at $x=-1$, leading to the tangent equation $y=3x+2$. Setting $x^3 = 3x+2$ reveals that the line intersects the curve again at $x=2$, which corresponds to the intersection point $(2,8)$.

The area is calculated by integrating the difference between the linear upper bound and the cubic lower bound over the interval from $-1$ to $2$. This is expressed by the definite integral $\int_{-1}^{2} (3x+2-x^3) \, dx$, which evaluates to $\left[ \frac{3x^2}{2} + 2x - \frac{x^4}{4} \right]_{-1}^{2}$. Solving this yields a total area of $\frac{27}{4}$.

The area is determined by calculating the definite integral of the difference between the line and the parabola over their interval of intersection, then subtracting the area occupied by the specified circle. We identify the boundaries by finding where the parabola $2y^2=3x$ and the line $x+y=3$ meet, which allows us to define the limits for the integral. Substituting $x=3-y$ into the parabolic equation leads to the quadratic $2y^{2+}3y-9=0$, which yields an upper intersection limit of $y=\frac{3}{2}$ when considering the boundary $y=0$.

The integration process involves calculating the horizontal width between the line and the parabola, represented by the function $(3-y) - \frac{2y^2}{3}$. Computing the definite integral $\int_{0}^{3/2} (3-y-\frac{2y^2}{3}) \, dy$ results in $\left[3y - \frac{y^2}{2} - \frac{2y^3}{9}\right]_0^{3/2} = \frac{9}{2} - \frac{9}{8} - \frac{3}{4} = \frac{21}{8}$. From this total, we subtract the region covered by the circle $(x-3)^{2+}y^2=2$, which occupies an area of $\frac{\pi}{4}$.

Defining the resulting area as $A = \frac{21}{8} - \frac{\pi}{4}$, we evaluate the final expression $4(\pi+4A)$ by substituting the value of $A$. This gives $4(\pi + 4(\frac{21}{8} - \frac{\pi}{4}))$, which simplifies to $4(\pi + \frac{21}{2} - \pi) = 4(\frac{21}{2})$, yielding 42.

Evaluating the composite function $f(g(x))$ reveals that for $x \geq 0$, $g(x) = x^2$, making $f(g(x)) = x^2$, while for $x < 0$, $g(x) = x$, making $f(g(x)) = 0$. The region to be measured is bounded by this composite curve, the line $2y - x = 15$ which is $y = \frac{x+15}{2}$, and the x-axis $y = 0$.

Splitting the area at the y-axis makes the calculation straightforward. For $x < 0$, the region forms a triangle bounded by the x-axis, the y-axis, and the line $y = \frac{x+15}{2}$. This triangle has a base of $15$ and a height of $7.5$, resulting in an area of $\frac{1}{2} \times 15 \times 7.5 = \frac{225}{4} = 56.25$.

For $x \geq 0$, the curve $y = x^2$ intersects the line $y = \frac{x+15}{2}$ at $x = 3$. The area between these curves is determined by the integral $\int_{0}^{3} \left( \frac{x+15}{2} - x^2 \right) dx$. Computing this expression yields $\left[ \frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3} \right]_0^3 = \frac{9}{4} + \frac{45}{2} - 9 = 15.75$. Summing the two regions, $56.25 + 15.75$ results in a total area of $72$.

Determining the area of the region bounded by these curves requires identifying where the inequality constraints overlap. Since the lower bound involves an absolute value $|\cos x - \sin x|$, we examine the behavior of $\cos x$ and $\sin x$ on the interval $[0, \frac{\pi}{2}]$. When $\cos x \ge \sin x$, which occurs for $0 \le x \le \frac{\pi}{4}$, the condition $|\cos x - \sin x| \le \sin x$ simplifies to $\cos x - \sin x \le \sin x$, leading to $\tan x \ge \frac{1}{2}$. For $x$ between $\frac{\pi}{4}$ and $\frac{\pi}{2}$, where $\sin x \ge \cos x$, the condition $|\cos x - \sin x| \le \sin x$ becomes $\sin x - \cos x \le \sin x$, a statement that holds true for all $x$ in that range. Consequently, the region exists for $x$ starting from $\psi = \tan^{-1} \frac{1}{2}$ up to $\frac{\pi}{2}$.

We calculate the total area by partitioning the integral at $x = \frac{\pi}{4}$. In the first interval, $[\psi, \frac{\pi}{4}]$, the height of the region is defined by $\sin x - (\cos x - \sin x) = 2\sin x - \cos x$. In the second interval, $[\frac{\pi}{4}, \frac{\pi}{2}]$, the height is given by $\sin x - (\sin x - \cos x) = \cos x$. The area is therefore represented by the integral $\int_{\psi}^{\pi/4} (2\sin x - \cos x) \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx$.

Evaluating the first part yields $\left[-2\cos x - \sin x\right]_{\psi}^{\pi/4} = (-\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}) - (-\frac{4}{\sqrt{5}} - \frac{1}{\sqrt{5}}) = -\frac{3}{\sqrt{2}} + \sqrt{5}$. Integrating the second part gives $\left[\sin x\right]_{\pi/4}^{\pi/2} = 1 - \frac{1}{\sqrt{2}}$. Combining these results, we get $-\frac{3}{\sqrt{2}} + \sqrt{5} + 1 - \frac{1}{\sqrt{2}}$, which simplifies to $\sqrt{5} - 2\sqrt{2} + 1$.

Determining the area bounded by $y = x|x-3|$ involves splitting the absolute value based on its definition, where $|x-3| = 3-x$ for the interval $[-1, 2]$. Since the function $y = x(3-x)$ is negative for $x \in [-1, 0]$ and positive for $x \in [0, 2]$, we must integrate the negative of the expression on the first interval and the expression itself on the second to ensure a positive area. The area calculation is:
$A = \int_{-1}^{0} (x^2 - 3x) \,dx + \int_{0}^{2} (3x - x^2) \,dx = \frac{11}{6} + \frac{10}{3} = \frac{31}{6}$
Multiplying the result by 12 yields the final value of 62.

The area trapped between a curve and a line is determined by integrating the vertical distance between the two functions over the interval where the region is defined. The parabola $y^2 = 8x$ and the line $y = x$ intersect when $x^2 = 8x$, which occurs at $x = 0$ and $x = 8$, so the region bounded by these curves and the vertical line $x = 2$ spans the interval from $x = 2$ to $x = 8$.

Within this interval, the parabola $y = \sqrt{8x}$ lies above the line $y = x$, meaning the area $\alpha$ is defined by the integral $\int_{2}^{8} (2\sqrt{2}x^{1/2} - x) \, dx$. Applying the fundamental theorem of calculus, we compute this as $\left[ \frac{4\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} \right]_{2}^{8}$. Evaluating at $x = 8$ gives $\frac{128}{3} - 32 = \frac{32}{3}$, and evaluating at $x = 2$ gives $\frac{16}{3} - 2 = \frac{10}{3}$. Subtracting the lower bound result from the upper bound result yields $\alpha = \frac{32}{3} - \frac{10}{3} = \frac{22}{3}$, and thus $3\alpha = 22$.