Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

Determining the area requires identifying the periodic nature of function composition. Repeatedly composing $f(x) = \frac{x-1}{x+1}$ shows that $f^4(x) = x$. By taking the remainder of 26 divided by 4, which is 2, we find that $f^{26}(x) = f^2(x) = -\frac{1}{x}$. Given the relationship $g(x) + f^{26}(x) = 0$, the function is defined as $g(x) = \frac{1}{x}$.

The geometry of the region involves the curves $y = \frac{1}{x}$ and $y = x - \frac{3}{2}$, which is the simplified form of $2y = 2x - 3$. The line intersects the x-axis at $x = \frac{3}{2}$ and meets the curve $y = \frac{1}{x}$ at $x = 2$. Consequently, the total area consists of two parts: the integral of the linear function from $x = \frac{3}{2}$ to $x = 2$, and the integral of the rational function from $x = 2$ to $x = 4$. Integrating the linear segment yields $\frac{1}{8}$, and integrating the rational segment yields $\log_{e} 2$, resulting in a final area of $\frac{1}{8} + \log_{e} 2$.

The area of a region bounded by curves $y_2(x)$ and $y_1(x)$ between $x=a$ and $x=b$ is defined by $\int_{a}^{b} (y_2 - y_1) dx$. The intersection points of $y = 1+x^2$ and $y = 3-x$ are determined by $1+x^2 = 3-x$, which simplifies to $x^{2+}x-2=0$ or $(x+2)(x-1) = 0$, giving the limits $x=-2$ and $x=1$. The line $x=-1$ divides this region into two segments corresponding to the intervals $[-2, -1]$ and $[-1, 1]$.

Evaluating the area $A_1$ for the interval $[-2, -1]$ yields:
$A_1 = \int_{-2}^{-1} (3-x - (1+x^2)) dx = \int_{-2}^{-1} (2-x-x^2) dx = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{-1}$
Substituting the limits results in $A_1 = \left(-2 - \frac{1}{2} + \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right) = -\frac{13}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6}$.

Evaluating the area $A_2$ for the interval $[-1, 1]$ yields:
$A_2 = \int_{-1}^{1} (2-x-x^2) dx = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^{1} = \left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-2 - \frac{1}{2} + \frac{1}{3}\right) = \frac{7}{6} - \left(-\frac{13}{6}\right) = \frac{20}{6} = \frac{10}{3}$

The ratio of the areas $A_1 : A_2$ is $\frac{7}{6} : \frac{20}{6}$, which simplifies to $7:20$, implying $m=7$ and $n=20$. Since $\gcd(7, 20) = 1$, the sum $m+n$ equals $7+20 = 27$.

Finding the area of a region bounded by two curves begins by identifying their points of intersection and determining which curve acts as the upper boundary on the interval defined by those points. For the hyperbola $16x^{2}-9 y^{2}=144$ and the line $8x-3 y=24$, we solve for the intersection by substituting the line's equation, rewritten as $y = \frac{8x-24}{3}$, into the hyperbola's equation, which simplifies to the quadratic $x^{2}-8x+15=0$, identifying the intersection points at $x=3$ and $x=5$. Within this interval, the hyperbola lies above the line, so the area $A$ is given by the definite integral $\int_{3}^{5} \left( \frac{4}{3} \sqrt{x^{2}-9} - \frac{8x-24}{3} \right) dx$.

Evaluating the integral of the hyperbola term requires the standard form $\int \sqrt{x^{2}-a^{2}} dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \log_{e}(x+\sqrt{x^{2}-a^{2}})$, which, when multiplied by the constant $\frac{4}{3}$ and evaluated from $3$ to $5$, yields $\frac{40}{3} - 12 \log_{e} 3 + 6 \log_{e} 3$. Subtracting the integral of the linear component, which evaluates to $\frac{16}{3}$ over the same interval, provides the final area $A = 8 - 6 \log_{e} 3$. Substituting this value into the expression $3(A+6 \log_{e} 3)$ simplifies the calculation to $3(8 - 6 \log_{e} 3 + 6 \log_{e} 3)$, resulting in a final value of $24$.

The region is defined by $x \ge 0, y \ge 0$ and bounded by $y \ge 1-2x$ and $y \le 4-x^2$. The line $y=1-2x$ intersects the x-axis at $x=0.5$, and the parabola $y=4-x^2$ intersects the x-axis at $x=2$. For $0 \le x \le 0.5$, the region lies between the curves $y=1-2x$ and $y=4-x^2$, while for $0.5 \le x \le 2$, the region is bounded by $y=0$ and $y=4-x^2$.

The area $A$ is expressed by the sum of these two definite integrals:
$A = \int_0^{0.5} ((4-x^2) - (1-2x)) \, dx + \int_{0.5}^2 (4-x^2) \, dx$
Evaluating the first integral yields:
$\int_0^{0.5} (3 + 2x - x^2) \, dx = \left[ 3x + x^2 - \frac{x^3}{3} \right]_0^{0.5} = 1.5 + 0.25 - \frac{0.125}{3} = \frac{7}{4} - \frac{1}{24} = \frac{41}{24}$
Evaluating the second integral yields:
$\int_{0.5}^2 (4-x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{0.5}^2 = \left( 8 - \frac{8}{3} \right) - \left( 2 - \frac{0.125}{3} \right) = \frac{16}{3} - 2 + \frac{1}{24} = \frac{10}{3} + \frac{1}{24} = \frac{81}{24}$
Summing the results gives $A = \frac{41}{24} + \frac{81}{24} = \frac{122}{24} = \frac{61}{12}$. Given $\alpha = 61$ and $\beta = 12$, both being coprime, the value of $\alpha + \beta$ is $61 + 12 = 73$.

Calculating the area enclosed by $y = x^{2}$, $y = 27/x$, and the boundary $y = 1$ requires identifying the intersection point where $x^{2} = 27/x$, which simplifies to $x^{3} = 27$, giving $x = 3$. Because the region is bound between $y = 1$ and these two curves, it splits into two distinct segments at this intersection point. From $x = 1$ to $x = 3$, the area lies between the parabola $y = x^{2}$ and the base line $y = 1$, while from $x = 3$ to $x = 27$, the region is bounded by the hyperbola $y = 27/x$ and the same line $y = 1$.

Summing the two integrals provides the total area:
$\int_{1}^{3} (x^{2} - 1) \, dx + \int_{3}^{27} \left( \frac{27}{x} - 1 \right) \, dx$
The first integral evaluates to $\left[ \frac{x^{3}}{3} - x \right]_{1}^{3}$, which is $(9 - 3) - (1/3 - 1)$, resulting in $20/3$. The second integral becomes $[27 \ln x - x]_{3}^{27}$, giving $(27 \ln 27 - 27) - (27 \ln 3 - 3)$. Simplifying $27 \ln 27$ to $81 \ln 3$ yields $54 \ln 3 - 24$. Combining these results, $20/3 - 24$ simplifies to $-52/3$, leading to the final expression $54 \ln 3 - \frac{52}{3}$.

The area $f(\alpha)$ of the region in the first quadrant bounded by $x=0, x=1, y=\sqrt{x}$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^2$ is defined by the definite integral $f(\alpha) = \int_0^1 |(|\alpha x-5|-|1-\alpha x|+\alpha x^2) - \sqrt{x}| \, dx$. For $\alpha=0$, the function simplifies to $y=|0-5|-|1-0|+0=4$, resulting in the integral f(0) = \int_0^1 (4-\sqrt{x}) \, dx = \[4x - \frac{2}{3}x^{3/2}]_0^1 = 4 - \frac{2}{3} = \frac{10}{3}$.

For $\alpha=1$, the function becomes $y=|x-5|-|1-x|+x^2$. On the interval $x \in [0, 1]$, $x-5 < 0$ and $1-x \geq 0$, so $|x-5|=5-x$ and $|1-x|=1-x$, which simplifies the function to $y=(5-x)-(1-x)+x^2 = 4+x^2$. The area is f(1) = \int_0^1 (4+x^{2-}\sqrt{x}) \, dx = \[4x + \frac{x^3}{3} - \frac{2}{3}x^{3/2}]_0^1 = 4 + \frac{1}{3} - \frac{2}{3} = \frac{11}{3}$. Combining these results yields$f(0)+f(1) = \frac{10}{3} + \frac{11}{3} = \frac{21}{3} = 7$.

Determining the area enclosed between two curves defined as functions of $y$ involves integrating the difference between the rightmost curve and the leftmost curve over the interval defined by their points of intersection. The given curves are $x = -3y^2$ and $x = 1 - 4y^2$. Setting these equations equal yields $-3y^2 = 1 - 4y^2$, which simplifies to $y^2 = 1$, identifying the limits of integration as $y = -1$ and $y = 1$.

Observing that $1 - 4y^2$ represents the right boundary and $-3y^2$ represents the left boundary within this interval, the area calculation is established by the definite integral:
$A = \int_{-1}^{1} ((1 - 4y^2) - (-3y^2)) \, dy = \int_{-1}^{1} (1 - y^2) \, dy$
Because the integrand is an even function, the area is calculated by doubling the integral over the positive half of the interval:
$A = 2 \int_{0}^{1} (1 - y^2) \, dy = 2 \left[ y - \frac{y^3}{3} \right]_0^1 = 2 \left( 1 - \frac{1}{3} \right) = \frac{4}{3}$

Because the functions $y \leq \pi - |x|$ and $y \leq |x \sin x|$ are symmetric about the y-axis, we calculate the area for $x \geq 0$ and double the result. The region in the first quadrant is defined by the minimum of $x \sin x$ and $\pi - x$, which requires splitting the integral at the intersection point $x = \frac{\pi}{2}$ where both curves equal $\frac{\pi}{2}$.

Integrating $x \sin x$ from 0 to $\frac{\pi}{2}$ by parts results in 1, while the area bounded by $\pi - x$ from $\frac{\pi}{2}$ to $\pi$ results in $\frac{\pi^2}{8}$.

Summing these two areas and doubling the result gives the final calculation:
$2 \left( \int_{0}^{\frac{\pi}{2}} x \sin x \, dx + \int_{\frac{\pi}{2}}^{\pi} (\pi - x) \, dx \right) = 2 + \frac{\pi^2}{4}$

The area of the ellipse $x^2 + 4y^2 = 4$, which can be written in standard form as $\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1$, is given by $\pi \times 2 \times 1 = 2\pi$. The curves $y = |x| - 1$ and $y = 1 - |x|$ define a rhombus with vertices at $(0, 1), (1, 0), (0, -1),$ and $(-1, 0)$. This rhombus has diagonals of lengths $2$ and $2$, resulting in an area of $\frac{1}{2} \times 2 \times 2 = 2$. Because the vertices of the rhombus lie on or inside the ellipse, subtracting the rhombus area from the ellipse area provides the result $2\pi - 2$, which simplifies to $2(\pi - 1)$.

The area $A$ is given by the integral $A = \int_0^{3\pi/2} |\max\{\sin x, \cos x\}| \, dx$. By partitioning the interval $[0, 3\pi/2]$ into sub-intervals where the function is defined and considering its sign relative to the $x$-axis, the area is calculated as the sum of the integrals $\int_0^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx + \int_{\pi}^{5\pi/4} (-\sin x) \, dx + \int_{5\pi/4}^{3\pi/2} (-\cos x) \, dx$. Evaluating these integrals yields $A = \frac{1}{\sqrt{2}} + (1 + \frac{1}{\sqrt{2}}) + (1 - \frac{1}{\sqrt{2}}) + (1 - \frac{1}{\sqrt{2}}) = 3$. Therefore, $A + A^2 = 3 + 3^2 = 12$.

The function $\varphi(x)$ is determined by the composition of the given inverse functions $f^{-1}$ and $g^{-1}$. Given that $g$ is a strictly increasing function mapping the set $\{1, e, e^{2}, e^{3}\}$ to $\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$, the value $\frac{1}{2}$ corresponds to the domain element $e^{2}$, establishing that $g^{-1}(\frac{1}{2}) = e^{2}$. Since $f$ is a strictly decreasing bijection from $\{1, 2, 3, 4\}$ to $\{1, e, e^{2}, e^{3}\}$, the element $e^{2}$ is mapped to by $2$ in the domain of $f$, meaning $f^{-1}(e^{2}) = 2$. Consequently, $\varphi(x) = 2^{x}$.

The area of the region $R$ is found by evaluating the definite integral $\int_{0}^{1} (2^{x} - x^{2}) \, dx$, where the curve $y = 2^{x}$ acts as the upper boundary and $y = x^{2}$ as the lower boundary. The antiderivative of $2^{x}$ is $\frac{2^{x}}{\log_{e}(2)}$ and the antiderivative of $x^{2}$ is $\frac{x^{3}}{3}$. Computing the difference at the boundaries $x=1$ and $x=0$ results in $\left( \frac{2}{\log_{e}(2)} - \frac{1}{3} \right) - \left( \frac{1}{\log_{e}(2)} - 0 \right)$, which simplifies to $\frac{1}{\log_{e}(2)} - \frac{1}{3}$, ultimately yielding $\frac{3-\log_{e}(2)}{3\log_{e}(2)}$.

Finding the area of a region bounded by multiple curves is best approached by expressing the boundaries in terms of a single variable, which simplifies the integration process. Given the constraints $0 \le y \le 6-x$, $y^{2} \ge 4x-3$, and $x \ge 0$, we can rearrange the inequalities to isolate $x$ as a function of $y$, yielding $x \le 6-y$ and $x \le \frac{y^{2+}3}{4}$, with the requirement $x \ge 0$. The two bounding curves intersect where $6-y = \frac{y^{2+}3}{4}$, which simplifies to the quadratic equation $y^{2+}4y-21=0$, and since $y \ge 0$, we find the intersection at $y=3$.

The total area is determined by integrating with respect to $y$, splitting the integral at the intersection point $y=3$. From $y=0$ to $y=3$, the boundary is $x = \frac{y^{2+}3}{4}$, and from $y=3$ to $y=6$, the boundary is $x = 6-y$. The first integral evaluates to $\int_{0}^{3} \frac{y^{2+}3}{4} dy = \frac{1}{4} \left[ \frac{y^3}{3} + 3y \right]_0^3 = \frac{1}{4}(9+9) = \frac{9}{2}$. The second integral evaluates to $\int_{3}^{6} (6-y) dy = \left[ 6y - \frac{y^2}{2} \right]_3^6 = (36-18) - \left(18 - \frac{9}{2}\right) = 18 - 13.5 = \frac{9}{2}$. Combining these two parts, the total area of the region is $\frac{9}{2} + \frac{9}{2} = 9$.

To find the area bounded by two curves, we integrate the difference between the upper curve and the lower curve over the interval defined by their points of intersection. We begin by equating the two boundary functions $y = x^2 - 8x$ and $y = -x$ to determine these limits. Setting $x^2 - 8x = -x$ leads to the equation $x^2 - 7x = 0$, which factors as $x(x - 7) = 0$, identifying our intersection points at $x = 0$ and $x = 7$.

Within this interval, the line $y = -x$ lies above the parabola $y = x^2 - 8x$, so the area is given by the integral of the difference between these functions:
$\int_{0}^{7} (-x - (x^2 - 8x)) \, dx$
Simplifying the expression inside the integral yields $7x - x^2$, allowing us to write the integral as:
$\int_{0}^{7} (7x - x^2) \, dx$
Performing the integration term by term results in $\frac{7x^2}{2} - \frac{x^3}{3}$, which we then evaluate at the boundaries of $0$ and $7$. Calculating this at the upper limit gives $\frac{7(49)}{2} - \frac{343}{3}$, or $\frac{343}{2} - \frac{343}{3}$. By converting these terms to a common denominator of $6$, we reach the final area of $\frac{343}{6}$.

The area $A$ between two curves $f(x)$ and $g(x)$ over an interval $[a, b]$ is given by the definite integral $\int_a^b |f(x) - g(x)| dx$. For the parabolas $y=4x^2$ and $y=x^{2+}27$, the intersection points are found by setting $4x^2 = x^{2+}27$, which yields $3x^2 = 27$ or $x = \pm 3$. The area $A_1$ is calculated as \int_{-3}^3 (x^{2+}27-4x^2) dx = \int_{-3}^3 (27-3x^2) dx = \[27x - x^3]_{-3}^3 = (81-27) - (-81+27) = 108$.

For the region bounded by $y=4x^2$ and $y=\alpha x$, the intersection points occur where $4x^2 = \alpha x$, giving $x=0$ and $x=\alpha/4$. The area $A_2$ is \int_0^{\alpha/4} (\alpha x - 4x^2) dx = \[\frac{\alpha x^2}{2} - \frac{4x^3}{3}]_0^{\alpha/4} = \frac{\alpha^3}{32} - \frac{4\alpha^3}{3 \cdot 64} = \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96}$. Setting the condition$A_1 = 6A_2$provides$108 = 6 \cdot \frac{\alpha^3}{96}$, which simplifies to$108 = \frac{\alpha^3}{16}$. Solving for$\alpha$gives$\alpha^3 = 1728$, resulting \in$\alpha = 12$.

The intersection points of the ellipse $4x^2 + y^2 = 8$ and the parabola $y^2 = 4x$ are found by substituting $y^2 = 4x$ into the ellipse equation to obtain $4x^2 + 4x - 8 = 0$, which simplifies to $(x+2)(x-1) = 0$. Since $x \ge 0$ for the parabola, the intersection occurs at $x = 1$, which corresponds to $y = \pm 2$.

The area of the region is computed by integrating the difference between the boundary functions $x = \frac{1}{2}\sqrt{8-y^2}$ and $x = \frac{y^2}{4}$ with respect to $y$ over the interval $[-2, 2]$. Utilizing symmetry, the area is $2 \int_{0}^{2} (\frac{1}{2}\sqrt{8-y^2} - \frac{y^2}{4}) \, dy$, which simplifies to $\int_{0}^{2} \sqrt{8-y^2} \, dy - \int_{0}^{2} \frac{y^2}{2} \, dy$.

Evaluating the first term using the integral formula $\int \sqrt{a^{2-}u^2} \, du = \frac{u}{2} \sqrt{a^{2-}u^2} + \frac{a^2}{2} \sin^{-1}(\frac{u}{a})$ with $a^2=8$ gives $[\frac{y}{2}\sqrt{8-y^2} + 4 \sin^{-1}(\frac{y}{2\sqrt{2}})]_0^2 = (1 \cdot \sqrt{4} + 4 \cdot \frac{\pi}{4}) - 0 = 2 + \pi$. The second term is calculated as \int_{0}^{2} \frac{y^2}{2} \, dy = \[\frac{y^3}{6}]_0^2 = \frac{8}{6} = \frac{4}{3}$. Subtracting the two values yields$\pi + 2 - \frac{4}{3} = \pi + \frac{2}{3}$.

The area enclosed by two continuous functions $f(x)$ and $g(x)$ over an interval $[a, b]$ is given by the definite integral $\int_a^b |f(x) - g(x)| \, dx$. For the first region $A_1$, the boundary curves are $y=x^{2+}2$ and $y=8-x$, which intersect at $x=2$. The bounded area in the first quadrant is calculated as
$A_1 = \int_{0}^{2} ((8-x) - (x^{2+}2)) \, dx = \int_{0}^{2} (6-x-x^2) \, dx = \left[6x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{0}^{2} = \frac{22}{3}$
For the second region $A_2$, the curves are $y=x^{2+}2$ and $y=\sqrt{x}$ (the positive branch of $y^2=x$) between $x=0$ and $x=2$. Evaluating the integral gives
$A_2 = \int_{0}^{2} ((x^{2+}2) - \sqrt{x}) \, dx = \left[\frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2}\right]_{0}^{2} = \frac{8}{3} + 4 - \frac{4\sqrt{2}}{3} = \frac{20-4\sqrt{2}}{3}$
Subtracting these areas yields
$A_1 - A_2 = \frac{22}{3} - \frac{20-4\sqrt{2}}{3} = \frac{2+4\sqrt{2}}{3} = \frac{2}{3}(1+2\sqrt{2})$

The intersection area is determined by integrating the difference between the upper boundary of the lower circle, $y = \sqrt{4-x^2}$, and the lower boundary of the upper circle, $y = 2 - \sqrt{4-x^2}$, over the interval defined by their intersection points. Solving the system $x^{2+}y^2=4$ and $x^{2+}(y-2)^2=4$ yields $y=1$, which leads to $x = \pm\sqrt{3}$ as the limits of integration.

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

Exploiting symmetry across the y-axis, the integral becomes $4 \int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1) \, dx$. Using the integral formula $\int \sqrt{a^{2-}x^2} \, dx = \frac{x}{2}\sqrt{a^{2-}x^2} + \frac{a^2}{2}\arcsin(\frac{x}{a})$, the calculation is as follows:

$4 \left[ \frac{x}{2}\sqrt{4-x^2} + 2\arcsin\left(\frac{x}{2}\right) - x \right]_{0}^{\sqrt{3}} = 4 \left( \frac{\sqrt{3}}{2} + 2\left(\frac{\pi}{3}\right) - \sqrt{3} \right)$

Simplifying the expression results in $4(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{8\pi}{3} - 2\sqrt{3}$, which factors to $\frac{2}{3}(4\pi - 3\sqrt{3})$.

The function $f(x)$ is identified by observing that the given equation $f(x)+3f\left(\frac{\pi}{2}-x\right)=\sin x$ constitutes one part of a linear system. Substituting $\frac{\pi}{2}-x$ for $x$ produces the second equation $f\left(\frac{\pi}{2}-x\right)+3f(x)=\cos x$. Solving these simultaneously leads to $f(x) = \frac{1}{8}(3\cos x - \sin x)$. Applying the standard amplitude formula for trigonometric combinations, the maximum value $\alpha$ is found to be $\frac{\sqrt{10}}{8}$.

The region bounded by $g(x)=x^{2}$ and $h(x)=\beta x^{3}$ is enclosed between the intersection points $x=0$ and $x=\frac{1}{\beta}$. Evaluating the definite integral $\int_{0}^{\frac{1}{\beta}} (x^{2} - \beta x^{3}) \, dx$ results in the area $\frac{1}{12\beta^{3}}$. Equating this expression to $\alpha^{2} = \frac{10}{64}$ produces the relation $\frac{1}{12\beta^{3}} = \frac{5}{32}$, which yields the final result $30\beta^{3} = 16$.