Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The geometric region enclosed by the circle $x^{2+}y^2=1$ and the curves $|y|=1-x^2$ is determined by taking the total area of the unit circle and subtracting the four symmetric parabolic segments. Evaluating the definite integral $\int_{0}^{1} (1-x^2) \, dx$ yields the value $\frac{2}{3}$, and multiplying this by $4$ indicates that the combined area of these segments is $\frac{8}{3}$. Subtracting this from the circle's total area $\pi$ results in $\alpha = \pi - \frac{8}{3}$. Multiplying the entire expression by $9$ leads to $9\pi - 24$, allowing for the identification of the integers $\beta = 9$ and $\gamma = -24$ within the form $\beta\pi + \gamma$. Finding the absolute difference $|\beta - \gamma|$ requires calculating $|9 - (-24)|$, which results in $33$.

The region is bounded by two curves defined in terms of $y$: $x = \frac{1}{1+y^2}$ and $x = \frac{y^2}{2}$. Because these functions express $x$ directly as a function of $y$, the most efficient way to calculate the area is to integrate with respect to $y$ across the interval defined by their intersection points. We find these points by equating the two expressions:

$\frac{1}{1+y^2} = \frac{y^2}{2}$

Multiplying across results in $y^4 + y^2 - 2 = 0$, which can be factored as $(y^{2+}2)(y^{2-}1)=0$. Since the term $(y^{2+}2)$ has no real solutions, the intersection points are determined by $y^2 = 1$, or $y = \pm 1$. Observing the relative positions of the curves, the expression $\frac{1}{1+y^2}$ maintains a greater $x$-value than the parabola $\frac{y^2}{2}$ throughout the interval $[-1, 1]$, making it the right-hand boundary. The area is found by integrating the difference between these functions over the identified range:

$\text{Area} = \int_{-1}^{1} \left( \frac{1}{1+y^2} - \frac{y^2}{2} \right) dy$

Integrating term by term yields $[\tan^{-1} y - \frac{y^3}{6}]_{-1}^{1}$. Substituting the upper and lower limits, the expression becomes $(\tan^{-1}(1) - \frac{1}{6}) - (\tan^{-1}(-1) - \frac{-1}{6})$. This simplifies to $(\frac{\pi}{4} - \frac{1}{6}) - (-\frac{\pi}{4} + \frac{1}{6})$, which results in a total area of $\frac{\pi}{2} - \frac{1}{3}$.

The area $A$ of the bounded region is determined by integrating the vertical difference between the upper boundary and the lower boundary over the specified interval. Based on the given constraints, the region is bounded by the curve $y = -3\sqrt{x}$ and the line $y = 3x - 6$ between the limits of $x = 0$ and $x = 1$. The total area $A$ is expressed by the definite integral $\int_{0}^{1} (-3\sqrt{x} - (3x - 6)) \, dx$.

This integral simplifies to $\int_{0}^{1} (-3x^{1/2} - 3x + 6) \, dx$. Integrating term by term, we find the antiderivative to be $-2x^{3/2} - \frac{3}{2}x^2 + 6x$. Evaluating this expression at the upper limit of $1$ results in $-2(1) - \frac{3}{2}(1) + 6(1)$, which is $-2 - 1.5 + 6 = 2.5$. Since the lower limit of $0$ contributes nothing to the evaluation, the area $A$ is $5/2$. Consequently, multiplying this value by $6$ gives $6 \times (5/2) = 15$.

The area between two curves is calculated by integrating the difference between the upper boundary and the lower boundary over the interval defined by their points of intersection. Setting the two equations $4 - \frac{x^2}{4}$ and $\frac{x-4}{2}$ equal reveals that they intersect at $x = -6$ and $x = 4$. Within this interval, the parabola lies above the linear function, so the area $\alpha$ is found by evaluating the integral $\int_{-6}^{4} (4 - \frac{x^2}{4} - \frac{x-4}{2}) dx$.

Simplifying the integrand leads to $6 - \frac{x^2}{4} - \frac{x}{2}$, which is then integrated as follows:
$\int_{-6}^{4} \left(6 - \frac{x^2}{4} - \frac{x}{2}\right) dx = \left[ 6x - \frac{x^3}{12} - \frac{x^2}{4} \right]_{-6}^{4}$
Evaluating this expression at the upper limit of $4$ yields $\frac{44}{3}$, while the value at the lower limit of $-6$ is $-27$. Calculating the net area $\alpha$ by subtracting the lower value from the upper value gives $\frac{44}{3} - (-27) = \frac{125}{3}$. Multiplying this area by $6$ provides the final result of $6 \times \frac{125}{3} = 250$.

Determining the larger area between $x^{2+}y^2=25$ and $y=|x-1|$ begins by finding where the curves meet, which occurs by solving $x^{2+}(x-1)^2=25$ to identify intersection points at $x=-3$ and $x=4$. The area of the smaller portion between these boundaries is found using the definite integral $\int_{-3}^{4} (\sqrt{25-x^2} - |x-1|) \, dx$. Subtracting this integral value from the total circle area of $25\pi$ leaves the area of the larger portion, which evaluates to the form $\frac{1}{4}(75\pi + 2)$. Comparing this to the given expression identifies $b=75$ and $c=2$, which sums to a final result of 77.

The area $A$ of the region bounded by $f(x) = 4\sqrt{x}$ and $g(x) = |x-5|$ is determined by evaluating the integral $\int (f(x) - g(x)) dx$ over the range where $f(x) \geq g(x)$. To identify the limits of integration, we solve $4\sqrt{x} = |x-5|$, which yields the intersection points $x=1$ and $x=25$.

Due to the absolute value function, the area is computed by splitting the integral at $x=5$:
$A = \int_{1}^{5} (4\sqrt{x} - (5-x)) dx + \int_{5}^{25} (4\sqrt{x} - (x-5)) dx$
Evaluating the first integral gives:
$\int_{1}^{5} (4x^{1/2} + x - 5) dx = \left[ \frac{8}{3}x^{3/2} + \frac{x^2}{2} - 5x \right]_{1}^{5} = \left(\frac{40\sqrt{5}}{3} - \frac{25}{2}\right) - \left(\frac{8}{3} + \frac{1}{2} - 5\right) = \frac{40\sqrt{5}}{3} - \frac{32}{3}$
Evaluating the second integral gives:
$\int_{5}^{25} (4x^{1/2} - x + 5) dx = \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{2} + 5x \right]_{5}^{25} = \left(\frac{1000}{3} - \frac{375}{2}\right) - \left(\frac{40\sqrt{5}}{3} + \frac{25}{2}\right) = \frac{400}{3} - \frac{40\sqrt{5}}{3}$
Summing these results, we find $A = \frac{40\sqrt{5}}{3} - \frac{32}{3} + \frac{400}{3} - \frac{40\sqrt{5}}{3} = \frac{368}{3}$. Consequently, $3A = 3 \times \frac{368}{3} = 368$.

The area enclosed between two curves is found by integrating their difference over the interval defined by their intersection and the specified boundary. In the region where $x \leq 0$, the expression $|e^x - 1|$ simplifies to $1 - e^x$, and setting this equal to $e^x$ gives $2e^x = 1$, identifying the intersection at $x = \ln(1/2)$. With the $y$-axis defined as $x=0$, the area is calculated by integrating the difference function $(2e^x - 1)$ from the limit $x = \ln(1/2)$ to $0$. Computing the antiderivative $[2e^x - x]_{\ln(1/2)}^{0}$ evaluates to $2 - (1 + \log_{e} 2)$, which simplifies to $1 - \log_{e} 2$.

Because the absolute value function changes behavior at $x = 0$, we determine the area by splitting the integral across the interval $[-1, 1]$. When $x$ is between $-1$ and $0$, the expression $e^{|x|} - e^{-x}$ simplifies to zero since $|x| = -x$, which reduces the height of the region to the constant $a$. For the positive portion of the interval, where $0 \leq x \leq 1$, the height of the region is defined by the function $a + e^x - e^{-x}$.

Evaluating the first integral over $[-1, 0]$ yields a value of $a$, while calculating the second integral over $[0, 1]$ involves evaluating the antiderivative $ax + e^x + e^{-x}$ at the boundaries, resulting in $a + e + \frac{1}{e} - 2$. Summing these two segments gives a total area of $2a + e + \frac{1}{e} - 2$. Equating this to the provided area of $\frac{e^{2+}8e+1}{e}$, which simplifies to $e + 8 + \frac{1}{e}$, allows us to remove the common terms $e$ and $\frac{1}{e}$ from both sides. This leaves the simplified equation $2a - 2 = 8$, which confirms that the value of $a$ is $5$.

Determining the area trapped between two curves requires integrating the difference between the upper boundary and the lower boundary across the interval defined by their points of intersection. Because both the circle $(x-2\sqrt{3})^{2+}y^2=12$ and the parabola $y^2=2\sqrt{3}x$ are symmetric about the x-axis, calculating twice the area of the upper half provides the total area of the region.

Intersection points arise where the parabola equation satisfies the circle equation; substituting $y^2=2\sqrt{3}x$ into the circle equation gives $x^{2-}4\sqrt{3}x+12+2\sqrt{3}x=12$, which simplifies to $x^{2-}2\sqrt{3}x=0$. This determines the integration limits to be $0$ and $2\sqrt{3}$. The area is found by evaluating the integral $2\int_{0}^{2\sqrt{3}} (\sqrt{12-(x-2\sqrt{3})^2}-\sqrt{2\sqrt{3}x}) \, dx$ representing the difference between the upper semicircular arc and the upper parabolic curve. Solving this definite integral yields $2(3\pi-8)$, resulting \in a final area of $6\pi-16$.

The area of a region bounded by a curve defined by a maximum operator involves determining which component of the function dictates the upper boundary across the entire interval. The given curve $y=\max \{|x|, x|x-2|\}$ behaves differently across distinct sub-intervals within $x \in [-2, 4]$, requiring us to split the domain. In the range $x \in [-2, 0]$, $|x|$ is the positive upper boundary, which forms a triangle with base 2 and height 2, contributing an area of 2 square units.

For $x \in [0, 4]$, we compare $|x|$ and $x|x-2|$ to identify which is larger. From $x=0$ to $x=1$, the function $x|x-2|$, which simplifies to $2x-x^2$ because the expression $|x-2|$ equals $2-x$, is greater than $x$, leading to the integral $\int_{0}^{1} (2x-x^2) dx$. As $x$ increases from $1$ to $3$, $x$ becomes the larger value, necessitating the integral $\int_{1}^{3} x dx$. In the final segment from $x=3$ to $x=4$, the function $x|x-2|$ again exceeds $x$ and simplifies to $x^{2-}2x$, requiring the integral $\int_{3}^{4} (x^{2-}2x) dx$.

Summing these distinct portions involves the evaluation of the total integral:
$\text{Area} = 2 + \int_{0}^{1} (2x-x^2) dx + \int_{1}^{3} x dx + \int_{3}^{4} (x^{2-}2x) dx$
Evaluating these terms results in $2 + \left[x^2 - \frac{x^3}{3}\right]_0^1 + \left[\frac{x^2}{2}\right]_1^3 + \left[\frac{x^3}{3} - x^2\right]_3^4$, which simplifies to $2 + \frac{2}{3} + 4 + \frac{16}{3} = 12$. The total area of the bounded region is 12 square units.

Determining the enclosed area between two curves relies on integrating the difference between the upper boundary and the lower boundary over the interval of their intersection. The curves $y=x^{2-}4x+4$ and $y^2=16-8x$ intersect at $x=0$ and $x=2$. Within this interval, the curve defined by $y=\sqrt{16-8x}$ remains above the curve $y=x^{2-}4x+4$, so the area is given by the integral
$\int_{0}^{2} (\sqrt{16-8x} - (x^{2-}4x+4)) dx$

Integrating the first term, $\sqrt{16-8x}$, involves the power rule with the chain rule, resulting in the antiderivative $-\frac{1}{12}(16-8x)^{3/2}$. Integrating the second part, $-(x^{2-}4x+4)$, produces $-\frac{x^3}{3} + 2x^2 - 4x$. By evaluating the combined expression
$\left[-\frac{1}{12}(16-8x)^{3/2} - \frac{x^3}{3} + 2x^2 - 4x\right]_0^2$
at the upper limit $2$ and the lower limit $0$, we obtain a total area of $\frac{8}{3}$.

The inherent symmetry of the region about the $y$-axis simplifies the problem, as calculating the area for $x \ge 0$ and multiplying by two provides the full result. Within the specified range of $|x| \le 3$, the boundaries $y = 2x + 1$ and $y = x^2 + 1$ intersect at $x=0$ and $x=2$, dividing the positive domain into two distinct segments. For values of $x$ between 0 and 2, the parabola $y = x^2 + 1$ serves as the effective upper bound for $y$, while for values of $x$ between 2 and 3, the line $y = 2x + 1$ restricts the region.

Integrating these functions over their respective intervals gives the area of the right half, represented by $\int_{0}^{2} (x^{2+}1) \, dx + \int_{2}^{3} (2x+1) \, dx$. Computing the first integral results in $\frac{14}{3}$, and the second term evaluates to 6, yielding a combined half-area of $\frac{32}{3}$. Multiplying this value by two confirms the total area of the region is $\frac{64}{3}$.

The boundaries of the region intersect at $x = -1$, $x = 1$, and $x = 2$, which allows the total area to be partitioned into two intervals for integration. For the first interval from $x = -1$ to $x = 1$, the region is bounded above by $y = \frac{x^{2+}3}{2}$ and below by $y = 1-x$, leading to the following integral:
$\int_{-1}^{1} \left( \frac{x^{2+}2x+1}{2} \right) dx = \frac{4}{3}$
For the second interval from $x = 1$ to $x = 2$, the upper boundary is $y = 3-x$ and the lower boundary is $y = x-1$, resulting in the following integral:
$\int_{1}^{2} (4-2x) dx = 1$
Summing these two areas gives the total area $A = \frac{7}{3}$, so $6A = 14$.

The inequality $|x-y| \le y$ defines a region where $y \ge x/2$ and $x \ge 0$, which, when combined with $y \le 4\sqrt{x}$, constrains the area between the curve $y = 4\sqrt{x}$ and the line $y = x/2$. Equating these two boundary functions identifies the intersection points at 0 and 64. Calculating the area involves evaluating the definite integral of the difference between these boundaries: $\int_{0}^{64} \left(4\sqrt{x} - \frac{x}{2}\right) dx = \left. \frac{8}{3} x^{3/2} - \frac{x^2}{4} \right|_{0}^{64}$ Substituting the upper limit yields $\frac{8}{3}(512) - \frac{4096}{4}$, which simplifies to $\frac{4096}{3} - 1024 = \frac{1024}{3}$.

To calculate the area of the region, we first identify the intersection points of the boundary curves $y = x^2$ and $y = |4-x^2|$ within the constraint $y \le 4$ and $x \ge 0$. The curves intersect at $x = \sqrt{2}$ where $x^2 = 4-x^2$, at $x = 2$ where $x^2 = 4$, and at $x = 2\sqrt{2}$ where $x^{2-}4 = 4$. Based on these boundaries, we split the integration into two intervals. From $x = \sqrt{2}$ to $x = 2$, the region is bounded above by $y = x^2$ and below by $y = 4-x^2$, giving the integral $\int_{\sqrt{2}}^{2} (2x^2 - 4) \, dx$. From $x = 2$ to $x = 2\sqrt{2}$, the region is bounded above by the line $y = 4$ and below by $y = x^{2-}4$, resulting in the integral $\int_{2}^{2\sqrt{2}} (8 - x^2) \, dx$.

Evaluating the first integral yields $[\frac{2x^3}{3} - 4x]_{\sqrt{2}}^{2} = (\frac{16}{3} - 8) - (\frac{4\sqrt{2}}{3} - 4\sqrt{2}) = \frac{8\sqrt{2}-8}{3}$. Calculating the second integral gives $[8x - \frac{x^3}{3}]_{2}^{2\sqrt{2}} = (16\sqrt{2} - \frac{16\sqrt{2}}{3}) - (16 - \frac{8}{3}) = \frac{32\sqrt{2}-40}{3}$. Summing these parts, the total area is $\frac{40\sqrt{2}}{3} - 16$. By comparing this to the expression $\frac{80\sqrt{2}}{\alpha} - \beta$, we rewrite $\frac{40\sqrt{2}}{3}$ as $\frac{80\sqrt{2}}{6}$, which identifies $\alpha = 6$ and $\beta = 16$. Consequently, $\alpha + \beta = 6 + 16 = 22$.

The region defined by $1+x^2 \le y \le \min \{x+7, 11-3x\}$ is bounded by the parabola $y=x^{2+}1$ and two linear functions that intersect each other at $x=1$. By determining where these lines meet the parabola, we find that the integration boundaries for the region extend from $x=-2$ to $x=2$, with the linear upper bound changing from $x+7$ to $11-3x$ at the intersection point of $x=1$.

The area $A$ is calculated by integrating the difference between the upper boundary function and the lower parabola $y=x^{2+}1$ over the two intervals: $A = \int_{-2}^1 ((x+7) - (x^{2+}1))\,dx + \int_{1}^2 ((11-3x) - (x^{2+}1))\,dx$. Solving these definite integrals yields $A = \frac{50}{3}$, which results in $3A = 50$.

Differentiating the integral equation $f(x)=1-2x+\int_{0}^{x}e^{x-t}f(t)dt$ with respect to $x$ using Leibniz's rule yields $f'(x)=-2+f(x)+\int_{0}^{x}e^{x-t}f(t)dt$. Since the integral term $\int_{0}^{x}e^{x-t}f(t)dt$ is equivalent to $f(x)-1+2x$ from the original equation, substituting this expression produces the first-order linear differential equation $f'(x)=2f(x)+2x-3$. Multiplying this equation by the integrating factor $e^{-2x}$ results in \frac{d}{dx}\[f(x)e^{-2x}]$=(2x-3)e^{-2x}$, and integrating both sides yields$f(x)e^{-2x}=(1-x)e^{-2x}+C$. Given the initial condition$f(0)=1$, the constant of integration$C$evaluates to$0$, simplifying the function to$f(x)=1-x$. The area of the region bounded by$y=1-x$and the coordinate axes, specifically between$x=0$and the intercept$x=1$, is calculated by$\int_{0}^{1}(1-x)dx$, which results in$\frac{1}{2}$.

To find the area of the region bounded by the parabola $y^2 = 2x$ and the line $y = 4x-1$, it is most efficient to express the equations as functions of $y$, yielding $x = y^2/2$ and $x = (y+1)/4$. Determining the intersection points by equating these expressions leads to $y^2/2 = (y+1)/4$, which simplifies to the quadratic equation $2y^2 - y - 1 = 0$, giving the integration boundaries of $y = 1$ and $y = -\frac{1}{2}$.

The area is computed by evaluating the integral $\int_{-\frac{1}{2}}^{1} \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy$, where the line serves as the right-hand boundary and the parabola as the left-hand boundary. Calculating the definite integral $\left[ \frac{1}{4} \left( \frac{y^2}{2} + y \right) - \frac{y^3}{6} \right]_{-\frac{1}{2}}^{1}$ by substituting the upper and lower limits results in a final area of $\frac{9}{32}$ square units.

Determining the area enclosed by two curves requires integrating the difference between the upper and lower functions within the interval defined by their points of intersection. For the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$, equating them gives $1+3x-2x^2 = \frac{1}{x}$, which rearranges to the cubic equation $2x^{3-}3x^{2-}x+1=0$. Substituting the given intersection point $x=\frac{1}{2}$ confirms it is a solution, and algebraic division shows the upper limit of the interval is $x=\frac{1+\sqrt{5}}{2}$.

The area is calculated by the definite integral $\int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}} (1+3x-2x^{2-}\frac{1}{x}) dx$, which has the antiderivative $x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x$. Applying the limits requires evaluating this expression at the upper bound $\frac{1+\sqrt{5}}{2}$ and the lower bound $\frac{1}{2}$. Evaluating at the upper limit yields $\frac{17+7\sqrt{5}}{12} - \ln(\frac{1+\sqrt{5}}{2})$, and evaluating at the lower limit results in $\frac{19}{24} + \ln(\frac{1}{2})$.

Subtracting the lower bound result from the upper bound gives $\left(\frac{34+14\sqrt{5}}{24} - \ln(\frac{1+\sqrt{5}}{2})\right) - \left(\frac{19}{24} + \ln(\frac{1}{2})\right)$. Since $\ln(\frac{1+\sqrt{5}}{2}) - \ln(\frac{1}{2})$ simplifies to $\ln(\frac{1+\sqrt{5}}{2} \cdot 2) = \ln(1+\sqrt{5})$, the final expression is $\frac{14\sqrt{5}+15}{24} - \ln(1+\sqrt{5})$. Comparing this to $\frac{1}{24}(\ell\sqrt{5}+m)-n\ln(1+\sqrt{5})$, the values are identified as $\ell=14$, $m=15$, and $n=1$. Summing these constants yields $14+15+1=30$.

The area bounded by two intersecting curves is found by identifying their points of intersection to establish appropriate limits and leveraging the symmetry of the shapes to simplify the calculation. Substituting $y^2=4x$ into the circle equation $x^{2+}y^2=5$ yields the quadratic $x^{2+}4x-5=0$, which solves to $x=1$ for the region of interest. Since both the parabola and the circle are symmetric about the x-axis, the total area is twice the area enclosed in the first quadrant. This area comprises two distinct segments: the region under the parabola from $x=0$ to $x=1$ and the region under the circular arc from $x=1$ to $x=\sqrt{5}$.

Calculating these segments involves the integrals $\int_{0}^{1} \sqrt{4x} \, dx$ and $\int_{1}^{\sqrt{5}} \sqrt{5-x^2} \, dx$. The first integral evaluates to $\frac{4}{3}$, while the second results in $\frac{5\pi}{4} - 1 - \frac{5}{2}\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$. Summing these two components gives $\frac{1}{3} + \frac{5\pi}{4} - \frac{5}{2}\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$, which represents the area in the first quadrant. Doubling this total provides $\frac{2}{3} + \frac{5\pi}{2} - 5\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$. Using the trigonometric identity $\frac{\pi}{2} - \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) = \cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$, the expression becomes $\frac{2}{3} + 5\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$. Converting the inverse cosine term to an inverse sine term yields the final result of $\frac{2}{3}+5\sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$.