Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The parabola passing through the points $(-1,0), (0,1),$ and $(1,0)$ is given by the equation $p(x) = 1-x^2$. The region defined by the circle $(x+1)^{2+}(y-1)^2 \le 1$ and the inequality $y \le p(x)$ lies within the interval $x \in [-1, 0]$, where the upper boundary is the parabola and the lower boundary is the upper arc of the circle. This area $A$ is equivalent to the area of the quarter-circle sector, $\frac{\pi}{4}$, minus the area between the square $[ -1, 0 ] \times [ 0, 1 ]$ and the parabola, which is calculated as $1 - \int_{-1}^0 (1-x^2) dx = 1 - \frac{2}{3} = \frac{1}{3}$. Consequently, $A = \frac{\pi}{4} - \frac{1}{3}$. Evaluating the expression $12(\pi - 4A)$ by substituting this value gives $12(\pi - 4(\frac{\pi}{4} - \frac{1}{3})) = 12(\pi - \pi + \frac{4}{3})$, which simplifies to $16$.

The area trapped between two curves is found by integrating the vertical distance between them over the interval where they intersect. For the parabolas $2y=5x^2$ and $x^{2-}y+6=0$, the functions are $y=\frac{5}{2}x^2$ and $y=x^{2+}6$. Their intersection occurs where $x^{2+}6=\frac{5}{2}x^2$, which simplifies to $x^2=4$, defining integration limits from -2 to 2. Calculating the area between these parabolas gives:

$2\int_{0}^{2} \left(x^{2+}6 - \frac{5}{2}x^2\right)dx = 2\int_{0}^{2} \left(6 - \frac{3}{2}x^2\right)dx = 16$

Next, we find the area bounded by $2y=5x^2$ and $y=\alpha x$. The intersection of these curves occurs where $\frac{5}{2}x^2=\alpha x$, leading to points at $x=0$ and $x=\frac{2\alpha}{5}$. Integrating the difference between the line and the parabola yields:

$\int_{0}^{\frac{2\alpha}{5}} \left(\alpha x - \frac{5}{2}x^2\right)dx = \left[\frac{\alpha x^2}{2} - \frac{5x^3}{6}\right]_0^{\frac{2\alpha}{5}} = \frac{2\alpha^3}{75}$

Equating this result to 16, we have $\frac{2\alpha^3}{75} = 16$, which confirms that the value of $\alpha^3$ is 600.

Calculating the area bounded by $f(x)=\max\{\sin x, \cos x\}$ and the $x$-axis over the interval $[-\pi, \pi]$ requires partitioning the domain based on the intersection points $x = -3\pi/4$ and $x = \pi/4$, as well as the zero at $x = -\pi/2$ where the function transitions relative to the $x$-axis. Because the function takes negative values in the first two segments, we integrate the absolute values over $(-\pi, -3\pi/4)$ and $(-3\pi/4, -\pi/2)$ while using standard integration for the positive regions $(-\pi/2, \pi/4)$ and $(\pi/4, \pi)$. The total area is given by the sum $\left|\int_{-\pi}^{-3\pi/4} \sin x \, dx\right| + \left|\int_{-3\pi/4}^{-\pi/2} \cos x \, dx\right| + \int_{-\pi/2}^{\pi/4} \cos x \, dx + \int_{\pi/4}^{\pi} \sin x \, dx$, which evaluates to 4.

Integration with respect to $y$ serves as the most efficient approach to calculate the area bounded by the curves $y=\log_e(x+e^2)$, $x=\log_e(2/y)$, and $x=\log_e 2$ given the condition $y \ge 1$. Expressing $x$ in terms of $y$ for the first curve yields $x=e^y-e^2$. Comparing the functions reveals that the region is situated between the left boundary $x=e^y-e^2$ and the right boundary $x=\log_e(2/y)$, with $y$ spanning from the lower bound $y=1$ up to the intersection at $y=2$.

The area is determined by evaluating the integral $\int_{1}^{2} (\log_e(2/y) - e^y + e^2) dy$, which simplifies to $\int_{1}^{2} (\log_e 2 - \log_e y - e^y + e^2) dy$. Computing the antiderivative gives $[y \log_e 2 - (y \log_e y - y) - e^y + e^2 y]_1^2$. Applying the upper limit of $2$ results in $2 \log_e 2 - (2 \log_e 2 - 2) - e^2 + 2e^2 = 2 + e^2$, while applying the lower limit of $1$ results in $\log_e 2 - (1 \log_e 1 - 1) - e + e^2 = \log_e 2 + 1 - e + e^2$. Subtracting these values yields $1+e-\log_e 2$.

Calculating the area between curves defined by parabolas opening horizontally is most efficient when expressing $x$ as a function of $y$ and integrating with respect to the vertical axis. Rearranging the equations $y^2=2x-1$ and $y^2=4x-3$ yields $x=\frac{y^{2+}1}{2}$ and $x=\frac{y^{2+}3}{4}$, which intersect at $y=\pm 1$. Because the region is symmetric across the x-axis, the total area equals twice the integral from $y=0$ to $y=1$ of the difference between the outer and inner curves. This calculation is represented as:

$2 \int_{0}^{1} \left( \frac{y^{2+}3}{4} - \frac{y^{2+}1}{2} \right) dy = 2 \int_{0}^{1} \left( \frac{1}{4} - \frac{y^2}{4} \right) dy$

Evaluating this integral results in $2 \left[ \frac{1}{4} - \frac{1}{12} \right]$, which simplifies to $\frac{1}{3}$.

The function $y=|x^{2-}9|$ is symmetric about the y-axis, allowing us to find the area in the first quadrant and multiply the result by two. The curve intersects the line $y=3$ where $|x^{2-}9|=3$, which occurs at $x=\sqrt{6}$ and $x=\sqrt{12}$ in the first quadrant. Because the absolute value function changes definition at $x=3$, we must split the bounded region into two separate integrals along the x-axis.

Between $x=\sqrt{6}$ and $x=3$, the curve is represented by $y=9-x^2$, so the area is given by the integral $\int_{\sqrt{6}}^3 (3 - (9 - x^2)) \,dx$, which simplifies to $\int_{\sqrt{6}}^3 (x^2 - 6) \,dx$ and evaluates to $4\sqrt{6}-9$. Between $x=3$ and $x=\sqrt{12}$, the curve is represented by $y=x^{2-}9$, making the area $\int_3^{\sqrt{12}} (3 - (x^2 - 9)) \,dx$, or $\int_3^{\sqrt{12}} (12 - x^2) \,dx$, which evaluates to $16\sqrt{3}-27$. Summing these two areas and multiplying by two to account for both sides of the y-axis yields the final area calculation:

$2(4\sqrt{6}-9 + 16\sqrt{3}-27) = 8(4\sqrt{3}+\sqrt{6}-9)$

The slope of the curve $4x^3 - 3xy^2 + 6x^2 - 5xy - 8y^2 + 9x + 14 = 0$ is found via implicit differentiation with respect to $x$, yielding $(12x^2 - 3y^2 + 12x - 5y + 9) + (-6xy - 5x - 16y)\frac{dy}{dx} = 0$. Substituting the coordinates $(-2, 3)$ results in $(12(4) - 3(9) + 12(-2) - 5(3) + 9) + (-6(-2)(3) - 5(-2) - 16(3))\frac{dy}{dx} = 0$, which simplifies to $-9 - 2\frac{dy}{dx} = 0$, giving the tangent slope $m = -\frac{9}{2}$. The tangent line is $y - 3 = -\frac{9}{2}(x + 2)$, which simplifies to $9x + 2y + 12 = 0$, while the normal line is $y - 3 = \frac{2}{9}(x + 2)$, which simplifies to $2x - 9y + 31 = 0$. Setting $y=0$ for both equations identifies the x-intercepts at $x = -\frac{4}{3}$ and $x = -\frac{31}{2}$. These intercepts and the point of tangency $(-2, 3)$ form a triangle with a base of length $|-\frac{4}{3} - (-\frac{31}{2})| = \frac{85}{6}$ and a height of $3$. The area $A = \frac{1}{2} \times \frac{85}{6} \times 3 = \frac{85}{4}$, therefore $8A = 8 \times \frac{85}{4} = 170$.

To calculate the area of the region bounded by the curves $x=0$, $x=y^a$, $y=1$, and $y=3$, we integrate the function $x=y^a$ with respect to $y$ over the interval from $1$ to $3$. This integration yields the expression $\int_{1}^{3} y^a \, dy = \left[ \frac{y^{a+1}}{a+1} \right]_1^3 = \frac{3^{a+1} - 1}{a+1}$.

Equating this expression to the given area of $\frac{364}{3}$ results in the equation $\frac{3^{a+1} - 1}{a+1} = \frac{364}{3}$. Testing the given options for the odd natural number $a$, we substitute $a=5$ to obtain $\frac{3^{5+1} - 1}{5+1} = \frac{3^6 - 1}{6} = \frac{729 - 1}{6} = \frac{728}{6} = \frac{364}{3}$. This confirms that $a=5$ is the value that satisfies the area requirement.

The area bounded by two parabolas, one opening to the right and the other to the left, is best calculated by identifying their intersection points and integrating the horizontal distance between them along the $y$-axis. First, equating $y^2=8x$ and $y^2=16(3-x)$ reveals that the curves intersect where $8x=48-16x$, which simplifies to $24x=48$ or $x=2$. Substituting this $x$-value back into either equation yields $y^2=16$, confirming that the enclosed region spans from $y=-4$ to $y=4$.

By solving each equation for $x$, we express the boundaries as $x=\frac{y^2}{8}$ for the first parabola and $x=\frac{48-y^2}{16}$ for the second. The total area is the integral of the horizontal distance between these curves, calculated as $\int_{-4}^{4} (\frac{48-y^2}{16} - \frac{y^2}{8}) \, dy$. Due to the symmetry of the region across the $x$-axis, this is equivalent to $2 \int_{0}^{4} (\frac{48-y^2}{16} - \frac{2y^2}{16}) \, dy$, which simplifies to $\frac{1}{8} \int_{0}^{4} (48-3y^2) \, dy$. Evaluating this integral gives \frac{1}{8} \[48y - y^3]_{0}^{4}$, resulting \in$\frac{1}{8} (192 - 64)$, which equals 16.

The area of a region bounded by curves is determined by integrating the difference between the upper boundary and the lower boundary across the specified interval. By identifying the intersection points of the parabola $y^2 = 8x$ and the line $y = \sqrt{2}x$, we equate $(\sqrt{2}x)^2 = 8x$ to find $2x^2 = 8x$, which simplifies to $x=0$ and $x=4$. Since the problem specifies $x \ge 1$, we integrate from $1$ to $4$, where the parabola functions as the upper curve $\sqrt{8x}$ and the line functions as the lower curve $\sqrt{2}x$.

The integral is set up as follows:
$\int_{1}^{4} (\sqrt{8x} - \sqrt{2}x) \, dx$
Evaluating the antiderivatives results in:
$\left[ \frac{2\sqrt{8}}{3} x^{3/2} - \frac{x^2}{\sqrt{2}} \right]_{1}^{4}$
Substituting the upper limit $x = 4$ yields:
$\frac{2\sqrt{8}}{3} (8) - \frac{16}{\sqrt{2}} = \frac{16\sqrt{8}}{3} - \frac{16}{\sqrt{2}} = \frac{32\sqrt{2}}{3} - 8\sqrt{2} = \frac{8\sqrt{2}}{3}$
Substituting the lower limit $x = 1$ yields:
$\frac{2\sqrt{8}}{3} (1) - \frac{1}{\sqrt{2}} = \frac{4\sqrt{2}}{3} - \frac{\sqrt{2}}{2} = \frac{8\sqrt{2} - 3\sqrt{2}}{6} = \frac{5\sqrt{2}}{6}$
Subtracting these values gives the final area of:
$\frac{16\sqrt{2}}{6} - \frac{5\sqrt{2}}{6} = \frac{11\sqrt{2}}{6}$

The area enclosed between the parabola $y^2 = 8x+4$ and the circle $x^{2+}y^{2+}4\sqrt{3}x-4=0$ is determined by finding the intersection points and summing the areas of the resulting parabolic and circular segments. Solving these equations simultaneously shows that the curves intersect at the points $(0, 2)$ and $(0, -2)$, which naturally divides the region. The parabolic segment extends from its vertex at $x = -1/2$ to $x = 0$, and its area is calculated as $\frac{2}{3} \times 4 \times \frac{1}{2} = \frac{4}{3}$.

For the circular portion, the circle has a center at $(-2\sqrt{3}, 0)$ and a radius of $4$. The relevant angle $\theta$ is defined by $\cos \theta = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$, which corresponds to $\theta = 30^{\circ}$ or $\frac{\pi}{6}$ radians. The area of this part is found by calculating the area of the circular sector as $4^2 \times \frac{\pi}{6} = \frac{8\pi}{3}$ and subtracting the triangular region with an area of $\frac{1}{2} \times 4 \times 2\sqrt{3} = 4\sqrt{3}$. Combining the parabolic and circular components gives a total area of $\frac{4}{3} + \frac{8\pi}{3} - 4\sqrt{3}$, which simplifies to $\frac{1}{3}(4-12\sqrt{3}+8\pi)$.

The area is defined by the region bounded by the parabola $y^2=8x$ and the line $y=\sqrt{2}x$, excluding a specific triangular portion. By integrating the difference between the two curves with respect to $y$ from 0 to $4\sqrt{2}$, we determine the total enclosed area as $\int_{0}^{4\sqrt{2}} \left(\frac{y}{\sqrt{2}} - \frac{y^2}{8}\right) dy$. Evaluating this integral gives $\left[\frac{y^2}{2\sqrt{2}} - \frac{y^3}{24}\right]_{0}^{4\sqrt{2}}$, which simplifies to $8\sqrt{2} - \frac{16\sqrt{2}}{3}$, resulting in a total area of $\frac{8\sqrt{2}}{3}$. The excluded region is a triangle with vertices at $A(2, 2\sqrt{2})$, $B(1, 2\sqrt{2})$, and $C(1, \sqrt{2})$, where the base $AB$ has a length of 1 and the height $BC$ has a length of $\sqrt{2}$. Using the geometric formula for the area of a triangle, $\frac{1}{2} \times \text{base} \times \text{height}$ yields $\frac{\sqrt{2}}{2}$. Subtracting this triangular area from the total integral area gives $\frac{8\sqrt{2}}{3} - \frac{\sqrt{2}}{2}$, which equates to $\frac{16\sqrt{2}-3\sqrt{2}}{6}$, arriving at the final result of $\frac{13\sqrt{2}}{6}$.

The area of a region bounded by two functions is determined by evaluating the definite integral of the difference between the upper curve and the lower curve over the interval defined by their intersection points. The upper boundary is the semicircular arc $y = \sqrt{5-x^2}$ with radius $\sqrt{5}$, and the lower boundary is the V-shaped function $y = |x-1|$. Solving the equation $\sqrt{5-x^2} = |x-1|$ shows that the curves intersect at $x = -1$ and $x = 2$, identifying the integration limits as $[-1, 2]$.

The integral of the semicircular portion, $\int_{-1}^{2} \sqrt{5-x^2} \, dx$, is calculated using the standard integral form $\frac{x}{2} \sqrt{5-x^2} + \frac{5}{2} \sin^{-1} \left( \frac{x}{\sqrt{5}} \right)$, which evaluates to $2 + \frac{5\pi}{4}$. For the lower boundary, the area under the V-shaped curve is formed by two triangles, with the first covering the interval from $x=-1$ to $x=1$ having an area of $2$, and the second from $x=1$ to $x=2$ having an area of $0.5$, resulting in a combined triangular area of $\frac{5}{2}$. Subtracting this triangular area from the area under the semicircle leaves a final result of $\frac{5\pi}{4} - \frac{1}{2}$.

The area enclosed by the curves $y = |x^2 - 1|$ and $y = 1$ can be found by applying definite integration, leveraging the symmetry of the functions about the $y$-axis to simplify the calculation. Since the graphs are symmetric, the total area is twice the integral from $x = 0$ to $x = \sqrt{2}$. The intersection points are determined by setting $|x^2 - 1| = 1$, which confirms the boundaries of integration at $x = 0$ and $x = \sqrt{2}$.

The function $|x^2 - 1|$ changes behavior at $x = 1$. In the interval $[0, 1]$, the expression inside the modulus is negative, so $|x^2 - 1|$ becomes $1 - x^2$, while in the interval $[1, \sqrt{2}]$, the expression is positive, so it becomes $x^2 - 1$. Subtracting these from $y = 1$ leads to integrands of $1 - (1 - x^2) = x^2$ and $1 - (x^2 - 1) = 2 - x^2$, respectively. Evaluating the integral $2 \left[ \int_{0}^{1} x^2 \, dx + \int_{1}^{\sqrt{2}} (2 - x^2) \, dx \right]$ yields $2 \left[ \left[ \frac{x^3}{3} \right]_0^1 + \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt{2}} \right]$, which simplifies to $\frac{8}{3}(\sqrt{2} - 1)$.

Determining the area enclosed by the boundaries $y = 4x^2$, $x^2 = 9y$, and $y = 4$ is most effectively handled by expressing $x$ in terms of $y$ and utilizing symmetry about the $y$-axis. By rewriting the curves as $x = 3\sqrt{y}$ and $x = \frac{1}{2}\sqrt{y}$, the width of the region at any vertical position $y$ is simply the difference between these two functions. Integrating this width from $y = 0$ to $y = 4$ and doubling the result to account for both sides of the $y$-axis gives the following integral:

$A = 2 \int_0^4 \left(3\sqrt{y} - \frac{1}{2}\sqrt{y}\right) \, dy = 2 \int_0^4 \frac{5}{2} \sqrt{y} \, dy$

Evaluating this expression as $5 \left[ \frac{2}{3} y^{3/2} \right]_0^4$ leads to a final value of $\frac{10}{3} (8)$, resulting in a total area of $\frac{80}{3}$.

When a circle is tangent to the $x$-axis, its distance from the centre $(\alpha, \beta)$ to the $x$-axis is equal to its radius, which means the radius of this variable circle is $\beta$. The condition for this circle to touch the fixed circle $x^{2+}(y-1)^2=1$ externally is that the distance between their centres must equal the sum of their radii, leading to the relation $\sqrt{\alpha^2 + (\beta-1)^2} = \beta + 1$. Squaring both sides simplifies this geometric condition to $\alpha^2 + \beta^2 - 2\beta + 1 = \beta^2 + 2\beta + 1$, confirming that the locus of the centre is the parabola $x^2 = 4y$.

The area bounded by this parabola and the line $y=4$ is determined by integrating the horizontal width of the parabola with respect to $y$ across the interval $[0, 4]$. Doubling this value to account for both sides of the parabola leads to the calculation:
$2 \int_{0}^{4} 2\sqrt{y} \, dy = 4 \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{4}$
Evaluating this expression results in $4 \cdot \frac{2}{3} \cdot 8$, giving a final area of $\frac{64}{3}$ square units.

The area enclosed between two curves is calculated by evaluating the definite integral of the function defining the right boundary minus the function defining the left boundary over the interval of their intersection. By expressing the parabola as $x = \frac{y^2}{2}$ and the line as $x = 4 - y$, the points of intersection are found to be $y = -4$ and $y = 2$. Integrating the difference $(4 - y - \frac{y^2}{2})$ with respect to $y$ over this interval yields the expression $\left[4y - \frac{y^2}{2} - \frac{y^3}{6}\right]_{-4}^{2}$. Computing this definite integral leads to a final enclosed area of 18 square units.

Determining the curve $y=y(x)$ begins by translating the area relationship $A_1 = 2A_2$ into a mathematical expression. Given that $A_1 + A_2 = xy - 8$, substituting $A_1 = 2A_2$ into the total area equation yields $\frac{3}{2}A_1 = xy - 8$, which simplifies to $A_1 = \frac{2}{3}(xy - 8)$. Since $A_1$ represents the area $\int_{4}^{x} y \, dt$, equating this integral to the expression $\frac{2}{3}(xy - 8)$ and differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus produces the differential equation $y = \frac{2}{3}(y + xy')$. Rearranging the terms leads to $3y = 2y + 2xy'$, or $y = 2xy'$, which separates into $\int \frac{dy}{y} = \int \frac{dx}{2x}$. Integrating this gives $\ln y = \frac{1}{2} \ln x + \ln C$, and applying the condition that the curve passes through $(4, 2)$ determines that $C = 1$, identifying the curve as $y^2 = x$.

To find the normal, first note that the line $2x - 12y = 15$ has a slope of $\frac{1}{6}$, meaning any perpendicular normal must have a slope of $m = -6$. For the parabola $y^2 = 4ax$ where $4a = 1$, the standard equation of a normal with slope $m$ is $y = mx - 2am - am^3$. Substituting $m = -6$ and $a = \frac{1}{4}$ results in the equation $y = -6x - 2\left(\frac{1}{4}\right)(-6) - \frac{1}{4}(-6)^3$, which simplifies to $y = -6x + 57$. Testing the provided coordinate pairs in this equation shows that $(6, 21)$, $(8, 9)$, and $(12, -15)$ all satisfy $y = -6x + 57$. However, substituting $x = 10$ into the normal equation yields $y = -6(10) + 57 = -3$, confirming that the point $(10, -4)$ does not lie on this line.

Finding the relative areas of regions bounded by intersecting curves relies on determining the boundary points and applying definite integrals. The total area of $S$, bounded by $y=x^3$ and $y^2=x$ from $x=0$ to $x=1$, is computed as:
$\int_{0}^{1} (\sqrt{x} - x^3) \, dx = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}$
The curve $y=2|x|$ partitions this region. Since it intersects $y^2=x$ at $x=1/4$, the sub-region $R_1$ is bounded by $y^2=x$ and $y=2x$ within the interval $[0, 1/4]$:
$R_1 = \int_{0}^{1/4} (\sqrt{x} - 2x) \, dx = \frac{1}{12} - \frac{1}{16} = \frac{1}{48}$
By subtracting $R_1$ from the total area, the remaining area is defined by $R_2 = \frac{5}{12} - \frac{1}{48} = \frac{19}{48}$. The resulting ratio is $\frac{R_2}{R_1} = 19$.

The inequality $x^{2/3} + y^{2/3} \le 1$ defines the region enclosed by an astroid. The constraints $y \ge 0$ and $x+y \ge 0$ restrict this region to the portion residing in the first quadrant, combined with a segment in the second quadrant bounded below by the line $y = -x$. The intersection of the astroid $x^{2/3} + y^{2/3} = 1$ with the line $y = -x$ occurs at $x = -(1/2)^{3/2}$, which establishes the leftmost boundary of the integration domain.

Calculating the area requires integrating the height of the region across the interval $[-(1/2)^{3/2}, 1]$. The area $A$ is expressed as the integral $\int_{-(1/2)^{3/2}}^{1} (1-x^{2/3})^{3/2} dx + \int_{-(1/2)^{3/2}}^{0} x dx$. Using the substitution $x = \sin^3 \theta$, which implies $dx = 3 \sin^2 \theta \cos \theta d\theta$, the integration limits transform to $[-\pi/4, \pi/2]$ for the curve segment, allowing for the evaluation of the definite integral. Performing this calculation results in $A = \frac{36\pi}{256}$. Consequently, multiplying this area $A$ by $\frac{256}{\pi}$ yields 36.