Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The area of a region bounded by a symmetric curve can be determined by integrating the upper bounding function across the given interval and doubling the result to account for the full vertical span. Integrating $y = \sqrt{x}$ from 0 to $\alpha$ and multiplying by two yields the area formula
$A(\alpha) = 2\int_{0}^{\alpha} \sqrt{x} \, dx = \frac{4}{3}\alpha^{3/2}$
With the ratio $A(\lambda)/A(4) = 2/5$, we establish the equation $\frac{\frac{4}{3}\lambda^{3/2}}{\frac{4}{3}4^{3/2}} = \frac{2}{5}$, which reduces to $\frac{\lambda^{3/2}}{8} = \frac{2}{5}$. Multiplying both sides by 8 gives $\lambda^{3/2} = \frac{16}{5}$, and raising both sides to the power of $\frac{2}{3}$ results in $\lambda = (\frac{16}{5})^{2/3}$, which simplifies algebraically to $4\left( \frac{4}{25} \right)^{1/3}$.

The definition of the region relies on the absolute value function, which requires splitting the interval of integration at $x=0$ because the behavior of the expression $x|x|$ changes sign. For the interval from -1 to 0, the term $|x|$ becomes $-x$, simplifying the boundary function to $-x^2 + 1$, whereas for the interval from 0 to 1, the term $|x|$ becomes $x$, resulting in the function $x^2 + 1$. Evaluating the integral $\int_{-1}^{0} (-x^2 + 1) dx$ produces a value of 2/3, and the integral $\int_{0}^{1} (x^2 + 1) dx$ produces 4/3. Combining these two areas provides the total region area of 2 square units.

The region defined by $x^2 \le y \le x+2$ is bounded by the parabola $y = x^2$ and the line $y = x+2$, where the total area is given by the integral of the upper function minus the lower function across the interval formed by their intersection points. Finding these intersections by setting $x^2 = x + 2$ produces the quadratic equation $x^2 - x - 2 = 0$, which yields the limits $x = -1$ and $x = 2$. Calculating the integral $\int_{-1}^{2} (x + 2 - x^2) \, dx$ leads to the expression $[\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}$, resulting in $\frac{10}{3} - (-\frac{7}{6}) = \frac{9}{2}$. Alternatively, one can use the shortcut for the area of a parabolic segment, $\frac{D^{3/2}}{6a^2}$, where the discriminant of the quadratic $x^2 - x - 2 = 0$ is $D = 9$, yielding $\frac{27}{6}$, which simplifies to $\frac{9}{2}$.

The region is defined by the constraints $x \in [0, 3]$ and $y$ being restricted by both 4 and the curve $y = x^2 + 3x$. Solving $x^2 + 3x = 4$ indicates that for $0 \leq x \leq 1$, the area is bounded by the curve, and for $1 < x \leq 3$, the area is bounded by the constant line $y = 4$. The total area is the sum of the integral of the curve over the first interval and the area of the rectangle formed over the second interval.

$\int_{0}^{1} (x^2 + 3x) \, dx + \int_{1}^{3} 4 \, dx$

Evaluating the integral $\int_{0}^{1} (x^2 + 3x) \, dx$ yields $\frac{1}{3} + \frac{3}{2} = \frac{11}{6}$, while the rectangle from $x = 1$ to $x = 3$ has a width of 2 and a height of 4, resulting in an area of 8. Combining these portions gives $\frac{11}{6} + 8 = \frac{59}{6}$.

Determining the area of this region requires identifying the intersection of the parabola $y^2=4x$ and the line $y=1-x$ in the first quadrant to establish the boundaries for integration. Substituting $y=1-x$ into the parabolic equation gives $(1-x)^2=4x$, which expands to the quadratic equation $x^{2-}6x+1=0$. Solving this yields $x = 3 \pm 2\sqrt{2}$, and given the boundary conditions $x+y \le 1$ and $x \ge 0$, the intersection point within the region is $x = 3-2\sqrt{2}$.

The total area is calculated by summing two integral segments: the area under the parabola from $x=0$ to $3-2\sqrt{2}$ and the area under the line from $x=3-2\sqrt{2}$ to $x=1$. This is expressed as $\int_{0}^{3-2\sqrt{2}} 2\sqrt{x} \, dx + \int_{3-2\sqrt{2}}^{1} (1-x) \, dx$. Evaluating the first integral yields $\frac{4}{3}(3-2\sqrt{2})^{3/2}$, which simplifies to $\frac{20\sqrt{2}-28}{3}$, and the second integral evaluates to $6-4\sqrt{2}$. Combining these values results in $\frac{8\sqrt{2}-10}{3}$, which can be written as $\frac{8}{3}\sqrt{2} - \frac{10}{3}$. With $a = \frac{8}{3}$ and $b = -\frac{10}{3}$, subtracting $b$ from $a$ yields $6$.

Calculating the area of a region defined by horizontal boundaries requires integrating the difference between the curves $x = y + 4$ and $x = \frac{y^2}{2}$ with respect to $y$ across their points of intersection. Equating $\frac{y^2}{2} = y + 4$ leads to the quadratic equation $y^2 - 2y - 8 = 0$, which identifies the limits of integration at $y = -2$ and $y = 4$. Integrating the difference $(y + 4 - \frac{y^2}{2})$ with respect to $y$ results in the antiderivative $\frac{y^2}{2} + 4y - \frac{y^3}{6}$. Evaluating this expression at the boundaries yields a total area of 18.

The total area of the region is determined by the sum of definite integrals over the intervals $[0, 1]$ and $[1, t]$, as the curves $y=x^2$ and $y=\frac{1}{x}$ intersect at $x=1$. The area is expressed as:

$\int_{0}^{1} x^2 \, dx + \int_{1}^{t} \frac{1}{x} \, dx = 1$

Evaluating the first integral provides $\left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$, while the second integral yields $[\ln x]_1^t = \ln t - \ln 1 = \ln t$. Setting the sum equal to $1$, we obtain $\frac{1}{3} + \ln t = 1$, which simplifies to $\ln t = \frac{2}{3}$, and consequently, $t = e^{\frac{2}{3}}$.

The composite function $(g \circ f)(x)$ is determined by substituting $f(x) = \sqrt{x}$ into $g(x) = \cos x^2$, which simplifies to $\cos(\sqrt{x})^2 = \cos x$. To establish the integration limits, we solve the quadratic equation $18x^{2-}9\pi x+\pi^2=0$. Factoring the expression as $(6x-\pi)(3x-\pi)=0$ reveals the roots $\frac{\pi}{6}$ and $\frac{\pi}{3}$. Given the condition $\alpha < \beta$, we identify $\alpha = \frac{\pi}{6}$ and $\beta = \frac{\pi}{3}$ as our lower and upper bounds.

The area bounded by this curve and the vertical lines is the definite integral of $\cos x$ from $\alpha$ to $\beta$:

$\int_{\pi/6}^{\pi/3} \cos x \, dx = [\sin x]_{\pi/6}^{\pi/3} = \sin\left(\frac{\pi}{3}\right) - \sin\left(\frac{\pi}{6}\right)$

Substituting the known values $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$ into the expression yields $\frac{\sqrt{3}}{2} - \frac{1}{2}$, which simplifies to $\frac{\sqrt{3}-1}{2}$ square units.

The region is bounded by the curves $y = \sqrt{x}$ and $y = x-2$ along with the axes $x=0$ and $y=0$. By expressing the curves in terms of $y$, we have $x = y^2$ and $x = y+2$, respectively. These curves intersect where $y^2 = y+2$, which simplifies to $y^2 - y - 2 = 0$, yielding $y=2$ as the relevant intersection point in the first quadrant, as $y \ge 0$.

The area is calculated by integrating the difference between the right boundary, $x = y+2$, and the left boundary, $x = y^2$, with respect to $y$ from $0$ to $2$. The integral is given by $\int_0^2 (y+2 - y^2) \, dy$, which evaluates to $[\frac{y^2}{2} + 2y - \frac{y^3}{3}]_0^2$. Substituting the upper limit gives $\frac{2^2}{2} + 2(2) - \frac{2^3}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$.

The area of a region enclosed by functions is found by integrating the vertical distance between the upper and lower boundary curves. In this problem, the lower boundary is defined by the parabola $y = \frac{x^2}{4}$ for the relevant range of $x$, while the upper boundary consists of two distinct segments: the curve $y = 1+\sqrt{x}$ for $x$ values between $0$ and $1$, and the line $y = 3-x$ for $x$ values between $1$ and $2$. Integrating these upper boundary functions over their respective intervals and subtracting the integral of the lower parabolic boundary from $0$ to $2$ yields the following expression:

$\int_{0}^{1} (1+\sqrt{x}) dx + \int_{1}^{2} (3-x) dx - \int_{0}^{2} \frac{x^2}{4} dx$

Computing these definite integrals results in the final value of $\frac{5}{2}$.

The area enclosed between the circle $x^{2+}y^2=4$ and the parabola $y^2=3x$ is identified by finding their points of intersection, which occurs where $x^{2+}3x-4=0$. Solving this quadratic equation yields $x=1$ (since $x=-4$ lies outside the parabolic domain), meaning the curves meet at $x=1$ where $y=\pm\sqrt{3}$. Because the shape is symmetric about the x-axis, the total area is twice the integral of the upper portion, which spans from $x=0$ to $x=2$. This boundary requires splitting the integral at the intersection point $x=1$, as the upper curve is defined by the parabola $y=\sqrt{3x}$ for $0 \leq x \leq 1$ and by the circle $y=\sqrt{4-x^2}$ for $1 \leq x \leq 2$.

Computing the parabolic segment provides $2 \int_{0}^{1} \sqrt{3x} \, dx = 2\sqrt{3} \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1} = \frac{4}{3}\sqrt{3}$. For the circular segment, the integral is $2 \int_{1}^{2} \sqrt{4-x^2} \, dx$. Applying the standard integral form $\int \sqrt{a^{2-}x^2} \, dx = \frac{x}{2}\sqrt{a^{2-}x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$ with $a=2$ and evaluating from $1$ to $2$ gives $2 \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{1}^{2}$. Substituting the bounds results in $2 \left[ (0 + 2(\frac{\pi}{2})) - (\frac{\sqrt{3}}{2} + 2(\frac{\pi}{6})) \right]$, which simplifies to $2\pi - \sqrt{3} - \frac{2\pi}{3}$, or $\frac{4\pi}{3} - \sqrt{3}$. Combining the parabola contribution $\frac{4}{3}\sqrt{3}$ with the circular contribution $\frac{4\pi}{3} - \sqrt{3}$ yields $\frac{4\sqrt{3}-3\sqrt{3}}{3} + \frac{4\pi}{3}$, simplifying to the final area of $\frac{1}{\sqrt{3}} + \frac{4\pi}{3}$.

Finding the area of this region relies on partitioning the shape into two distinct, manageable sections by identifying the intersection points of the line $y = 1-x$, the parabola $y = x^2 - 5x + 4$, and the horizontal axis $y = 0$. These curves meet at $(1, 0)$, $(3, -2)$, and $(4, 0)$.

The portion spanning from $x = 1$ to $x = 3$ forms a triangle bounded by the line $y = 1-x$, the x-axis, and the vertical line at $x = 3$. Given that the base along the x-axis is $2$ units and the vertical height at $x = 3$ is $2$ units, the area of this triangular section is $\frac{1}{2} \cdot 2 \cdot 2 = 2$.

For the section between $x = 3$ and $x = 4$, the region lies between the x-axis and the parabola $y = x^2 - 5x + 4$. The area is determined by evaluating the definite integral of the upper boundary minus the lower boundary, which simplifies to $\int_3^4 (0 - (x^2 - 5x + 4)) \, dx = \int_3^4 (-x^2 + 5x - 4) \, dx$. This integral evaluates as follows:
$\left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 4x \right]_3^4 = \left( -\frac{64}{3} + 40 - 16 \right) - \left( -9 + 22.5 - 12 \right) = \frac{8}{3} - 1.5 = \frac{7}{6}$
Combining the triangular area with the result of the integral provides the total area of $2 + \frac{7}{6} = \frac{19}{6}$.

Visualizing the region defined by the given inequalities reveals two primary curves: a circle centered at $(2,0)$ with a radius of $2$, described by $x^{2+}y^2=4x$, and a parabola opening along the positive x-axis defined by $y^2=2x$. These two curves intersect at the origin $(0,0)$ and the point $(2,2)$ in the first quadrant. To determine the area of the region bounded by these curves, we integrate the difference between the upper boundary of the circle and the upper boundary of the parabola with respect to $x$ over the interval $[0,2]$.

The area under the circular arc is calculated using the integral $\int_{0}^{2} \sqrt{4x - x^2} \, dx$, which geometrically corresponds to a quarter circle of area $\pi$. Subtracting the area enclosed by the parabola, found by evaluating $\int_{0}^{2} \sqrt{2x} \, dx$, provides the final measurement of the region. This integral evaluates to \sqrt{2} \cdot \frac{2}{3} \[x^{3/2}]_{0}^{2} = \frac{8}{3}$, so subtracting the parabolic area from the circular segment results \in a total area of$\pi - \frac{8}{3}$.

Calculating the area between the curve $y^2 = 2x$ and the line $y = 4x-1$ is best approached by expressing both relations as functions of $y$, yielding $x = \frac{y^2}{2}$ and $x = \frac{y+1}{4}$. The points of intersection are found by setting these expressions equal, resulting in the equation $\frac{y^2}{2} = \frac{y+1}{4}$, which simplifies to $2y^2 - y - 1 = 0$ and provides roots at $y = -1/2$ and $y = 1$. The required region is defined by the integral of the difference between these boundaries over the interval from $-1/2$ to $1$.

The setup for the area integration is:
$\int_{-1/2}^{1} \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy$

Evaluating this expression involves computing:
$\frac{1}{4} \left( \frac{y^2}{2} + y \right) \Big|_{-1/2}^{1} - \frac{1}{2} \left( \frac{y^3}{3} \right) \Big|_{-1/2}^{1} = \frac{1}{4} \left( \frac{1}{2} + 1 - \left( \frac{1}{8} - \frac{1}{2} \right) \right) - \frac{1}{2} \left( \frac{1}{3} \left( 1 + \frac{1}{8} \right) \right) = \frac{9}{32}$

Visualizing the overlapping shapes defined by $x^{2+}y^2 \leq 1$ and $y^2 \leq 1-x$ reveals that the region can be split into two distinct parts along the vertical axis at $x=0$. The curves intersect where $x^2 + 1 - x = 1$, which simplifies to $x^2 - x = 0$, giving intersection points at $x=0$ and $x=1$. The total area comprises the left half of the unit circle and the portion bounded by the parabola for $0 \leq x \leq 1$.

The area of the left semi-circle is $\frac{\pi}{2}$, while the area bounded by the parabola $y = \pm\sqrt{1-x}$ from $x=0$ to $x=1$ is calculated as $2\int_{0}^{1}\sqrt{1-x} \, dx$. Evaluating this integral yields $\frac{4}{3}$, as the integral of $(1-x)^{1/2}$ is $-\frac{2}{3}(1-x)^{3/2}$. Adding these two components together results in a total area of $\frac{\pi}{2}+\frac{4}{3}$.

Integrating with respect to the variable $y$ is the most efficient way to determine the area bounded by the curves $y = \sqrt{x}$ and $2y - x + 3 = 0$ in the first quadrant. Expressing the boundaries as functions of $y$, we obtain $x = y^2$ for the parabola and $x = 2y + 3$ for the line. The intersection of these curves occurs where $y^2 = 2y + 3$, or $y^2 - 2y - 3 = 0$, which factors into $(y - 3)(y + 1) = 0$. Since the region lies in the first quadrant, the interval of integration spans from the $x$-axis at $y = 0$ up to the intersection height of $y = 3$. The area is defined by the integral $\int_{0}^{3} ((2y + 3) - y^2) dy$. Evaluating the antiderivative $[y^2 + 3y - \frac{y^3}{3}]_{0}^{3}$ and substituting the limits gives $9 + 9 - 9$, resulting in a final area of $9$.

The area enclosed by the parabolas $x^2 = 9y$ and $x^2 = \frac{y}{4}$ bounded by the line $y=2$ can be found by expressing the curves as $x = \pm 3\sqrt{y}$ and $x = \pm \frac{\sqrt{y}}{2}$. Since both parabolas are symmetric about the y-axis, calculating the area requires doubling the integral of the region on one side, where the right boundary is $x = 3\sqrt{y}$ and the left boundary is $x = \frac{\sqrt{y}}{2}$.

This leads to the setup $2 \int_0^2 \left( 3\sqrt{y} - \frac{\sqrt{y}}{2} \right) dy$, which simplifies to $5 \int_0^2 \sqrt{y} dy$. Integrating $y^{1/2}$ yields $\frac{2}{3}y^{3/2}$, so evaluating this from $0$ to $2$ gives $5 \times \frac{2}{3} \times (2\sqrt{2})$. Thus, the total area is $\frac{20\sqrt{2}}{3}$.

Determining the area under these curves relies on identifying the intersection point at $x=1$, where the upper boundary transitions from the linear function to the reciprocal curve. The region is composed of two adjacent segments: the triangle formed by $y=x$ over the interval $[0, 1]$ and the area defined by $y=\frac{1}{x}$ over the interval $[1, e]$. The total area is obtained by summing the definite integrals for these segments:

$\int_{0}^{1} x \, dx + \int_{1}^{e} \frac{1}{x} \, dx = \left[\frac{x^2}{2}\right]_{0}^{1} + \left[\ln x\right]_{1}^{e} = \frac{1}{2} + 1 = \frac{3}{2}$

square units.

The total area between two functions $f(x)$ and $g(x)$ is calculated by integrating the absolute difference $|f(x) - g(x)|$ over the specified range, which requires splitting the integral at points where the curves intersect to ensure the upper function is always subtracted by the lower function. Within the interval from $x=0$ to $x=\frac{3\pi}{2}$, the curves $y = \cos x$ and $y = \sin x$ cross at $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$. Thus, the total area is given by the sum of three distinct integrals: $\int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) dx + \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}} (\cos x - \sin x) dx$. Evaluating these leads to the antiderivatives $(\sin x + \cos x)$ for the first and third parts and $(-\cos x - \sin x)$ for the middle part. Computing the definite values results in $(\sqrt{2}-1) + 2\sqrt{2} + (\sqrt{2}-1)$, which sums to $4\sqrt{2} - 2$.

To find the area enclosed by the parabola $(y-2)^2=x-1$, its tangent at $(2,3)$, and the $x$-axis, we treat $x$ as a function of $y$ to simplify the integration. The parabola can be expressed as $x = (y-2)^2 + 1$. By differentiating the parabola's equation with respect to $y$, we find that at the point of tangency $(2,3)$, the slope $\frac{dx}{dy}$ is $2(3-2) = 2$. This allows us to determine the tangent line equation as $x - 2 = 2(y - 3)$, which simplifies to $x = 2y - 4$.

Since the region is bounded between these two curves and extends from the $x$-axis where $y=0$ up to the intersection point at $y=3$, we calculate the area by integrating the difference between the right-hand boundary (the parabola) and the left-hand boundary (the tangent line). Setting up the definite integral gives:

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

Simplifying the integrand by expanding $(y-2)^2 + 1 - 2y + 4$ yields $y^2 - 6y + 9$, which is simply $(y-3)^2$. Evaluating the integral of $(y-3)^2$ with respect to $y$ over the interval $[0, 3]$ results in:

$\left[ \frac{(y-3)^3}{3} \right]_0^3 = 0 - \left( \frac{-27}{3} \right) = 9$

The area of the described region is 9 square units.