Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

Determining the area enclosed by two curves requires evaluating the integral of the upper function minus the lower function across the specified interval. Because the expression $y = |\cos x - \sin x|$ contains an absolute value, it changes definition at the point where $\cos x = \sin x$, which is $x = \frac{\pi}{4}$. On the sub-interval $[0, \frac{\pi}{4}]$, the expression $\cos x$ is greater than or equal to $\sin x$, so the integrand simplifies to $(\sin x + \cos x) - (\cos x - \sin x) = 2\sin x$. On the sub-interval $[\frac{\pi}{4}, \frac{\pi}{2}]$, $\sin x$ is greater than or equal to $\cos x$, leading to an integrand of $(\sin x + \cos x) - (\sin x - \cos x) = 2\cos x$.

The total area is found by summing the results of the two definite integrals:
$\int_{0}^{\frac{\pi}{4}} 2\sin x \,dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cos x \,dx$
Evaluating these, the first integral becomes $2[-\cos x]_{0}^{\frac{\pi}{4}} = 2(1 - \frac{1}{\sqrt{2}})$, and the second integral becomes $2[\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 2(1 - \frac{1}{\sqrt{2}})$. Adding these segments together results in $2(1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}}) = 2(2 - \sqrt{2})$. Factoring this expression leads to the final result of $2\sqrt{2}(\sqrt{2}-1)$.

Calculating the area under a piecewise function requires identifying the interval-specific expression for each segment by determining which component function dominates. Over the domain $[-3, 0]$, the equality $x+6 = x^2$ reveals a transition point at $x=-2$, where $x+6$ provides the smaller value on $[-3, -2]$ and $x^2$ is smaller on $[-2, 0]$. In the remaining interval $[0, 1]$, the expression $\sqrt{x}$ consistently exceeds $x^2$, defining the curve for this region.

The total area $A$ is the sum of three definite integrals:
$A = \int_{-3}^{-2} (x+6) \, dx + \int_{-2}^{0} x^2 \, dx + \int_{0}^{1} \sqrt{x} \, dx$
Evaluating these individually, the first integral $\left[\frac{x^2}{2} + 6x\right]_{-3}^{-2}$ equals $\frac{7}{2}$, the second integral $\left[\frac{x^3}{3}\right]_{-2}^{0}$ equals $\frac{8}{3}$, and the third integral $\left[\frac{2}{3}x^{3/2}\right]_0^1$ equals $\frac{2}{3}$. Summing these results gives $A = \frac{7}{2} + \frac{8}{3} + \frac{2}{3} = \frac{41}{6}$, and multiplying this value by 6 yields 41.

The area of the region bounded by $x^{2} \leq y \leq 2x$ is best determined by integrating with respect to the variable $y$, which requires expressing the boundary curves as $x = \sqrt{y}$ and $x = y/2$. The curves intersect where $x^{2} = 2x$, defining the range of $y$ from $0$ to $4$. The total area of this region is given by the definite integral $\int_{0}^{4} (\sqrt{y} - \frac{y}{2}) dy$, which evaluates to $\frac{4}{3}$.

Since the horizontal line $y = \alpha$ divides the region into two equal parts, the integral of the boundary difference from $0$ to $\alpha$ must equal half of the total area, which is $\frac{2}{3}$. Setting up the equation $\int_{0}^{\alpha} (\sqrt{y} - \frac{y}{2}) dy = \frac{2}{3}$ leads to the evaluation $[\frac{2}{3}y^{3/2} - \frac{y^{2}}{4}]_{0}^{\alpha} = \frac{2}{3}$, which simplifies to $\frac{2}{3}\alpha^{3/2} - \frac{\alpha^{2}}{4} = \frac{2}{3}$. Multiplying the entire equation by 12 clears the denominators to produce $8\alpha^{3/2} - 3\alpha^{2} = 8$, which rearranges to the final relation $3\alpha^{2} - 8\alpha^{3/2} + 8 = 0$.

To determine the area of the region, we first identify the point where the curves intersect by solving $x^2 + 1 = x + 1$, which shows the intersection occurs at $x=1$. Between the x-values of $\frac{1}{2}$ and $1$, the upper boundary is defined by the parabola $y = x^2 + 1$, whereas between $1$ and $2$, the upper boundary is defined by the line $y = x + 1$.

The total area is found by adding the integral $\int_{1/2}^{1} (x^2 + 1) \, dx$ to the area of the trapezoid formed under the line from $1$ to $2$. The integration results in $\frac{19}{24}$, and the geometric area of the trapezoid with parallel sides of length $2$ and $3$ and a height of $1$ evaluates to $\frac{5}{2}$. Summing these parts, $\frac{19}{24} + \frac{5}{2}$, leads to the total area of $\frac{79}{24}$.

The total area enclosed by the curves $C_1 : y^2 = ax$ and $C_2 : x^2 = ay$ is determined by integrating the difference between the upper function, $y = \sqrt{ax}$, and the lower function, $y = \frac{x^2}{a}$, from the origin to the point of intersection $P(a, a)$. Calculating this definite integral, $\int_{0}^{a} (\sqrt{ax} - \frac{x^2}{a}) dx$, results in a total area of $\frac{a^2}{3}$. Because the line $x=b$ bisects this area, the definite integral of the difference function from $0$ to $b$ must equal half of the total area, which is $\frac{a^2}{6}$. This provides the equation $\int_{0}^{b} (\sqrt{ax} - \frac{x^2}{a}) dx = \frac{a^2}{6}$, which simplifies to $\frac{2\sqrt{a}}{3}b^{3/2} - \frac{b^3}{3a} = \frac{a^2}{6}$.

To find the value of $b$, consider the triangle $\Delta OQR$ formed by the origin, the intersection point $Q$ of the line $x=b$ with the chord $OP$, and the intersection point $R$ of the line $x=b$ with the x-axis. The chord connecting the origin $(0,0)$ and $P(a, a)$ lies on the line $y=x$, so point $Q$ is located at $(b, b)$. This creates a right-angled triangle with a base of $b$ and a height of $b$, so its area is $\frac{1}{2}b^2$. Given that the area of $\Delta OQR$ is $\frac{1}{2}$, it follows that $b^2 = 1$, or $b = 1$. Substituting $b=1$ into the previously derived area equation yields $\frac{2\sqrt{a}}{3} - \frac{1}{3a} = \frac{a^2}{6}$. Multiplying this entire equation by $6a$ gives $4a^{3/2} - 2 = a^3$, which can be rewritten as $a^3 + 2 = 4a^{3/2}$. Squaring both sides of this equation results in $a^6 + 4a^3 + 4 = 16a^3$, which simplifies to $a^6 - 12a^3 + 4 = 0$.

The condition $|x|+|y| \le 1$ represents a diamond-shaped region centered at the origin, with vertices at $(\pm 1, 0)$ and $(0, \pm 1)$. When combined with the inequality $2y^2 \ge |x|$, the region is restricted to the space bounded by the parabolas $x = 2y^2$ and $x = -2y^2$. Because the geometric shape is symmetric across both the x and y axes, we can calculate the area of the region lying in the first quadrant and multiply that result by four to determine the total area.

In the first quadrant, the region is bounded by the line $x+y=1$ and the parabola $x=2y^2$. Substituting $x = 1-y$ into the equation $x=2y^2$ yields the quadratic $2y^2 + y - 1 = 0$, which factors as $(2y-1)(y+1)=0$. This confirms the intersection point occurs at $y = \frac{1}{2}$, which corresponds to $x = \frac{1}{2}$.

The area in the first quadrant is found by integrating with respect to $y$, splitting the region at the intersection point. The area under the parabola from $y=0$ to $y=\frac{1}{2}$ plus the area under the line from $y=\frac{1}{2}$ to $y=1$ covers the entire segment:
$\int_{0}^{\frac{1}{2}} 2y^2 \, dy + \int_{\frac{1}{2}}^{1} (1-y) \, dy = \left[ \frac{2y^3}{3} \right]_{0}^{\frac{1}{2}} + \left[ y - \frac{y^2}{2} \right]_{\frac{1}{2}}^{1} = \frac{1}{12} + \frac{1}{8} = \frac{5}{24}$
Multiplying this quadrant area by four gives a total area of $\frac{5}{6}$.

Finding the area of the circle that remains after removing the region enclosed by the parabola $y^2=x$ and the line $y=x$ requires subtracting the common area from the total area of the circle. The circle $x^{2+}y^2=2$ has a radius squared of $2$, giving it a total area of $2\pi$.

To determine the shared area, we find the intersection points of the parabola $y^2=x$ and the line $y=x$ by setting $x^2=x$, which yields $x=0$ and $x=1$. Calculating the definite integral of the difference between the upper curve $\sqrt{x}$ and the lower curve $x$ over this interval gives $\int_{0}^{1}(\sqrt{x}-x)dx = \left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$. Subtracting this portion from the total area of the circle results in $2\pi - \frac{1}{6}$, which simplifies to $\frac{12\pi-1}{6}$ or $\frac{1}{6}(12\pi-1)$.

To calculate the area enclosed by two curves, integrate the difference between the upper function and the lower function across the interval defined by their intersection points. Setting $x^{2}-1$ equal to $1-x^{2}$ simplifies to $2x^{2} = 2$, confirming the limits of integration at $x=-1$ and $x=1$.

With $1-x^{2}$ positioned as the upper curve, the area is defined by the integral $\int_{-1}^{1} ((1-x^{2}) - (x^{2}-1)) dx$, which simplifies to $\int_{-1}^{1} (2-2x^{2}) dx$. Taking advantage of symmetry, this expression becomes $4 \int_{0}^{1} (1-x^{2}) dx$. Evaluating the integral results in 4\[x - \frac{x^{3}}{3}]_{0}^{1}$, which simplifies to$4(\frac{2}{3})$, yielding a final area of$\frac{8}{3}$.

Finding the area enclosed by two curves involves calculating the definite integral of the upper function minus the lower function across the interval defined by their points of intersection. By setting $y = 4x^2$ equal to $y = 8x + 12$, we arrive at the quadratic equation $4x^2 - 8x - 12 = 0$, which simplifies to $x^2 - 2x - 3 = 0$. Factoring this expression gives the roots $x = -1$ and $x = 3$, which serve as the boundaries for our integration.

Within this interval, the line remains above the parabola, requiring us to integrate the difference function $(8x + 12 - 4x^2)$ with respect to $x$ from -1 to 3. Evaluating the definite integral $\int_{-1}^{3} (8x + 12 - 4x^2) dx$ results in the antiderivative $[4x^2 + 12x - \frac{4}{3}x^3]_{-1}^{3}$. Applying the fundamental theorem of calculus by substituting the upper and lower limits, we calculate the difference as $36 - (-\frac{20}{3})$, which yields a final area of $\frac{128}{3}$.

Finding the area of a region nested between two curves requires calculating the area enclosed by the outer boundary and subtracting the area enclosed by the inner boundary. The ellipse defined by $\frac{x^2}{4} + \frac{y^2}{9} = 1$ has semi-axes of lengths 2 and 3, resulting in an area calculated as $\pi \times 2 \times 3 = 6\pi$. The inner boundary, described by $\frac{|x|}{2} + \frac{|y|}{3} = 1$, forms a rhombus with diagonal lengths of 4 and 6, which yields an area of $\frac{1}{2} \times 4 \times 6 = 12$. Subtracting the rhombus area from the ellipse area gives the final result:

$6\pi - 12 = 6(\pi - 2)$

Finding the area of the region $A$ bounded by two functions involves integrating the difference between the upper curve and the lower curve across the specified interval $[0, 2]$. The upper boundary is given by $y = 2\sqrt{x}$. The lower boundary, $y = (x-1)[x]$, is a piecewise function defined by the behavior of the greatest integer function $[x]$. On the interval $[0, 1)$, the value of $[x]$ is 0, so the function simplifies to $y = 0$. On the interval $[1, 2]$, the value of $[x]$ is 1, simplifying the function to $y = x - 1$.

Evaluating the definite integral of the upper curve, $\int_{0}^{2} 2\sqrt{x} \, dx$, yields $\frac{8}{3}\sqrt{2}$. The area contributed by the lower boundary occurs only in the interval $[1, 2]$, representing the area under the linear segment $y = x - 1$. Geometrically, this region forms a right-angled triangle with a base of 1 and a height of 1, resulting in an area of $\frac{1}{2}$. Subtracting this area from the total area derived from the upper curve leaves a final result of $\frac{8}{3}\sqrt{2} - \frac{1}{2}$.

Finding the area bounded by two curves requires identifying the intersection points where the boundaries meet, which is done by setting the equations equal to each other. By equating $x^2 = 3 - 2x$, we obtain the quadratic equation $x^2 + 2x - 3 = 0$, which factors into $(x+3)(x-1) = 0$, establishing our integration limits at $x = -3$ and $x = 1$. Within this interval, the linear expression $3 - 2x$ exceeds the parabolic expression $x^2$, so the region is determined by the integral of their difference.

The area is evaluated by the integral $\int_{-3}^{1} (3 - 2x - x^2) \, dx$. Applying the fundamental theorem of calculus with the antiderivative $3x - x^2 - \frac{x^3}{3}$ allows for the substitution of the limits. At the upper bound $x = 1$, the value is $\frac{5}{3}$, and at the lower bound $x = -3$, the value is -9. Subtracting the lower bound result from the upper bound result leads to $\frac{5}{3} - (-9)$, which simplifies to $\frac{32}{3}$.

To find the area bounded by two functions, we calculate the definite integral of the difference between the upper function and the lower function across the specified boundaries. In the interval defined by $x = \frac{1}{2}$ and $x = \frac{\sqrt{3}}{2}$, the function $f(x)$ acts as $1-x$ and the function $g(x)$ is defined as $(x-\frac{1}{2})^2$. Since $1-x$ represents the upper boundary in this range, we calculate the area using the integral:

$\int_{1/2}^{\sqrt{3}/2} \left( (1-x) - \left(x-\frac{1}{2}\right)^2 \right) \, dx$

Using the substitution $u = x - \frac{1}{2}$, where $dx = du$, the integration limits shift from $[\frac{1}{2}, \frac{\sqrt{3}}{2}]$ to $[0, \frac{\sqrt{3}-1}{2}]$. Substituting these into the integral, the expression $1-x$ becomes $1 - (u + \frac{1}{2})$, which simplifies to $\frac{1}{2} - u$. The integrand thus becomes $\frac{1}{2} - u - u^2$. Calculating the antiderivative gives $\frac{1}{2}u - \frac{1}{2}u^2 - \frac{1}{3}u^3$. Evaluating this expression at the upper limit of $\frac{\sqrt{3}-1}{2}$ yields the final area of $\frac{\sqrt{3}}{4} - \frac{1}{3}$.

Finding the area enclosed by $y=2^{x}$ and $y=|x+1|$ in the first quadrant requires identifying the behavior of these functions where $x \ge 0$. Since the absolute value function $|x+1|$ simplifies to $x+1$ for these non-negative values, the curves are effectively $y=2^{x}$ and $y=x+1$. These functions intersect at $x=0$ and $x=1$, which define the horizontal boundaries for integration. Within this interval, the line $y=x+1$ lies above the exponential curve $y=2^{x}$, so the area is given by the integral $\int_{0}^{1} (x+1 - 2^{x}) \, dx$. Computing this leads to the antiderivative $\frac{x^2}{2} + x - \frac{2^x}{\ln 2}$. Evaluating this expression at the limits results in $(\frac{1}{2} + 1 - \frac{2}{\ln 2}) - (0 + 0 - \frac{1}{\ln 2})$, which simplifies to $\frac{3}{2} - \frac{1}{\ln 2}$.

The area of a region trapped between a curve and a line is determined by calculating the definite integral of the difference between the upper and lower boundaries between their intersection points. By substituting the parabola $x^2 = 4y$ into the line $x = 4y - 2$, we form the equation $x^2 - x - 2 = 0$, which reveals intersection points at $x = -1$ and $x = 2$.

Expressing both relations in terms of $y$ identifies the line as $y = \frac{x+2}{4}$ and the parabola as $y = \frac{x^2}{4}$. Evaluating the definite integral of the difference between these functions across this interval provides the final area:

$\int_{-1}^{2} \left(\frac{x+2}{4} - \frac{x^2}{4}\right) \, dx = \frac{9}{8}$

The area bounded by the parabola $y = x^2 + 1$ and the coordinate axes in the first quadrant is determined by finding the total area under the curve and subtracting the area of the triangle formed by the tangent line. Differentiating the parabola equation gives a slope of $dy/dx = 2x$, which equals 4 at the point (2, 5). The tangent line equation, $y - 5 = 4(x - 2)$, simplifies to $y = 4x - 3$ and crosses the x-axis at x = 3/4. Calculating the definite integral $\int_0^2 (x^2 + 1) \, dx$ provides 14/3. The triangle formed between the x-axis, the tangent line, and the vertical line at x = 2 has a base of 5/4 and a height of 5, resulting in an area of 25/8. Subtracting this triangular area from the integral yields 37/24 square units.

The area bounded between two curves is determined by calculating the definite integral of the difference between the upper and lower functions over the interval from $x = 0$ to $x = 3$. This can be solved by evaluating the integral of the parabola $y = x^2 + 2$ to find the total area under that curve and then subtracting the area of the trapezoidal region formed by the line $y = x + 1$. The calculation proceeds as follows:

$\int_{0}^{3} (x^2 + 2) \, dx - \frac{1}{2} \times 5 \times 3 = 15 - \frac{15}{2} = \frac{15}{2}$

Subtracting the trapezoidal area from the parabolic integral results in a final area of $\frac{15}{2}$ sq. units.

The area bounded by two curves is computed by integrating the difference between the upper and lower functions within the interval defined by their points of intersection. Setting the line equation $y=\lambda x$ equal to the parabola $y^2=4\lambda x$ identifies the intersection points at $0$ and $\frac{4}{\lambda}$. The integral $\int_{0}^{\frac{4}{\lambda}} (\sqrt{4\lambda x} - \lambda x) \, dx$ evaluates to the expression $\frac{8}{3\lambda}$. Equating this to the provided area of $\frac{1}{9}$ results in the equation $\frac{8}{3\lambda} = \frac{1}{9}$, establishing that $\lambda = 24$.

Calculating the area bounded by a curve, a tangent, and the y-axis depends on identifying the boundaries and integrating the difference between the upper and lower functions. For the parabola $y = x^2 - 1$ at the point (2, 3), the slope of the tangent is given by the derivative $y' = 2x$, which equals 4 at $x = 2$. Using the point-slope form, the tangent equation becomes $y = 4x - 5$.

Since the y-axis acts as the left boundary $x = 0$ and the tangent touches the curve at $x = 2$, these values serve as the limits of integration. Within this interval, the parabola $y = x^2 - 1$ is positioned above the tangent line $y = 4x - 5$. The enclosed area is found by evaluating the definite integral $\int_{0}^{2} ((x^2 - 1) - (4x - 5)) \, dx$, which simplifies to $\int_{0}^{2} (x - 2)^2 \, dx$. Applying the fundamental theorem of calculus, the result is $\left[ \frac{(x - 2)^3}{3} \right]_{0}^{2}$, which simplifies to $0 - (-8/3)$ and yields a total area of 8/3 square units.

The area bounded by two parabolas defined as $x^2 = 4ay$ and $y^2 = 4bx$ is given by the expression $\frac{16ab}{3}$. Rearranging the provided equations $y = kx^2$ and $x = ky^2$ into standard parabolic form results in $x^2 = \frac{1}{k}y$ and $y^2 = \frac{1}{k}x$, allowing for the identification of $4a = \frac{1}{k}$ and $4b = \frac{1}{k}$. These relations indicate that $a = \frac{1}{4k}$ and $b = \frac{1}{4k}$, which, when substituted into the area formula, creates the following equation:

$\frac{16(\frac{1}{4k} \cdot \frac{1}{4k})}{3} = 1$

This simplifies directly to $\frac{1}{3k^2} = 1$. Isolating $k^2$ yields $k^2 = \frac{1}{3}$, and because the condition $k > 0$ is specified, the final result is $k = \frac{1}{\sqrt{3}}$.