Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The curve defined by $y=3-\left|x-\frac{1}{2}\right|-\left|x+1\right|$ forms a trapezoidal region when bounded by the $x$-axis. In the interval $-1 \leq x \leq \frac{1}{2}$, the function simplifies to the constant value $y=\frac{3}{2}$, which establishes the top parallel edge of the trapezoid with a length of $\frac{3}{2}$. Outside this interval, the graph slopes linearly until it reaches the $x$-axis at $x=-\frac{7}{4}$ and $x=\frac{5}{4}$, effectively creating a base on the $x$-axis with a total length of $3$. By applying the geometric area formula for a trapezoid, we calculate the area of this enclosed region as:
$Area = \frac{1}{2} \left(3 + \frac{3}{2}\right) \cdot \frac{3}{2} = \frac{27}{8}$

The region is bounded below by the parabola $y=x^2$ and above by the minimum of two lines, $y=x+2$ and $y=4-3x$. Identifying the boundaries requires locating the intersection points: the parabola $y=x^2$ meets $y=x+2$ at $x=-1$, meets $y=4-3x$ at $x=1$, and the two lines intersect each other at $x=0.5$. Because the upper boundary shifts from $x+2$ to $4-3x$ at $x=0.5$, calculate the total area by evaluating the integral in two separate sections.

Integrating the difference between the first line and the parabola from $x=-1$ to $x=0.5$ gives:
$\int_{-1}^{\frac{1}{2}} (x+2-x^2) dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{\frac{1}{2}} = \left(\frac{1}{8} + 1 - \frac{1}{24}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{9}{4}$
Integrating the difference between the second line and the parabola from $x=0.5$ to $x=1$ gives:
$\int_{\frac{1}{2}}^{1} (4-3x-x^2) dx = \left[ 4x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{\frac{1}{2}}^{1} = \left(4 - \frac{3}{2} - \frac{1}{3}\right) - \left(2 - \frac{3}{8} - \frac{1}{24}\right) = \frac{7}{12}$
Summing these areas, the total region area is $\frac{9}{4} + \frac{7}{12} = \frac{27}{12} + \frac{7}{12} = \frac{34}{12} = \frac{17}{6}$.

The region defined by $|x+\alpha| \le y \le 2-|x|$ is contained within a large triangle with vertices at $(-2, 0)$, $(2, 0)$, and $(0, 2)$. This larger triangle has a total area of $\frac{1}{2} \times 4 \times 2$, which equals 4. To find the enclosed area, we subtract the two smaller triangular regions formed by the lower boundary $y = |x+\alpha|$ from the large triangle, resulting in the expression $4 - \left( \frac{(2-\alpha)^2}{4} + \frac{(2+\alpha)^2}{4} \right)$. Setting this expression equal to $\frac{3}{2}$ leads to the simplified equation $2\alpha^2 + 8 = 10$, which confirms that $\alpha = 1$ since $\alpha > 0$. Applying this value to the second region defined by $|x| \le 1$ and $0 \le y \le x+2$, we identify a trapezoidal area. This trapezoid has a width of 2 between the boundaries $-1$ and $1$, with vertical sides of length 1 and 3 at those bounds. The area is $\frac{1+3}{2} \times 2$, which equals 4.

The area of a circle with radius $r$ is given by $\pi r^2$, and the area of an ellipse with semi-axes $a$ and $b$ is given by $\pi ab$. Given the provided areas, we establish two relationships: $\pi a^2 - \pi ab = 30\pi$ and $\pi ab - \pi b^2 = 18\pi$. Dividing each equation by $\pi$ simplifies them to $a^2 - ab = 30$ and $ab - b^2 = 18$. Subtracting the second equation from the first yields $a^2 - ab - (ab - b^2) = 30 - 18$, which simplifies to $a^2 - 2ab + b^2 = 12$. Because this result is the algebraic expansion of $(a-b)^2$, the value is $12$.

The region $A_1$ is defined by the intersection of the parabola $|x| = y^2$ and the lines described by $|x|+2y = 8$. Due to symmetry across the $y$-axis, the total area is double the integral over the positive $x$ domain, where the region is bounded above by $y = \frac{8-x}{2}$ and below by $y = \sqrt{x}$. Calculating the intersection point $x = 4$, the area $A_1$ is obtained by integrating:
$A_1 = 2 \int_{0}^{4} \left( \frac{8-x}{2} - \sqrt{x} \right) dx = \frac{40}{3}$

The region $A_2$ defined by $|x|+|y| \le k$ represents a square rotated by $45^\circ$ with diagonals of length $2k$, resulting in an area of $A_2 = 2k^2$. Substituting these values into the given relationship $27(\text{Area } A_1) = 5(\text{Area } A_2)$, the expression becomes $27 \cdot \frac{40}{3} = 5 \cdot 2k^2$. Simplifying this equation yields $360 = 10k^2$, which results in $k^2 = 36$ and therefore $k = 6$.

The equation $4y^2 = x^2(4-x)(x-2)$ describes a region bounded by a curve that is real only when $(4-x)(x-2) \geq 0$. Solving this inequality indicates that the curve exists solely for $x$ in the interval $[2, 4]$. Within this range, we can express the vertical boundary as $y = \pm \frac{x}{2} \sqrt{(4-x)(x-2)}$, and since the area is symmetric about the x-axis, we calculate the total area by doubling the integral of the positive branch, yielding $A = \int_2^4 x \sqrt{(4-x)(x-2)} \, dx$.

To evaluate this integral, we employ the definite integral property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$, where $a=2$ and $b=4$. Replacing $x$ with $6-x$ transforms the integral into $A = \int_2^4 (6-x) \sqrt{(4-(6-x))((6-x)-2)} \, dx$, which simplifies to $\int_2^4 (6-x) \sqrt{(x-2)(4-x)} \, dx$. Adding the original expression for $A$ to this new version results in $2A = 6 \int_2^4 \sqrt{(4-x)(x-2)} \, dx$, which reduces to $A = 3 \int_2^4 \sqrt{(4-x)(x-2)} \, dx$.

Completing the square inside the radical, we rewrite $(4-x)(x-2)$ as $-x^2 + 6x - 8$, which equals $1 - (x-3)^2$. The integral becomes $3 \int_2^4 \sqrt{1 - (x-3)^2} \, dx$. By substituting $u = x-3$, the limits shift to $-1$ and $1$, resulting in $3 \int_{-1}^1 \sqrt{1-u^2} \, du$. This final integral represents the area of a semicircle with radius 1, so the calculation yields $3 \cdot \frac{\pi(1)^2}{2}$, which is $\frac{3\pi}{2}$.

The graph of $y = | |x-1| - 2 |$ forms a triangular region bounded by the X-axis, which is identified by finding the roots where the function equals zero. Solving the equation $| |x-1| - 2 | = 0$ yields the x-intercepts $x = -1$ and $x = 3$, creating a base width of 4. The peak of the triangle occurs at the vertex $x = 1$, corresponding to a vertical height of 2. Applying the geometric area formula $Area = \frac{1}{2} \times \text{base} \times \text{height}$, the calculation yields 4.

Finding the area enclosed by two curves requires integrating the difference between the upper function and the lower function over the interval defined by their points of intersection. Setting the expressions $x^2$ and $x+2$ equal leads to $x^2 - x - 2 = 0$, which factors as $(x-2)(x+1) = 0$, establishing the limits $x = -1$ and $x = 2$. Within this range, the line lies above the parabola, so the area is calculated using the definite integral $\int_{-1}^{2} (2 + x - x^2) dx$. Evaluating the antiderivative $\left[2x + \frac{x^2}{2} - \frac{x^3}{3}\right]$ at the bounds yields $(4 + 2 - \frac{8}{3})$ for the upper limit and $(-2 + \frac{1}{2} + \frac{1}{3})$ for the lower limit. Calculating the final difference between these values results in $\frac{10}{3} - (-\frac{7}{6})$, which equals $\frac{9}{2}$.

Finding the area between a curve and the x-axis requires integrating the absolute value of the function over the given interval, which involves identifying where the function crosses the x-axis because the sign of the function determines whether to add or subtract the integral. Calculating the local extrema $a$ and $b$ begins by setting the derivative $f'(x) = 6x^2 - 6x - 12$ to zero, factoring the expression into $6(x+1)(x-2) = 0$, which confirms $a = -1$ and $b = 2$.

Since the function crosses the x-axis at $x = 0$, the total area $A$ is the sum of the positive integral from $-1$ to $0$ and the absolute value of the negative integral from $0$ to $2$. The required area corresponds to the following equation:
$A = \int_{-1}^{0} (2x^3 - 3x^2 - 12x) dx + \int_{0}^{2} (-2x^3 + 3x^2 + 12x) dx$

Evaluating these definite integrals, the first part yields $\left[ \frac{x^4}{2} - x^3 - 6x^2 \right]_{-1}^{0} = \frac{9}{2}$, while the second part results in $\left[ -\frac{x^4}{2} + x^3 + 6x^2 \right]_{0}^{2} = 24$. Summing these values gives $A = \frac{57}{2}$, and multiplying by four results in $4A = 114$.

To determine the value of $m$ such that the line $y=mx$ bisects the area under the curve $y=1+4x-x^{2}$ from $x=0$ to $x=\frac{3}{2}$, we first compute the total area using the definite integral $\int_{0}^{\frac{3}{2}} (1+4x-x^{2}) dx$. Evaluating the antiderivative $[x+2x^{2}-\frac{x^{3}}{3}]$ at the given limits results in $\frac{3}{2} + 2(\frac{9}{4}) - \frac{1}{3}(\frac{27}{8})$, which simplifies to $\frac{3}{2} + \frac{9}{2} - \frac{9}{8} = \frac{39}{8}$.

Because the line $y=mx$ must divide this area exactly in half, the integral $\int_{0}^{\frac{3}{2}} mx \, dx$ must equal $\frac{1}{2}$ of the total area, or $\frac{39}{16}$. Calculating the integral for the line gives \frac{m}{2} \[x^{2}]_{0}^{\frac{3}{2}} = \frac{m}{2} × \frac{9}{4} = \frac{9m}{8}$. Setting these expressions equal,$\frac{9m}{8} = \frac{39}{16}$, leads to$m = \frac{39}{18}$. Multiplying this value by$12$provides the final result$12m = 26$.

Finding the area of a region bounded by two functions involves integrating the difference between the upper boundary and the lower boundary across the interval where they intersect. Converting the inequalities to their respective boundary functions, we have the line $y = \frac{3}{2}x + 6$ and the parabola $y = \frac{3}{4}x^2$. Equating these functions provides the intersection points:

$\frac{3x^2}{4} = \frac{3x}{2} + 6$

Solving the resulting quadratic equation $x^2 - 2x - 8 = 0$ gives $x = -2$ and $x = 4$, which define the limits of integration. The total area is calculated using the definite integral:

$\int_{-2}^{4} \left( \frac{3x}{2} + 6 - \frac{3x^2}{4} \right) dx$

Finding the antiderivative gives $\frac{3x^2}{4} + 6x - \frac{x^3}{4}$. Evaluating this expression at the upper limit of 4 yields 20, while evaluating it at the lower limit of -2 yields -7. Subtracting the value at the lower bound from the value at the upper bound results in 27.

The intersection points of the sine and cosine functions occur where $\sin x = \cos x$, which defines the interval boundaries at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. The area $A$ is represented by the definite integral:
$A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx$
Evaluating the antiderivative $[-\cos x - \sin x]$ from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ yields $\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right)$, which simplifies to $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$. Consequently, raising this result to the fourth power provides $A^{4} = (2\sqrt{2})^{4} = 64$.

The total area of the circle $x^{2}+y^{2}=36$ is $36\pi$, and the requested area is found by subtracting the region where the circle overlaps the parabola $y^{2}=9x$ from this total. The circle has a radius of $6$ centered at the origin, while the parabola opens to the right, intersecting the circle at $x=3$ since $x^{2}+9x-36=0$ has $x=3$ as the valid positive solution. The region enclosed by both curves is symmetric about the x-axis and consists of the area under the parabola from $x=0$ to $x=3$ and the area under the circle from $x=3$ to $x=6$.

Integrating the respective functions and doubling these results to account for the portion below the x-axis provides the total area inside both shapes. The area within the parabolic segment is calculated as $\int_{0}^{3} 2\sqrt{9x} \, dx$, which evaluates to $12\sqrt{3}$, and the integral for the circular portion is $2\int_{3}^{6} \sqrt{36-x^{2}} \, dx$, which equals $12\pi - 9\sqrt{3}$. Combining these two interior regions results in a total area of $12\pi + 3\sqrt{3}$. Subtracting this quantity from the total area of the circle leaves the final area outside the parabola as $24\pi - 3\sqrt{3}$.

The area bounded by the parabola defined by $(y-2)^{2}=(x-1)$ and the tangent at a specific ordinate is best determined by integrating with respect to $y$, as the parabola describes $x$ as a clear function of $y$. First, substitute the ordinate $y=3$ into the parabolic equation $(3-2)^{2}=x-1$ to find the corresponding $x$-coordinate, which yields $x=2$. Differentiating the equation $(y-2)^{2}=x-1$ with respect to $x$ provides $2(y-2)\frac{dy}{dx}=1$, which reveals that the slope of the tangent at $(2, 3)$ is $\frac{1}{2(3-2)}=\frac{1}{2}$. By applying the point-slope form, the tangent line equation becomes $y-3=\frac{1}{2}(x-2)$, which simplifies to the form $x=2y-4$.

To determine the area enclosed by the parabola, the tangent, and the X-axis, integrate the difference between the $x$-coordinates of the parabola and the tangent with respect to $y$ over the interval from $0$ to $3$. The setup for this definite integral is \int_{0}^{3} \[(y-2)^{2} + 1 - (2y - 4)]$ , dy$. Simplifying the expression inside the integral yields$\int_{0}^{3} (y^{2} - 4y + 4 + 1 - 2y + 4) , dy$, which further reduces to$\int_{0}^{3} (y^{2} - 6y + 9) , dy$. Evaluating the antiderivative gives$\left[ \frac{y^3}{3} - 3y^2 + 9y \right]_{0}^{3}$, resulting in a final value of$9$ square units.

The tangent line to the ellipse $x^{2}+4y^{2}=5$ at point $P(1,1)$ is determined by the equation $x+4y=5$, which can be written as $y=\frac{5-x}{4}$. For the region bounded by this tangent, the ellipse, and the vertical lines $x=1$ and $x=\sqrt{5}$, the area is calculated by integrating the difference between the tangent function and the ellipse function $y=\frac{\sqrt{5-x^{2}}}{2}$. This area is defined by the integral $\int_{1}^{\sqrt{5}} \left(\frac{5-x}{4}-\frac{\sqrt{5-x^{2}}}{2}\right) dx$. Performing this integration yields the expression $\frac{5}{4}\sqrt{5} - \frac{5}{4} - \frac{5}{4}\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$. Comparing this result to the standard form $\alpha\sqrt{5}+\beta+\gamma\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$, we identify the coefficients as $\alpha=\frac{5}{4}$, $\beta=-\frac{5}{4}$, and $\gamma=-\frac{5}{4}$. Consequently, evaluating the absolute value $\lvert\alpha+\beta+\gamma\rvert$ gives 1.

The area of the region is defined by the boundaries $y = 2x^{2}$ and $y = 4-2x$ within the interval where $x \geq 0$. Equating these two functions as $2x^{2} = 4-2x$ leads to the quadratic equation $x^{2} + x - 2 = 0$, which factors as $(x+2)(x-1) = 0$. Given the restriction $x \geq 0$, the curves intersect at $x=1$, meaning the region of interest spans from $x=0$ to $x=1$, where the line $4-2x$ lies above the parabola $2x^{2}$.

Calculating the area requires evaluating the definite integral:
$\int_{0}^{1} (4 - 2x - 2x^{2}) \, dx$
Applying the power rule for integration results in the expression $4x - x^{2} - \frac{2x^{3}}{3}$ evaluated at the boundaries. Substituting the limits yields $4(1) - (1)^{2} - \frac{2}{3}(1)^{3}$, which simplifies to $4 - 1 - \frac{2}{3}$. This calculation results in a final value of $\frac{7}{3}$.

Calculating the area of a region bounded by multiple parabolas relies on identifying the symmetry of the curves and using definite integrals to compute the net space between the boundaries. Since the parabolas are symmetric about the $y$-axis within the upper half plane, the total area can be computed by finding the area of the portion in the first quadrant and doubling the result. This process involves integrating the boundary functions with respect to the coordinate axes to capture the enclosed region.

The total area is defined by the following expression:
$2\left[\int_{0}^{2}\frac{4-y^{2}}{4}dy-\int_{0}^{1}\frac{1-x^{2}}{2}dx\right]$
Evaluating the first definite integral gives $\frac{4}{3}$ and the second yields $\frac{1}{3}$, representing the differences between the specified curves. Subtracting these values provides $2\left[\frac{4}{3}-\frac{1}{3}\right]$, which simplifies to 2.

The area of a region bounded by two curves is found by integrating the difference between the upper and lower boundary functions over the intersection interval. Setting the equations $5x^2 = 2x^2 + 9$ equal reveals that the curves intersect at $x = \pm\sqrt{3}$. The total area is obtained by evaluating the integral of the difference function $9 - 3x^2$ over this range:
$\int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2) dx = 2 \int_{0}^{\sqrt{3}} (9 - 3x^2) dx = 2 \$[9x - x^3]_0^{\sqrt{3}}$
Evaluating this expression at the limits yields $2(9\sqrt{3} - 3\sqrt{3})$, which results in $12 \sqrt{3}$ square units.

Finding the area between curves involves integrating the difference between the functions over the interval bounded by their points of intersection. The curves $y = \sin x$ and $y = \cos x$ intersect in the first quadrant at $x = \frac{\pi}{4}$, where both functions equal $\frac{1}{\sqrt{2}}$. The area $A_1$ is calculated by integrating the upper curve $\cos x$ minus the lower curve $\sin x$ from $0$ to $\frac{\pi}{4}$, resulting in $\int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = [\sin x + \cos x]_{0}^{\frac{\pi}{4}} = \sqrt{2} - 1$. The area $A_2$ is defined by the integral of $\sin x$ from $0$ to $\frac{\pi}{4}$ added to the integral of $\cos x$ from $\frac{\pi}{4}$ to $\frac{\pi}{2}$, which evaluates to $(1 - \frac{1}{\sqrt{2}}) + (1 - \frac{1}{\sqrt{2}}) = 2 - \sqrt{2}$, or $\sqrt{2}(\sqrt{2} - 1)$. Taking the ratio $A_1 : A_2$ yields $(\sqrt{2} - 1) : \sqrt{2}(\sqrt{2} - 1)$, which simplifies to $1 : \sqrt{2}$. Summing the two areas provides $A_1 + A_2 = (\sqrt{2} - 1) + \sqrt{2}(\sqrt{2} - 1) = (\sqrt{2} - 1)(\sqrt{2} + 1) = 2 - 1 = 1$.

The area of the bounded region $R$ is determined by integrating the difference between the upper boundary curve $y = 2^{x}$ and the lower boundary defined by the piecewise function $\max\{0, \ln x\}$ over the interval $\frac{1}{2} \le x \le 2$. Since $\ln x$ is non-positive for $x \in [\frac{1}{2}, 1]$, the lower boundary is $0$ in that range, and it becomes $\ln x$ for $x \in [1, 2]$. Consequently, the area is calculated as the integral of $2^{x}$ from $\frac{1}{2}$ to $2$ minus the integral of $\ln x$ from $1$ to $2$.

Evaluating the first integral, $\int_{\frac{1}{2}}^{2} 2^{x} \, dx = \left[ \frac{2^{x}}{\ln 2} \right]_{\frac{1}{2}}^{2} = \frac{2^{2} - 2^{1/2}}{\ln 2} = \frac{4 - \sqrt{2}}{\ln 2}$. The second integral is $\int_{1}^{2} \ln x \, dx = \left[ x \ln x - x \right]_{1}^{2} = (2 \ln 2 - 2) - (0 - 1) = 2 \ln 2 - 1$. Combining these results yields $(4 - \sqrt{2})(\ln 2)^{-1} - 2 \ln 2 + 1$. Comparing this expression with the form $\alpha(\ln 2)^{-1} + \beta(\ln 2) + \gamma$, we identify the constants as $\alpha = 4 - \sqrt{2}$, $\beta = -2$, and $\gamma = 1$.

Finally, substituting these values into the expression $(\alpha + \beta - 2\gamma)^{2}$ results in $(4 - \sqrt{2} - 2 - 2(1))^{2}$. Simplifying this arithmetic leads to $(-\sqrt{2})^{2}$, which equals $2$.