Area Under The Curves – JEE Main MathematicsPractice Questions & PYQs

The area of a plane region bounded by two curves is found by integrating the difference between the rightmost function and the leftmost function over the interval defined by their intersection points, which are $(-2, 1)$ and $(-2, -1)$. Converting the given equations to $x = -2y^2$ and $x = 1 - 3y^2$ identifies the parabolas, and equating them yields the intersection boundaries $y = 1$ and $y = -1$. Since $x = 1 - 3y^2$ is to the right of $x = -2y^2$ on this interval, the area is given by the following integral:
$\int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy = \int_{-1}^{1} (1 - y^2) \, dy$
Evaluating this expression as $\left[ y - \frac{y^3}{3} \right]_{-1}^{1}$ yields $\left( 1 - \frac{1}{3} \right) - \left( -1 - \frac{-1}{3} \right) = \frac{4}{3}$ square units.

Calculating the area between two curves requires finding their intersection points and integrating the difference between the upper and lower functions over the resulting interval. The curves $y^2 = x$ and $y = |x|$ intersect at $x = 0$ and $x = 1$, where $y = \sqrt{x}$ lies above $y = x$ within the first quadrant. Integrating the difference $(\sqrt{x} - x)$ from $0$ to $1$ involves evaluating the expression $\left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} \right]_{0}^{1}$. Substituting the limits produces $\frac{2}{3} - \frac{1}{2}$, which simplifies to a final area of $\frac{1}{6}$.

The geometry of these two parabolas, $y^2=4x$ and $x^2=4y$, reveals a perfect symmetry across the diagonal line $y=x$ within the square region defined by the coordinate axes and the lines $x=4$ and $y=4$. Because these curves are reflections of one another, the three regions $S_1, S_2,$ and $S_3$ partition the square's total area of $16$ square units into three equal segments. To determine the area of the middle region $S_2$ trapped between the two curves, we evaluate the following integral:

$S_2 = \int_{0}^{4} \left(2\sqrt{x} - \frac{x^2}{4}\right) dx = \left[ \frac{4}{3}x^{3/2} - \frac{x^3}{12} \right]_{0}^{4} = \frac{16}{3}$

Due to the symmetry of the parabolas, the top area $S_1$ and bottom area $S_3$ are also identical in magnitude. Calculating the area for $S_3$, which is bounded by the $x$-axis and the curve $y=x^2/4$, we get:

$S_3 = \int_{0}^{4} \frac{x^2}{4} dx = \left[ \frac{x^3}{12} \right]_{0}^{4} = \frac{16}{3}$

Given that the total area of the square is $16$ square units, $S_1$ must also equal $\frac{16}{3}$ to satisfy the sum $S_1 + S_2 + S_3 = 16$. With each region measuring exactly $\frac{16}{3}$ square units, the ratio of the areas $S_1 : S_2 : S_3$ is $1:1:1$.

To calculate the area enclosed between the curve $y=\log_e(x+e)$ and the coordinate axes, first determine the intersection points with the axes. The y-intercept occurs where $x=0$, resulting in $y=\log_e(e)=1$. The x-intercept occurs where $y=0$, which requires $\log_e(x+e)=0$, leading to $x+e=1$ or $x=1-e$. Consequently, the area is represented by the definite integral from $1-e$ to $0$.

The calculation requires evaluating the integral $\int_{1-e}^{0} \log_e(x+e) \, dx$ Utilizing the substitution $t=x+e$ where $dx=dt$, the limits change to $1$ and $e$ respectively. This yields the integral $\int_{1}^{e} \log_e(t) \, dt$ Using the standard antiderivative $t\log_e(t)-t$, we compute $[t\log_e(t)-t]_1^e = (e\log_e(e)-e) - (1\log_e(1)-1)$ Simplifying this expression results \in $(e-e) - (0-1)$, which equals $1$.

The Fundamental Theorem of Calculus provides a direct path to uncovering an unknown function when its accumulated area is expressed as a function of its upper limit, because differentiating such an integral with respect to that upper variable returns the integrand itself. Applying this to the given equation $\int_{\frac{\pi}{4}}^{\beta} f(x) dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta$, we differentiate both sides with respect to $\beta$ to identify $f(\beta)$. Using the product rule for the first term $\beta \sin \beta$ results in $\sin \beta + \beta \cos \beta$, and differentiating the remaining terms $\frac{\pi}{4} \cos \beta$ and $\sqrt{2} \beta$ gives $-\frac{\pi}{4} \sin \beta$ and $\sqrt{2}$, respectively. Combining these parts leads to the function $f(\beta) = \beta \cos \beta + \sin \beta - \frac{\pi}{4} \sin \beta + \sqrt{2}$. Substituting $\beta = \frac{\pi}{2}$ into this expression and noting that $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$, we find the value $f(\frac{\pi}{2}) = 0 + 1 - \frac{\pi}{4} + \sqrt{2}$, which simplifies to $1 - \frac{\pi}{4} + \sqrt{2}$.

The absolute value function $y=|x-2|$ forms a V-shaped graph with its vertex at $x=2$. To calculate the area bounded by this curve, the $x$-axis, and the vertical lines $x=1$ and $x=3$, we can split the integral at the vertex to account for the sign change and exploit the symmetry of the two triangles formed. This area is expressed as $2 \int_{2}^{3} (x-2) \, dx$, which evaluates to $2 \left[ \frac{x^2}{2} - 2x \right]_{2}^{3}$ when integrating. Substituting the limits, we get $2 [ (\frac{9}{2} - 6) - (2 - 4) ]$, which simplifies to 1.

The area enclosed by $y = |x-1|$ and $y = 3 - |x|$ is determined by partitioning the region based on the points where the absolute value functions change, which occur at $x = -1, 0, 1,$ and $2$. Integrating the difference between the upper and lower functions across these three distinct intervals ensures the absolute values are handled correctly.

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

On the interval $[-1, 0]$, the difference between $3+x$ and $1-x$ is $2+2x$, which evaluates to $1$ square unit. Between $0$ and $1$, the curves simplify such that the upper function $3-x$ is $2$ units above the lower function $1-x$, yielding $2$ square units. For the interval $[1, 2]$, the upper boundary $3-x$ minus the lower boundary $x-1$ results in $4-2x$, which integrates to $1$ square unit. Aggregating these results produces a total area of $4$ square units.

Visualizing the graphs of these logarithmic functions reveals a symmetric diamond-shaped region centered around the origin, created by the interplay of the absolute value transformations. Each of the four functions reflects portions of the natural logarithm curve across the x-axis or y-axis, effectively partitioning the bounded space into four identical regions located in the intervals $(0, 1)$ and $(-1, 0)$. By analyzing the region within the first quadrant, we observe that the area is bounded by the functions in a way that simplifies to the integral of $-\ln x$ from $0$ to $1$. Evaluating this integral provides the area of one of the four symmetric segments:

$\int_0^1 -\ln x \, dx = -[x \ln x - x]_0^1 = 1$

Since the total bounded area is composed of four such identical segments, the calculation becomes $4 \times 1$, resulting in a final area of 4 square units.

The intersection points of these conics satisfy the curve $y^2 = 3x^2$, allowing us to substitute this relationship directly into the equations to solve for the unknown parameter $b$. Substituting $y^2 = 3x^2$ into $x^2 + y^2 = 4b$ leads to $4x^2 = 4b$, which simplifies to $x^2 = b$. Using this in the second equation $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ results in $\frac{b}{16} + \frac{3b}{b^2} = 1$, or $\frac{b}{16} + \frac{3}{b} = 1$. Solving the quadratic equation $b^2 - 16b + 48 = 0$ gives $b = 4$ and $b = 12$. Because the problem specifies that the conics are distinct, $b = 4$ is rejected as it leads to identical curves, leaving $b = 12$.

With $b = 12$, the intersection points are determined by $x^2 = 12$ and $y^2 = 36$, yielding coordinates of $(\pm 2\sqrt{3}, \pm 6)$. These four points define a rectangle with a width of $4\sqrt{3}$ and a height of $12$, resulting in an area of $48\sqrt{3}$. Multiplying this area by $3\sqrt{3}$ yields the final result of $3\sqrt{3} \times 48\sqrt{3} = 432$.

The regions $A$ and $B$ are defined by the interplay between the line $y=2x$ and the upper semi-circle defined by $(x-1)^2 + y^2 = 4$. For region $A$, we identify the area trapped between the line and the circular arc, which geometrically corresponds to a quarter-circle of radius 2 minus a right-angled triangle. Since the quarter-circle encompasses an area of $\pi$ and the triangle, defined by the vertices $(0,0)$, $(1,0)$, and $(1,2)$, has a base of 1 and height of 2, its area is 1, resulting in an area for $A$ of $\pi - 1$.

Conversely, region $B$ consists of the area under the line and the circular boundary up to the intersection point. This region is composed of the same triangle with area 1 and the quarter-circle segment, yielding a total area of $\pi + 1$. Dividing the area of $A$ by the area of $B$ gives the final ratio of $\frac{\pi-1}{\pi+1}$.