The area of the region enclosed by the parabola (y - 2)^2 = x - 1$(y - 2)^2 = x - 1$ , the line x - 2y + 4 = 0$x - 2y + 4 = 0$ and the positive coordinate axes is
Numerical Answer Type:
Enter a numerical valueAnswer: 5 to 5+4 marks
Solution & Explanation
### Related Formula
textArea enclosed integrating w.r.t y-axis: int (x_textright - x_textleft) dy$\text{Area enclosed integrating w.r.t y-axis: } \int (x_{\text{right}} - x_{\text{left}}) dy$
### Core Logic
Find intersection points between the parabola x = (y-2)^2 + 1$x = (y-2)^2 + 1$ and the line x = 2y - 4$x = 2y - 4$.
Set them equal:
(y-2)^2 + 1 = 2y - 4$(y-2)^2 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 4y + 4 + 1 = 2y - 4$y^2 - 6y + 9 = 0 Rightarrow (y-3)^2 = 0$y^2 - 6y + 9 = 0 \Rightarrow (y-3)^2 = 0$
The line is tangent to the parabola at y = 3$y = 3$. At y = 3$y = 3$, x = 2(3) - 4 = 2$x = 2(3) - 4 = 2$.
We need the area enclosed by the parabola, the line, and the *positive coordinate axes*.
Let's trace the boundary intercepts.
The line x = 2y - 4$x = 2y - 4$ cuts the y-axis (x=0$x=0$) at y=2$y=2$.
Parabola vertex is at (1, 2)$(1, 2)$. Parabola cuts y-axis (x=0$x=0$) at (y-2)^2 = -1$(y-2)^2 = -1$ (No real y-intercept).
So the curve (y-2)^2 = x-1$(y-2)^2 = x-1$ only exists for x ge 1$x \ge 1$.
### Step 1: Setting up the Integral w.r.t y
Area Under Curve diagram for Q29 - JEE Main 2024 Evening
Because the area is bound by the positive coordinate axes, we track integration strictly in the first quadrant.
The required area is the area between the y-axis, the x-axis, the line, and the curve. Wait, the problem says "area enclosed by the parabola... line... and positive coordinate axes".
Let's integrate with respect to y$y$.
The boundary curves are x = (y-2)^2 + 1$x = (y-2)^2 + 1$ (which forms the rightmost boundary) and the axes.
The area bounded by the curve with y-axis is from y=0$y=0$ to y=3$y=3$ minus the triangular region cut by the line below y=2$y=2$.
The line intersects the x-axis (y=0$y=0$) at x=-4$x=-4$ and y-axis (x=0$x=0$) at y=2$y=2$.
The positive coordinate axes enforce boundaries at x ge 0, y ge 0$x \ge 0, y \ge 0$.
So the exact area is the integral of the parabola's x$x$-value from y=0$y=0$ to y=3$y=3$, minus the area of the small triangle outside the region bounded by the line x=2y-4$x=2y-4$ and the axes in the positive quadrant.
The area under the parabola (to the left towards the y-axis) from y=0$y=0$ to y=3$y=3$ is int_0^3 x_textparabola dy$\int_0^3 x_{\text{parabola}} dy$.
The line bounds the region on the left from y=2$y=2$ to y=3$y=3$. Between y=0$y=0$ and y=2$y=2$, the area goes fully to the y-axis.
So, area = int_0^3 x_textparabola dy - textArea of Delta text bounded by line and y-axis$\int_0^3 x_{\text{parabola}} dy - \text{Area of } \Delta \text{ bounded by line and y-axis}$.
The triangle bounded by x=2y-4, x=0, y=2, y=3$x=2y-4, x=0, y=2, y=3$ is: base is x=2(3)-4 = 2$x=2(3)-4 = 2$ at y=3$y=3$, height is 1$1$ (from y=2$y=2$ to y=3$y=3$). Triangle area = frac12 times 2 times 1 = 1$\frac{1}{2} \times 2 \times 1 = 1$.
Hence, Enclosed Area = int_0^3 ((y-2)^2 + 1) dy - 1$\int_{0}^{3} ((y-2)^2 + 1) dy - 1$.
### Step 2: Evaluating the Integral
textArea = int_0^3 (y^2 - 4y + 5) dy - 1$\text{Area} = \int_{0}^{3} (y^2 - 4y + 5) dy - 1$= left[ fracy^33 - 2y^2 + 5y right]_0^3 - 1$= \left[ \frac{y^3}{3} - 2y^2 + 5y \right]_0^3 - 1$= left( frac273 - 2(9) + 5(3) right) - 0 - 1$= \left( \frac{27}{3} - 2(9) + 5(3) \right) - 0 - 1$= (9 - 18 + 15) - 1 = 6 - 1 = 5$= (9 - 18 + 15) - 1 = 6 - 1 = 5$
### Pattern Recognition
Instead of evaluating horizontal strips with a broken left-bound, evaluate the total area mapped to the y-axis int_0^3 x dy$\int_0^3 x dy$ and simply subtract the basic geometric triangle removed by the straight line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integral Calculus
Keywords:#Area bounded by parabola and line#JEE Main 2024 Evening Q29#Integral Calculus JEE Main 2024#Area Under Curve JEE Main 2024
More Integral Calculus Previous-Year Questions
Q57jee_main_2025_02_april_eveningProperties of Definite Integrals
Let (a, b)$(a, b)$ be the point of intersection of the curve x^2 = 2y$x^2 = 2y$ and the straight line y - 2x - 6 = 0$y - 2x - 6 = 0$ in the second quadrant. Then the integral I = int_a^b frac9x^21 + 5^x \, dx$I = \int_{a}^{b} \frac{9x^2}{1 + 5^x} \, dx$ is equal to:
A. 24
B. 27
C. 18
D. 21
Solution
### Related Formula
textKing's Property: int_a^b f(x) dx = int_a^b f(a+b-x) dx$\text{King's Property: } \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$
### Core Logic
First, we find the coordinates of intersection in the second quadrant to determine the integration limits a$a$ and b$b$.
### Step 1: Find points of intersection
Substitute y = fracx^22$y = \frac{x^2}{2}$ into the line equation y - 2x - 6 = 0$y - 2x - 6 = 0$:
fracx^22 - 2x - 6 = 0 implies x^2 - 4x - 12 = 0$\frac{x^2}{2} - 2x - 6 = 0 \implies x^2 - 4x - 12 = 0$(x - 6)(x + 2) = 0 implies x = 6 quad textor quad x = -2$(x - 6)(x + 2) = 0 \implies x = 6 \quad \text{or} \quad x = -2$
Since the point (a, b)$(a, b)$ lies in the second quadrant, x$x$ must be negative:
a = -2$a = -2$b = 2(a) + 6 = 2(-2) + 6 = 2$b = 2(a) + 6 = 2(-2) + 6 = 2$
Thus, the integration limits are a = -2$a = -2$ and b = 2$b = 2$.
### Step 2: Solve the Integral using King's Property
The integral is:
I = int_-2^2 frac9x^21 + 5^x dx quad text--- (1)$I = \int_{-2}^{2} \frac{9x^2}{1 + 5^x} dx \quad \text{--- (1)}$
Apply King's property, substituting x to -x$x \to -x$ (since -2 + 2 - x = -x$-2 + 2 - x = -x$):
I = int_-2^2 frac9(-x)^21 + 5^-x dx = int_-2^2 frac9x^2 cdot 5^x1 + 5^x dx quad text--- (2)$I = \int_{-2}^{2} \frac{9(-x)^2}{1 + 5^{-x}} dx = \int_{-2}^{2} \frac{9x^2 \cdot 5^x}{1 + 5^x} dx \quad \text{--- (2)}$
Adding equations (1) and (2):
2I = int_-2^2 9x^2 left( frac1 + 5^x1 + 5^x right) dx = int_-2^2 9x^2 dx$2I = \int_{-2}^{2} 9x^2 \left( \frac{1 + 5^x}{1 + 5^x} \right) dx = \int_{-2}^{2} 9x^2 dx$
Since 9x^2$9x^2$ is an even function:
2I = 2 int_0^2 9x^2 dx implies I = left[ 3x^3 right]_0^2 = 3(8) - 0 = 24$2I = 2 \int_{0}^{2} 9x^2 dx \implies I = \left[ 3x^3 \right]_{0}^{2} = 3(8) - 0 = 24$
### Pattern Recognition
Whenever you see an exponential denominator like 1 + c^x$1 + c^x$ inside a symmetric interval definite integral [-a, a]$[-a, a]$, applying King's property will almost always cancel the exponential factor cleanly when the remaining numerator is even.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Q63jee_main_2025_02_april_eveningIntegration by Rationalisation
4int_0^1left(frac1sqrt3 + x^2 + sqrt1 + x^2right)mathrmdx - 3log_eleft(sqrt3right)$4\int_{0}^{1}\left(\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}}\right)\mathrm{d}x - 3\log_{e}\left(\sqrt{3}\right)$ is equal to:
### Related Formula
int sqrta^2 + x^2 dx = fracx2 sqrta^2 + x^2 + fraca^22 lnleft| x + sqrta^2 + x^2 right|$\int \sqrt{a^2 + x^2} dx = \frac{x}{2} \sqrt{a^2 + x^2} + \frac{a^2}{2} \ln\left| x + \sqrt{a^2 + x^2} \right|$
### Core Logic
We first rationalise the denominator to split the integral into two standard integration terms.
### Step 1: Rationalise the integrand
Multiply the numerator and denominator by sqrt3+x^2 - sqrt1+x^2$\sqrt{3+x^2} - \sqrt{1+x^2}$:
frac1sqrt3 + x^2 + sqrt1 + x^2 = fracsqrt3 + x^2 - sqrt1 + x^2(3+x^2) - (1+x^2) = fracsqrt3 + x^2 - sqrt1 + x^22$\frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(3+x^2) - (1+x^2)} = \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2}$
Thus, the integral expression simplifies to:
I = 4 int_0^1 left( fracsqrt3 + x^2 - sqrt1 + x^22 right) dx - 3log_eleft(sqrt3right)$I = 4 \int_{0}^{1} \left( \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2} \right) dx - 3\log_{e}\left(\sqrt{3}\right)$I = 2 int_0^1 sqrt3 + x^2 dx - 2 int_0^1 sqrt1 + x^2 dx - frac32 log_e 3$I = 2 \int_{0}^{1} \sqrt{3 + x^2} dx - 2 \int_{0}^{1} \sqrt{1 + x^2} dx - \frac{3}{2} \log_{e} 3$
### Step 2: Evaluate the integrals
For the first integral:
2 int_0^1 sqrt3 + x^2 dx = 2 left[ fracx2 sqrt3 + x^2 + frac32 lnleft| x + sqrt3 + x^2 right| right]_0^1$2 \int_{0}^{1} \sqrt{3 + x^2} dx = 2 \left[ \frac{x}{2} \sqrt{3 + x^2} + \frac{3}{2} \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$= left[ x sqrt3 + x^2 + 3 lnleft| x + sqrt3 + x^2 right| right]_0^1$= \left[ x \sqrt{3 + x^2} + 3 \ln\left| x + \sqrt{3 + x^2} \right| \right]_{0}^{1}$= left( sqrt4 + 3 ln(1 + sqrt4) right) - left( 0 + 3 lnsqrt3 right)$= \left( \sqrt{4} + 3 \ln(1 + \sqrt{4}) \right) - \left( 0 + 3 \ln\sqrt{3} \right)$= 2 + 3 ln 3 - frac32 ln 3 = 2 + frac32 ln 3$= 2 + 3 \ln 3 - \frac{3}{2} \ln 3 = 2 + \frac{3}{2} \ln 3$
For the second integral:
-2 int_0^1 sqrt1 + x^2 dx = -2 left[ fracx2 sqrt1 + x^2 + frac12 lnleft| x + sqrt1 + x^2 right| right]_0^1$-2 \int_{0}^{1} \sqrt{1 + x^2} dx = -2 \left[ \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$= - left[ x sqrt1 + x^2 + lnleft| x + sqrt1 + x^2 right| right]_0^1$= - \left[ x \sqrt{1 + x^2} + \ln\left| x + \sqrt{1 + x^2} \right| \right]_{0}^{1}$= - left( sqrt2 + ln(1 + sqrt2) right)$= - \left( \sqrt{2} + \ln(1 + \sqrt{2}) \right)$
### Step 3: Sum the terms
Now compile all terms:
I = left( 2 + frac32 ln 3 right) - sqrt2 - ln(1 + sqrt2) - frac32 ln 3$I = \left( 2 + \frac{3}{2} \ln 3 \right) - \sqrt{2} - \ln(1 + \sqrt{2}) - \frac{3}{2} \ln 3$I = 2 - sqrt2 - ln(1 + sqrt2)$I = 2 - \sqrt{2} - \ln(1 + \sqrt{2})$
### Pattern Recognition
Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
Q71jee_main_2025_02_april_morningDefinite Integration with GIF
Let [cdot]$[\cdot]$ denote the greatest integer function. If int_0^e^3left[frac1e^x - 1right]mathrmdx = alpha -log_e2$\int_{0}^{e^{3}}\left[\frac{1}{e^{x - 1}}\right]\mathrm{d}x = \alpha -\log_{e}2$, then alpha^3$\alpha^3$ is equal to ________.
Numerical Answer.Answer: 8 to 8
Solution
### Related Formula
Greatest Integer Function boundaries:
[f(x)] = k quad textfor quad k le f(x) < k+1, \ k in mathbbZ$[f(x)] = k \quad \text{for} \quad k \le f(x) < k+1, \ k \in \mathbb{Z}$
### Core Logic
Analyze the value variations of f(x) = e^1-x$f(x) = e^{1-x}$ across the integration limits [0, e^3]$[0, e^3]$ to break down the integral into distinct piecewise continuous intervals.
### Step 1: Determine Step Function Transition Points
Let y = e^1-x$y = e^{1-x}$.
* At x = 0 implies y = e^1 approx 2.718$x = 0 \implies y = e^1 \approx 2.718$
* As x$x$ increases, e^1-x$e^{1-x}$ decreases monotonically.
* Find x$x$ where y = 2 implies e^1-x = 2 implies 1-x = ln 2 implies x = 1 - ln 2$y = 2 \implies e^{1-x} = 2 \implies 1-x = \ln 2 \implies x = 1 - \ln 2$.
* Find x$x$ where y = 1 implies e^1-x = 1 implies 1-x = 0 implies x = 1$y = 1 \implies e^{1-x} = 1 \implies 1-x = 0 \implies x = 1$.
* At the final boundary x = e^3 implies y = e^1-e^3$x = e^3 \implies y = e^{1-e^3}$, which is a very small positive decimal strictly inside (0,1)$(0,1)$.
### Step 2: Split the Definite Integral
Rewrite the integral based on the isolated interval blocks:
I = int_0^1-ln 2 2 \, mathrmdx + int_1-ln 2^1 1 \, mathrmdx + int_1^e^3 0 \, mathrmdx$I = \int_{0}^{1-\ln 2} 2 \, \mathrm{d}x + \int_{1-\ln 2}^{1} 1 \, \mathrm{d}x + \int_{1}^{e^3} 0 \, \mathrm{d}x$
### Step 3: Perform Integrations
I = 2[x]_0^1-ln 2 + 1[x]_1-ln 2^1 + 0$I = 2[x]_0^{1-\ln 2} + 1[x]_{1-\ln 2}^{1} + 0$I = 2(1 - ln 2 - 0) + 1(1 - (1 - ln 2)) = 2 - 2ln 2 + ln 2 = 2 - ln 2$I = 2(1 - \ln 2 - 0) + 1(1 - (1 - \ln 2)) = 2 - 2\ln 2 + \ln 2 = 2 - \ln 2$
### Step 4: Solve for Alpha Cubed
Compare the integrated value to alpha - ln 2$\alpha - \ln 2$:
alpha - ln 2 = 2 - ln 2 implies alpha = 2$\alpha - \ln 2 = 2 - \ln 2 \implies \alpha = 2$alpha^3 = 2^3 = 8$\alpha^3 = 2^3 = 8$
### Pattern Recognition
Always map the function values at the extreme boundary points first. Tracking the downward path from 2.71 rightarrow 2 rightarrow 1 rightarrow 0$2.71 \rightarrow 2 \rightarrow 1 \rightarrow 0$ reveals exactly where the integer thresholds are crossed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integrals
The integral int_0^pi frac8x \, dx4cos^2 x + sin^2 x$\int_0^\pi \frac{8x \, dx}{4\cos^2 x + \sin^2 x}$ is equal to
A.2pi^2$2\pi^2$
B.4pi^2$4\pi^2$
C.pi^2$\pi^2$
D.frac3pi^22$\frac{3\pi^2}{2}$
Solution
### Related Formula
Using the properties of definite integrals:
int_a^b f(x) \, dx = int_a^b f(a+b-x) \, dx$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$
Also, if f(2a-x) = f(x)$f(2a-x) = f(x)$, then:
int_0^2a f(x) \, dx = 2 int_0^a f(x) \, dx$\int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx$
### Core Logic
Let:
I = int_0^pi frac8x \, dx4cos^2 x + sin^2 x quad text--- (1)$I = \int_0^\pi \frac{8x \, dx}{4\cos^2 x + \sin^2 x} \quad \text{--- (1)}$
Applying x to pi-x$x \to \pi-x$:
I = int_0^pi frac8(pi - x) \, dx4cos^2 x + sin^2 x quad text--- (2)$I = \int_0^\pi \frac{8(\pi - x) \, dx}{4\cos^2 x + \sin^2 x} \quad \text{--- (2)}$
### Step 1: Eliminating the x$x$ term in numerator
Adding (1) and (2):
2I = 8pi int_0^pi fracdx4cos^2 x + sin^2 x$2I = 8\pi \int_0^\pi \frac{dx}{4\cos^2 x + \sin^2 x}$I = 4pi int_0^pi fracdx4cos^2 x + sin^2 x$I = 4\pi \int_0^\pi \frac{dx}{4\cos^2 x + \sin^2 x}$
Since the integrand is symmetric about x = pi/2$x = \pi/2$:
I = 8pi int_0^pi/2 fracdx4cos^2 x + sin^2 x$I = 8\pi \int_0^{\pi/2} \frac{dx}{4\cos^2 x + \sin^2 x}$
### Step 2: Integration using sec^2 x$\sec^2 x$ substitution
Divide numerator and denominator by cos^2 x$\cos^2 x$:
I = 8pi int_0^pi/2 fracsec^2 x \, dx4 + tan^2 x$I = 8\pi \int_0^{\pi/2} \frac{\sec^2 x \, dx}{4 + \tan^2 x}$
Let t = tan x implies dt = sec^2 x \, dx$t = \tan x \implies dt = \sec^2 x \, dx$
At x = 0$x = 0$, t = 0$t = 0$; at x = pi/2$x = \pi/2$, t to infty$t \to \infty$.
I = 8pi int_0^infty fracdtt^2 + 2^2$I = 8\pi \int_0^\infty \frac{dt}{t^2 + 2^2}$I = 8pi left[ frac12 tan^-1left(fract2right) right]_0^infty = 4pi left( fracpi2 - 0 right) = 2pi^2$I = 8\pi \left[ \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right) \right]_0^\infty = 4\pi \left( \frac{pi}{2} - 0 \right) = 2\pi^2$
### Pattern Recognition
The presence of a linear x$x$ factor in the numerator of a definite integral with symmetric trigonometric bounds is almost always eliminated using the a+b-x$a+b-x$ property. This reduces the integral to a standard substitution problem.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integrals
Q53jee_main_2025_29_jan_eveningArea Enclosed by Two Curves
Let the area enclosed between the curves |y| = 1 - x^2$|y| = 1 - x^2$ and x^2 + y^2 = 1$x^2 + y^2 = 1$ be alpha$\alpha$. If 9alpha = beta pi + gamma$9\alpha = \beta \pi + \gamma$, beta, gamma$\beta, \gamma$ are integers, then the value of |beta - gamma|$|\beta - \gamma|$ equals
A.27$27$
B.18$18$
C.15$15$
D.33$33$
Solution
### Related Formula
Area enclosed between two symmetric functions across four quadrants:
alpha = 4 times left( textArea of circle in 1^textst text quadrant - int_0^1 y_textparabola \, dx right)$\alpha = 4 \times \left( \text{Area of circle in } 1^{\text{st}} \text{ quadrant} - \int_{0}^{1} y_{\text{parabola}} \, dx \right)$
### Core Logic
The curves are symmetric across the axes. The region is enclosed between the unit circle x^2 + y^2 = 1$x^2 + y^2 = 1$ and the parabolas y = pm(1 - x^2)$y = \pm(1 - x^2)$.
Area Enclosed by Two Curves diagram for Q53 - JEE Main 2025 Evening
By symmetry, the area alpha$\alpha$ in the first quadrant is the area under the circle minus the area under the parabola from x=0$x=0$ to x=1$x=1$.
### Step 1: Calculate the Area Components
Area of a circle quadrant with r=1$r=1$ is fracpi4$\frac{\pi}{4}$.
Area under the parabola:
int_0^1 (1 - x^2) \, dx = left[ x - fracx^33 right]_0^1 = 1 - frac13 = frac23$\int_{0}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$
### Step 2: Evaluate the Total Area
alpha = 4 left[ fracpi4 - frac23 right] = pi - frac83$\alpha = 4 \left[ \frac{\pi}{4} - \frac{2}{3} \right] = \pi - \frac{8}{3}$
Multiplying both sides by 9:
9alpha = 9pi - 24$9\alpha = 9\pi - 24$
Comparing with 9alpha = betapi + gamma$9\alpha = \beta\pi + \gamma$, we get:
beta = 9, quad gamma = -24$\beta = 9, \quad \gamma = -24$
Hence,
|beta - gamma| = |9 - (-24)| = 33$|\beta - \gamma| = |9 - (-24)| = 33$
### Pattern Recognition
Symmetry helps divide the integration work by 4. Recognizing standard shapes (like a circle sector) allows you to bypass complex integration entirely for half of the problem.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Integral Calculus
More Integral Calculus Questions — jee_main_2024_30_january_evening
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