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 — Page 6
Q21jee_main_2024_31_jan_eveningProperties of Definite Integrals
left|frac120pi^3int_0^pi fracx^2sin xcos xsin^4x + cos^4x dxright|$\left|\frac{120}{\pi^3}\int_0^\pi \frac{x^2\sin x\cos x}{\sin^4x + \cos^4x} dx\right|$ is equal to
Numerical Answer.Answer: 15 to 15
Solution
### Related Formula
int_0^a x f(x) dx = fraca2 int_0^a f(x) dx quad textif f(a-x) = f(x)$\int_0^a x f(x) dx = \frac{a}{2} \int_0^a f(x) dx \quad \text{if } f(a-x) = f(x)$
### Core Logic
Let I = int_0^pi fracx^2sin xcos xsin^4x + cos^4x dx$I = \int_0^\pi \frac{x^2\sin x\cos x}{\sin^4x + \cos^4x} dx$.
Split the integral into int_0^pi/2 + int_pi/2^pi$\int_0^{\pi/2} + \int_{\pi/2}^\pi$.
For the second integral, substitute x = pi - t$x = \pi - t$:
I = int_0^pi/2 fracsin xcos xsin^4x + cos^4x (x^2 - (pi - x)^2) dx$I = \int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4x + \cos^4x} (x^2 - (\pi - x)^2) dx$I = int_0^pi/2 fracsin xcos xsin^4x + cos^4x (2pi x - pi^2) dx$I = \int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4x + \cos^4x} (2\pi x - \pi^2) dx$= 2pi int_0^pi/2 x f(x) dx - pi^2 int_0^pi/2 f(x) dx$= 2\pi \int_0^{\pi/2} x f(x) dx - \pi^2 \int_0^{\pi/2} f(x) dx$
where f(x) = fracsin x cos xsin^4 x + cos^4 x$f(x) = \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}$.
Since f(pi/2 - x) = f(x)$f(\pi/2 - x) = f(x)$, we have int_0^pi/2 x f(x) dx = fracpi4 int_0^pi/2 f(x) dx$\int_0^{\pi/2} x f(x) dx = \frac{\pi}{4} \int_0^{\pi/2} f(x) dx$.
I = 2pi left(fracpi4right) int_0^pi/2 f(x) dx - pi^2 int_0^pi/2 f(x) dx = -fracpi^22 int_0^pi/2 fracsin xcos xsin^4x + cos^4x dx$I = 2\pi \left(\frac{\pi}{4}\right) \int_0^{\pi/2} f(x) dx - \pi^2 \int_0^{\pi/2} f(x) dx = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin x\cos x}{\sin^4x + \cos^4x} dx$
To evaluate this simpler integral:
I = -fracpi^22 int_0^pi/2 fracsin xcos x1 - 2sin^2 xcos^2 x dx = -fracpi^22 int_0^pi/2 fracsin 2x2 - sin^2 2x dx$I = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin x\cos x}{1 - 2\sin^2 x\cos^2 x} dx = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin 2x}{2 - \sin^2 2x} dx$I = -fracpi^22 int_0^pi/2 fracsin 2x1 + cos^2 2x dx$I = -\frac{\pi^2}{2} \int_0^{\pi/2} \frac{\sin 2x}{1 + \cos^2 2x} dx$
Let cos 2x = t implies -2sin 2x dx = dt$\cos 2x = t \implies -2\sin 2x dx = dt$. Limits: 1$1$ to -1$-1$.
I = -fracpi^22 int_1^-1 frac-dt/21+t^2 = -fracpi^24 int_-1^1 fracdt1+t^2$I = -\frac{\pi^2}{2} \int_1^{-1} \frac{-dt/2}{1+t^2} = -\frac{\pi^2}{4} \int_{-1}^1 \frac{dt}{1+t^2}$I = -fracpi^24 [arctan t]_-1^1 = -fracpi^24 left(fracpi4 - left(-fracpi4right)right) = -fracpi^38$I = -\frac{\pi^2}{4} [\arctan t]_{-1}^1 = -\frac{\pi^2}{4} \left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = -\frac{\pi^3}{8}$
Finally:
left| frac120pi^3 left(-fracpi^38right) right| = 15$\left| \frac{120}{\pi^3} \left(-\frac{\pi^3}{8}\right) \right| = 15$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integrals
Q21jee_main_2024_31_jan_morningDefinite Integration by Substitution
If the integral 525 int_0^fracpi2 sin 2x cos^frac112x (1 + cos^frac52x)^frac12 dx$525 \int_{0}^{\frac{\pi}{2}} \sin 2x \cos^{\frac{11}{2}}x (1 + \cos^{\frac{5}{2}}x)^{\frac{1}{2}} dx$ is equal to (nsqrt2 - 64)$(n\sqrt{2} - 64)$, then n$n$ is equal to
Numerical Answer.Answer: 176 to 176
Solution
### Core Logic
I = int_0^fracpi2 sin 2x cdot (cos x)^frac112 (1 + (cos x)^frac52)^frac12 dx$I = \int_{0}^{\frac{\pi}{2}} \sin 2x \cdot (\cos x)^{\frac{11}{2}} (1 + (\cos x)^{\frac{5}{2}})^{\frac{1}{2}} dx$
Substitute cos x = t^2 implies sin x dx = -2t dt$\cos x = t^2 \implies \sin x dx = -2t dt$.
Since sin 2x = 2sin x cos x$\sin 2x = 2\sin x \cos x$, we have 2(t^2)(-2t dt) = -4t^3 dt$2(t^2)(-2t dt) = -4t^3 dt$.
Limits: x=0 to t=1$x=0 \to t=1$, x=fracpi2 to t=0$x=\frac{\pi}{2} \to t=0$.
I = 4 int_0^1 t^2 cdot (t^2)^frac112 sqrt1 + t^5 t dt = 4 int_0^1 t^14 sqrt1 + t^5 dt$I = 4 \int_{0}^{1} t^2 \cdot (t^2)^{\frac{11}{2}} \sqrt{1 + t^5} t dt = 4 \int_{0}^{1} t^{14} \sqrt{1 + t^5} dt$
### Step 1: Second Substitution
Put 1 + t^5 = k^2 implies 5t^4 dt = 2k dk$1 + t^5 = k^2 \implies 5t^4 dt = 2k dk$.
t^5 = k^2 - 1$t^5 = k^2 - 1$.
I = 4 int_1^sqrt2 (k^2 - 1)^2 cdot k cdot frac2k5 dk$I = 4 \int_{1}^{\sqrt{2}} (k^2 - 1)^2 \cdot k \cdot \frac{2k}{5} dk$I = frac85 int_1^sqrt2 (k^6 - 2k^4 + k^2) dk$I = \frac{8}{5} \int_{1}^{\sqrt{2}} (k^6 - 2k^4 + k^2) dk$
### Step 2: Evaluate the Integral
I = frac85 left[ frack^77 - frac2k^55 + frack^33 right]_1^sqrt2$I = \frac{8}{5} \left[ \frac{k^7}{7} - \frac{2k^5}{5} + \frac{k^3}{3} \right]_{1}^{\sqrt{2}}$I = frac85 left[ frac8sqrt27 - frac8sqrt25 + frac2sqrt23 - frac17 + frac25 - frac13 right]$I = \frac{8}{5} \left[ \frac{8\sqrt{2}}{7} - \frac{8\sqrt{2}}{5} + \frac{2\sqrt{2}}{3} - \frac{1}{7} + \frac{2}{5} - \frac{1}{3} \right]$I = frac85 left[ frac120sqrt2 - 168sqrt2 + 70sqrt2105 - frac15 - 42 + 35105 right]$I = \frac{8}{5} \left[ \frac{120\sqrt{2} - 168\sqrt{2} + 70\sqrt{2}}{105} - \frac{15 - 42 + 35}{105} \right]$I = frac85 left[ frac22sqrt2105 - frac8105 right]$I = \frac{8}{5} \left[ \frac{22\sqrt{2}}{105} - \frac{8}{105} \right]$
### Step 3: Equate with Given Form
Given 525 I = nsqrt2 - 64$525 I = n\sqrt{2} - 64$.
525 times frac85 left( frac22sqrt2 - 8105 right) = 8 times (22sqrt2 - 8) = 176sqrt2 - 64$525 \times \frac{8}{5} \left( \frac{22\sqrt{2} - 8}{105} \right) = 8 \times (22\sqrt{2} - 8) = 176\sqrt{2} - 64$
Comparing with (nsqrt2 - 64)$(n\sqrt{2} - 64)$, we get n = 176$n = 176$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integrals
Q30jee_main_2024_31_jan_morningProperties of Definite Integrals
Let f : mathbbR to mathbbR$f : \mathbb{R} \to \mathbb{R}$ be a function defined by f(x) = frac4^x4^x + 2$f(x) = \frac{4^x}{4^x + 2}$ and M = int_f(a)^f(1 - a) x sin^4(x(1 - x)) dx$M = \int_{f(a)}^{f(1 - a)} x \sin^4(x(1 - x)) dx$, N = int_f(a)^f(1 - a) sin^4(x(1 - x)) dx; a neq frac12$N = \int_{f(a)}^{f(1 - a)} \sin^4(x(1 - x)) dx; a \neq \frac{1}{2}$. If alpha M = beta N, alpha, beta in mathbbN$\alpha M = \beta N, \alpha, \beta \in \mathbb{N}$, then the least value of alpha^2 + beta^2$\alpha^2 + \beta^2$ is equal to
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
f(x) + f(1-x) = frac4^x4^x + 2 + frac4^1-x4^1-x + 2 = 1$f(x) + f(1-x) = \frac{4^x}{4^x + 2} + \frac{4^{1-x}}{4^{1-x} + 2} = 1$
### Core Logic
Using the property f(x) + f(1-x) = 1$f(x) + f(1-x) = 1$, we have f(a) + f(1-a) = 1$f(a) + f(1-a) = 1$.
Let limits be A = f(a)$A = f(a)$ and B = f(1-a)$B = f(1-a)$. Then A + B = 1$A + B = 1$.
M = int_A^B x sin^4(x(1 - x)) dx$M = \int_{A}^{B} x \sin^4(x(1 - x)) dx$
### Step 1: Apply King's Property
Apply the property int_A^B g(x) dx = int_A^B g(A + B - x) dx$\int_{A}^{B} g(x) dx = \int_{A}^{B} g(A + B - x) dx$.
M = int_A^B (1 - x) sin^4((1 - x)(1 - (1 - x))) dx$M = \int_{A}^{B} (1 - x) \sin^4((1 - x)(1 - (1 - x))) dx$M = int_A^B (1 - x) sin^4(x(1 - x)) dx$M = \int_{A}^{B} (1 - x) \sin^4(x(1 - x)) dx$M = int_A^B sin^4(x(1 - x)) dx - int_A^B x sin^4(x(1 - x)) dx$M = \int_{A}^{B} \sin^4(x(1 - x)) dx - \int_{A}^{B} x \sin^4(x(1 - x)) dx$M = N - M$M = N - M$
### Step 2: Conclusion
2M = N$2M = N$
Given alpha M = beta N$\alpha M = \beta N$, we get alpha = 2$\alpha = 2$ and beta = 1$\beta = 1$ (for least integral values).
Thus, alpha^2 + beta^2 = 2^2 + 1^2 = 5$\alpha^2 + \beta^2 = 2^2 + 1^2 = 5$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Integrals
Class 12 Maths: Relations and Functions
More Integral Calculus Questions — jee_main_2024_30_january_evening
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