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
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 ≥ 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 ≥ 0, y ≥ 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 ∫₀³ xparabola 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 = ∫₀³ xparabola dy - Area of Δ 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 = (1)/(2) × 2 × 1 = 1$\frac{1}{2} \times 2 \times 1 = 1$.
Hence, Enclosed Area = ∫₀³ ((y-2)² + 1) dy - 1$\int_{0}^{3} ((y-2)^2 + 1) dy - 1$.
Instead of evaluating horizontal strips with a broken left-bound, evaluate the total area mapped to the y-axis ∫₀³ x dy$\int_0^3 x dy$ and simply subtract the basic geometric triangle removed by the straight line.
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 7
Q75jee_main_2025_03_april_morningArea Under Bounded Curves
The area of the region bounded by the curve y = |x|, x|x - 2|$y = \max\left\{|x|, x|x - 2|\right\}$, the x-axis and the lines x = -2$x = -2$ and x = 4$x = 4$ is equal to:
Numerical Answer.Answer: 12 to 12
Solution
Related Formula
Definite integration geometry area: Split boundary zones around intersections where functional dominant switches occur.
Area Under Bounded Curves diagram for Q75 - JEE Main 2025 Morning
Core Logic
Analyze intersection points between y₁ = |x|$y_1 = |x|$ and y₂ = x|x-2|$y_2 = x|x-2|$ across required integration span regions:
For x in [-2, 0]$x \in [-2, 0]$: |x| = -x$|x| = -x$ and x|x-2| = -x(2-x) = x² - 2x$x|x-2| = -x(2-x) = x^2 - 2x$. Max curve tracks through distinct segments.
For positive sectors, compute intersections: x = x(2-x) x = 1$x = x(2-x) \implies x = 1$ or x=0$x=0$. Also check switch locations where graphs swap dominance.
Step 1: Setting up separate area integral blocks
Using geometric area partitions calculated across continuous regions:
Alternatively, splitting the boundary metrics via continuous definite limits yields identical whole tracking blocks matching exactly to 12 total units.
Class 12 Mathematics: Integrals (Application of Integrals)
Q68jee_main_2025_04_april_eveningProperties of Definite Integrals
Let f(x) + 2f((1)/(x)) = x² + 5$f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and 2 g (x) - 3 g ((1)/(2)) = x, x > 0$2 \mathrm {g} (\mathrm {x}) - 3 \mathrm {g} \left(\frac {1}{2}\right) = \mathrm {x}, \mathrm {x} > 0$. If α = ∫_ 1 ^ 2 f (x) d x$\alpha = \int_ {1} ^ {2} f (x) d x$, and β = ∫_ 1 ^ 2 g (x) d x$\beta = \int_ {1} ^ {2} g (x) d x$, then the value of 9α + β$9\alpha + \beta$ is:
A.1$1$
B.0$0$
C.10$10$
D.11$11$
Solution
Core Logic
We have two functional equations to solve before integrating.
Functional equations involving x → (1)/(x)$x \to \frac{1}{x}$ are easily solved by treating the swapped forms as a system of linear equations, allowing direct isolation of the underlying function.
Chapter Mix
Class 12 Mathematics: Definite Integrals
Class 12 Mathematics: Functional Equations
Q72jee_main_2025_04_april_eveningIntegration by Substitution
If ∫ (√(1 + x²) + x)¹⁰(√(1 + x²) - x)⁹dx = (1)/(m) (( 1 + x ^ 2 + x) ^ n (n 1 + x ^ 2 - x)) + C$\int \frac{\left(\sqrt{1 + x^2} + x\right)^{10}}{\left(\sqrt{1 + x^2} - x\right)^9}\mathrm{d}x = \frac {1}{m} \left(\left(\sqrt {1 + x ^ {2}} + x\right) ^ {n} \left(n \sqrt {1 + x ^ {2}} - x\right)\right) + C$ where C$C$ is the constant of integration and m,nin N$\mathbf{m},\mathbf{n}\in \mathbf{N}$, then m + n$\mathfrak{m} + \mathfrak{n}$ is equal to
Numerical Answer.Answer: 379 to 379
Solution
Core Logic
Let's simplify the integrand by rationalizing the denominator term block. Notice that:
Comparing this directly with the given answer format, we identify:
m = 360$m = 360$
n = 19$n = 19$
Computing m + n$m + n$:
m + n = 360 + 19 = 379$$m + n = 360 + 19 = 379$$
Pattern Recognition
Expressions containing conjugate factors like √(1+x²) ± x$\sqrt{1+x^2} \pm x$ frequently simplify under rationalization because their product equals 1. This dynamic quickly reduces fractional components into single power blocks.
Chapter Mix
Class 12 Mathematics: Indefinite Integrals
Q66jee_main_2025_04_april_morningProperties of Definite Integrals
The value of ∫₋₁¹ (1 + √(|x| - x))e^x + (√(|x| - x))e-xe^x + e-x dx$\int_{-1}^{1}\frac{\left(1 + \sqrt{|x| - x}\right)e^x + \left(\sqrt{|x| - x}\right)e^{-x}}{e^x + e^{-x}} \, \mathrm{d}x$ is equal to
We need to find the intersection points of the curves to understand the x+7, 11-3x$\min\{x+7, 11-3x\}$ boundary transition:
x+7 = 11-3x 4x = 4 x = 1$x+7 = 11-3x \implies 4x = 4 \implies x = 1$.
Hence, the line switches behavior at x=1$x=1$.
Intersecting 1+x²$1+x^2$ with x+7$x+7$:
x² - x - 6 = 0 (x-3)(x+2) = 0 x = -2 or x = 3$$x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0 \implies x = -2 \quad \text{or} \quad x = 3$$
Intersecting 1+x²$1+x^2$ with 11-3x$11-3x$:
x² + 3x - 10 = 0 (x+5)(x-2) = 0 x = 2 or x = -5$$x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0 \implies x = 2 \quad \text{or} \quad x = -5$$
Area Under Curves diagram for Q57 - JEE Main 2025 Evening
Step 1: Set up Integrals
The transition points show that from x = -2$x = -2$ to 1$1$, the upper line is x+7$x+7$, and from x = 1$x = 1$ to 2$2$, the upper line is 11-3x$11-3x$.
When a boundary contains a $\min\{\}$ or $\max\{\}$ component, always solve for their internal intersection first to identify the exact splitting point of your definite integrals.
Chapter Mix
Class 12 Mathematics: Integral Calculus
More Integral Calculus Questions — jee_main_2024_30_january_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
