Ask & Answer Question

Application of Integrals Ncert Solutions Maths class 12th Maths Class 12th Application of Integrals

19. Choose the correct answer:

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is

(A) 2 (π – 2)

(B) π – 2

(C) 2π – 1

(D) 2 (π + 2)

4 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

The equation of circle is $x^{2} + y^{2} = 4$ which has centre at (0,0) & radius,

$π = 2$

And the line $x + y = 2 = y = 2 - x$

The smaller area of circle is given by

Area (ABCA) area (BOAB) – area (BOA)

0 Upvotes 0 Downvotes

Similar Questions for you

Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

Raj Pandey

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) Þ c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

Raj Pandey

Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$

Þ a = 5

The area bounded by the curves y² = x and x² = -y is equal to

Vishal Baghel

$y^{2} = x, a = \frac{1}{4}$

$x^{2} = - y, b = \frac{1}{4}$

$A = \frac{16 | a b |}{3} = \frac{1}{3}$

limₓ→∞ [∫₀ˣ tan⁻¹t dt] / √x²+1 is equal to:

Vishal Baghel

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

Area of the region bounded by x = 0, y = 0, x = 2, y = 2, y ≤ e? and y ≥ ln x, is:

Vishal Baghel

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

19. Choose the correct answer:

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is

(A) 2 (π – 2)

(B) π – 2

(C) 2π – 1

(D) 2 (π + 2)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

19. Choose the correct answer: Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is (A) 2 (π – 2) (B) π – 2 (C) 2π – 1 (D) 2 (π + 2)

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

19. Choose the correct answer:

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is

(A) 2 (π – 2)

(B) π – 2

(C) 2π – 1

(D) 2 (π + 2)