Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 

› Ask & Answer › Question

Let S be the region bonded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1, and R2 _________.

If max {R1, R2} = R2, then R 2 R 1  is equal to _________.

0 4 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago

    R 2 = ∫ 0 1 ( x − x 3 ) d x = ( x 2 2 − x 4 4 ) 0 1 = 1 2 − 1 4 = 1 4

    R 1 = ∫ 0 1 4 ( x − 2 x ) d x = ( 2 x 3 / 2 3 − x 2 ) 0 1 4 = 1 1 2 − 1 1 6
    = 4 − 3 4 8 = 1 4 8
    R 1 + R 2 = ∫ 0 1 ( x − x 3 ) d x = ( 2 x 3 / 2 3 − x 4 4 ) 0 1 = 2 3 − 1 4 = 5 1 2
    R 1 + R 2 R 1 = 1 + R 2 R 1 = 5 1 2 1 4 8 = 2 0 ∴ R 2 R 1 = 1 9
     

     

Similar Questions for you

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

y 2 = x , a = 1 4

x 2 = − y , b = 1 4

A = 1 6 | a b | 3 = 1 3

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

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

Area of the figure bounded by the curves y = |x − 1| and y = 3 - |x| is
V
Vishal Baghel

y = |x − 1|, y = 3 – |x|

(A graph is shown with vertices A (1, 0), B (2, 1), C (0, 3), D (-1, 2). The lines are y = x - 1, y = 3 - x, y = 3 + x, and y = -x + 1)

AB = √2, BC = 2√2
⇒ Area = 4 sq. units

Let A₁ = {(x,y):|x|≤y², |x|+2y≤8} and A₂ = {(x,y):|x|+|y|≤k}: If 27 (Area A₁) = 5 (Area A₂), then k is equal to:
V
Vishal Baghel

Required area (above x-axis)
A? = 2∫? (8/2 - x - √x)dx
= 2 [16 - 16/4 - 8/3*2] = 40/3
and A? = 4 (1/2 k²) = 2k²


∴ 27 * (40/3) = 5 * (2k²)
=> k = 6
for above x-axis.

Let g(x) = ∫₀ˣ f(t)dt, where f is continuous function in such that 1/3 ≤ f(t) ≤ 1 for all t ∈ and 0 ≤ f(t) ≤ 1/2 for all t ∈ (1,3]. The largest possible interval in which g(3) lies is :
V
Vishal Baghel

We are given bounds for a function f (t) on two intervals and need to find the range of g (3) = ∫? ³ f (t) dt.
We split the integral: g (3) = ∫? ¹ f (t)dt + ∫? ³ f (t)dt.
For the first interval t ∈ [0, 1], we have 1/3 ≤ f (t) ≤ 1. Integrating from 0 to 1 gives:
∫? ¹ (1/3) dt ≤ ∫? ¹ f (t)dt ≤ ∫? ¹ 1 dt ⇒ 1/3 ≤ ∫? ¹ f (t)dt ≤ 1.
For the second interval t ∈ (1, 3], we have 0 ≤ f (t) ≤ 1/2. Integrating from 1 to 3 gives:
∫? ³ 0 dt ≤ ∫? ³ f (t)dt ≤ ∫? ³ (1/2) dt ⇒ 0 ≤ ∫? ³ f (t)dt ≤ (1/2) (3

...more

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

Sign Up on Shiksha

On Shiksha, get access to

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

Learn more about...

Share Your College Life Experience

Write a Review

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Ask Now
Community GuidelinesUser Points System
QuestionsDiscussionsTags
×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post