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Find the area bounded by the curves y = x|x| , x - axis , x = 1 & x = -1.

janakiram , 12 Years ago

Grade 12

5 Answers

Parvez ali

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Parvez Ali

askIITians Faculty

Last Activity: 12 Years ago

Sher Mohammad

y=x^2 for x>0

y=-x^2 for x<0

$\int_{-1}^0 -x^2 \, dx+\int_{0}^1 -x^2 \, dx=\frac{-1}{3}+\frac{1}{3}=0$

if we are just checking the magnitude of area,

$|\int_{-1}^0 -x^2 \, dx|+\int_{0}^1 -x^2 \, dx=\frac{1}{3}+\frac{1}{3}=2/3$

Sher Mohammad,

B.tech, IIT Delhi

Approved

Last Activity: 12 Years ago

bharat bajaj

Area bounded :
Integral ofy = x|x| , limits -1 to 1
It should be divided in two parts , -1 to 0 and 0 to 1
Integral 1= -x^2 , -1 to 0 = 1/3

Integral 2= x^2 , 0 to 1 = 1/3
Total area bounded : 2/3 sq. units

Thanks & Regards

Bharat Bajaj

askIITians Faculty
IIT Delhi

Last Activity: 12 Years ago

SADDAM NAFEES

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saddam nafees,

askIITians Faculty

Approved

Last Activity: 12 Years ago

SEKHAR

x|x| has "-" sign in (-1,0) and "+" sign in (1,0) Therefore, to get area of the curve x|x| integrate the function i.e., as -x^2 at (-1,0) and x^2 at (1,0) integrating the above two functions we get * integration of -x^2 at (-1,0) is -1/3 (and) * integration of x^2 at (1,0) is also -1/3 Therefore, adding the above two we get -2/3, Since the area cannot be negative, we take its modulus and it becomes 2/3. Hence, the area of the Curve x|x| is "2/3" from x=1 to x=-1.

Last Activity: 12 Years ago