Math, asked by StrongGirl, 8 months ago

Area bounded by curves y=x² -1 and y=1-x² is? A)1/3 B)8/3 C)4/3 D)2/3

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Answered by Mounikamaddula
6

Answer:

Answer:

  • Area bounded by the curves=8/3

Given:

The two curves are,

y =  {x}^{2}  - 1 and

y = 1 -  {x}^{2}

Equating both the equations,

 {x}^{2}  - 1 = 1 -  {x}^{2}

2 {x}^{2}  = 2

 {x}^{2}  = 1

x =  ± 1

Now,

The area between the curves

=2(^{1} ∫(1 -  {x}^{2}  - ( {x}^{2}  - 1))dx

 = 2( ^{1} ∫( - 2 {x}^{2}  + 2)dx

 = 2( ^{1} (  \frac{ { - 2x}^{3} }{3}  + 2x))

 = 2( \frac{ - 2}{3}  + 2)

 = 2(  \frac{ - 2 + 6}{3} )

 = 2( \frac{4}{3} )

 =  \frac{8}{3}

So, The area bounded by the curves is 8/3.

Answered by pulakmath007
31

\displaystyle\huge\red{\underline{\underline{Solution}}}

FORMULA TO BE IMPLEMENTED

The area bounded by the curve y = f(x), the lines x = a, x= b and x axis is given by

\displaystyle \int\limits_{a}^{b} f(x) \, dx  = \displaystyle \int\limits_{a}^{b} y \, dx

EVALUATION

The given curves are

y =  {x}^{2}  - 1 \:  \:  \:  \: ....(1)

y = 1 -  {x}^{2}  \:  \: ....(2)

The point of intersection of the above two curves are given by

 {x}^{2}  - 1 = 1 -  {x}^{2}

 \implies \: 2 {x}^{2}  = 2

 \implies \:  {x}^{2}  = 1

 \implies \:  {x}  =  \pm \: 1

From Equation (1)

Putting x = 1 we get

y = 0

Putting x = - 1 we get

y = 0

Similarly others points are obtained

So all the point of intersections are

(1,0), (-1,0),(0,1),(0,-1)

The shaded region is bounded by given two curves

Now the region is symmetric about both axis

So the required area is

  = \displaystyle4 \times  \int\limits_{0}^{1} ( 1-  {x}^{2} ) \, dx   \:  \: \:  \:  sq \: unit \:

 \displaystyle \:  = 4 \times  \bigg(x - \left \frac{x^3}{3} \bigg) \right|_0^1  \:   \: \: sq \: unit \:

 \displaystyle \:  = 4 \times \bigg(1 -  \frac{1}{3}  \bigg)  \:  \: sq \: unit \:

 \displaystyle \:  = 4 \times  \frac{2}{3}   \:  \: sq \: unit \:

 \displaystyle \:  =  \frac{8}{ 3}  \:  \:  \: sq \: unit \:

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