Question

find the area of the region between the curves y=3x^2-3 and y = -2x^+2. The area is

Answer 1

Given : y = 3x² - 3  and y = -2x² + 2

To Find : area of the region between the curves

Solution:

y = 3x² - 3  

y = -2x² + 2

3x² - 3   =  -2x² + 2

=> 5x² - 5 = 0

=> x² - 1 = 0

=> (x + 1)(x - 1) = 0

x = - 1 ,  1

between x = - 1 , 1    y = 3x² - 3   < 0   ( hence -ve sign will be taken )

and -2x² + 2 > 0

Hence Area is

 $-\int\limits^{1}_{-1} {(3x^2-3) \, dx + \int\limits^{1}_{-1} {(-2x^2+2) \, dx$

$=-\left[ x^3 -3x \right]_{-1}^1 + \left[- \frac{2x^3}{3} +2x \right]_{-1}^1$

= - [ 1 - 3  + 1 - 3]  + [ -2/3 + 2  -2/3 + 2]

=   4 + 8/3

= 20/3

area of the region between the curves  = 20/3 sq units

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