Question #0e2de

1 Answer
Archish R.
Dec 11, 2017

#104/15#

Explanation:

The two graphs intersect at (1,0) and (-1,0)

enter image source here 104/151

The red area is the area under the curve of #y=-4x^2+4#, or

#int_-1^1(-4x^2+4)dx#
#=int_-1^1(-4x^2)dx+int_-1^1(4)dx#
#=[4x-4/3x^3],(1,-1)#
#=(4(1)-4/3(1)^3)-(4(-1)-4/3(-1)^3)#
#=8/3-(-8/3)#
#=16/3#

The blue area is the area under the curve #x^4-1=y# or

#int_-1^1(x^4-1)dx#
#=int_-1^1(x^4)-int_-1^1(1)#
#=[x^5/5-x],(1,-1)#
#=((1)^5/5-(1))-((-1)^5/5-(-1))#
#=(-4/5)-(4/5)#
#=-8/5#
However, since we are taking the area, the integral must be positive.

Area under #y=-4x^2+4#: #16/3#
Area under #y=x^4-1#: #8/5#

Add them together to get: #16/3+8/5=104/15#

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