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Write a definite integral that yields the area of the region. (Do not evaluate the integral.)?

Answer 1

Emilia Aguilar

#int_(-4)^4 (4-absx)dx = 16#

The shaded area is the area between the #x# axis and the graph of the curve:

#y= 4-absx#

in the interval #x in (-4,4)#

Based on the properties of the integrals we therefore know that

#A = int_(-4)^4 (4-absx)dx#

We can also easily evaluate the integral using additivity:

#A = int_(-4)^0 (4-absx)dx +int_0^4 (4-absx)dx#

and as for # x in [-4,0]# we have #abs x = -x# and vice versa for #x in [0,4]# we have #abs x = x#:

#A = int_(-4)^0 (4+x)dx +int_0^4 (4-x)dx#

Substitute in the second integral:

#u = -x#

#int_0^4 (4-x)dx = -int_0^(-4) (4+u)du = int_(-4)^0 (4+u)du #

So:

#A = int_(-4)^0 (4+x)dx +int_(-4)^0 (4-u)du#

#A = 2int_(-4)^0 (4+x)dx#

#A= 2[(4+x)^2/2]_(-4)^0 = 16#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.