Question

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

Answer 1

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

Explanation:

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`