Piecewise (I think) definite integral

[FSUchess99]

In each part, evaluate the integral, given that:

f(x)= abs[x-2] when x is greater than or equal to 0
= x+2 when x is less than 0

part d: integrate f(x)dx from -4 to 6

I thought I had to divide this into the two integrable functions from -4 to 0 and 0 to 6, and then split the 0 to 6 absolute value function. The solutions manual divides it into four functions though; from -4 to -2, -2 to 0, 0 to 2, and 2 to 6.

THEN they solve it as a group of triangles, which I could sketch, but why would I? why not just evaluate?

question 1: Why can't I integrate x+2 from -4 to 0?

question 2: When I evaluate the 0 to 6 absolute value function, I can look at my notes and see you split it where x=2. from 0 to 2 its evaluated as 2-x because it would have been negative, and from 2 to 6 its evaluated as x-2. why is the function inverted when the absolute value would be negative?

thanks in advance for any help. I thought the solutions manual would help but it literally says this as the solution:

Triangle of height 2 and base 2, below axis, plus a triangle of height 2, base 2 above axis,
another of height 2 and base 2 above axis, and a triangle of height 4 and base 4, above axis.
Thus

f(x) = -2+2+2+8

 

[galactus]

In each part, evaluate the integral, given that:

\(\displaystyle f(x)=\left\{\begin{array}{cc}|x-2|, \;\ x\geq 0\\ x+2, \;\ x\leq 0\end{array}\)

THEN they solve it as a group of triangles, which I could sketch, but why would I? why not just evaluate?

You have to break it up because of the absolute value. There are some areas below the axis that would cancel out if you 'just evaluated'. What you actually need is the sum of the areas. Whether above or below the x-axis.

question 1: Why can't I integrate x+2 from -4 to 0?

Because if you integrate this way it will evaluate to 0. The triangle below the x-axis will give a negative area equal to the area of the triangle above the x-axis.

Do not do it this way:
\(\displaystyle \int_{-4}^{0}(x+2)dx=0\)

Do it this way:
\(\displaystyle \int_{-4}^{-2}(x+2)dx+\int_{-2}^{0}(x+2)dx=4\)

You want the total area, so the negative area is actually |-2|=2. Thus, the total area is 2+2=4.

See what I am getting at?.

question 2: When I evaluate the 0 to 6 absolute value function, I can look at my notes and see you split it where x=2. from 0 to 2 its evaluated as 2-x because it would have been negative, and from 2 to 6 its evaluated as x-2. why is the function inverted when the absolute value would be negative?

Because 2-x is the same as -x+2. As in \(\displaystyle \int_{0}^{2}(-x+2)dx=2\)

f(x) = -2+2+2+8

[/quote]

The areas are just 4 right triangles of area 2 each.

 

[tkhunny]

Q1: The integral on [-4,0] is zero. You get that with symmetry. If you intend to treat the area under the x-axis as positive, you should state that clearly.

 

[FSUchess99]

sorry if I stated the question incorrectly, but they're not asking for net signed area.

the instructions say "evaluate the integral, given that..." etc. so shouldn't I let the area left of the y axis cancel with itself due to symmetry and the area from x=0 to x=6 adds to 10?