Question

Which of the following integrals computes the area below? (See image)

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Answer 1

I really don't like this question but I am led to believe the answer is `II`

Explanation:

The answer cannot be `I` as this region is too large, imagine the space bound by the x-axis and the curve between `-5` and `-3`

The answer cannot be `III` as this region is too small, imagine the space bound by the x-axis and the curve between `-1` and `1`, this is only one part of region we are looking to find.

The answer cannot be `IV` as the region is again too small. It is equivalent to the area bound by `-2` and `-1`, this is only one part of region we are looking to find.

The reason I do not like the question is that you cannot evaluate positive and negative areas in the same integral.

As such you have to split the region and take the absolute value of the negative region.

`int_-2^1 = int_-2^-1 +|int_-1^1|`

`int_-1^1` is a negative so to make is positive we use `-int_-1^1`

So the interval we are looking for is

`int_-2^-1 f(x)dx- int_-1^1f(x)dx`

Breaking this into two parts gives us

`int_-2^-1 f(x)dx =F(x)|""_(-2)^(-1) = F(-1)-F(-2)`

`int_-1^1 f(x)dx =F(x)|""_(-1)^(1) = F(1)-F(-1)`

Now we can combine this

`int_-2^-1 f(x)dx- int_-1^1f(x)dx = F(-1)-F(-2) - (F(1)-F(-1))`

Simplify

`F(-1)-F(-2) - F(1) + F(-1))`

`2F(-1)-F(-2) - F(1)`

It has been a long time since my Fundamental Theorem of Calculus lectures!

Answer 2

Do we feel the graph should look like this?

Explanation:

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