Question

Can anyone explain why this two is being added before the integral? Does it have something to do with integral symmetry?

Answer 1

Yes it's exactly due to symmetry. Specifically, symmetry about the x axis.

Simply writing
`\displaystyle \int_(-8)^(-4)\left[√(x+8) \ \right]dx`, without the 2 out front, will only get you the area of the portion shown in red (see diagram below).

The blue region has equal area of the red region due to symmetry.

So,

`\displaystyle 2\int_(-8)^(-4)\left[√(x+8) \ \right]dx`

represents the red and blue regions combined.

The portion in green is
`\displaystyle \int_(-4)^(8)\left[√(x+8) - (x)/(2) \ \right]dx` which is the integral of the difference of the upper and lower curves over the interval
`-4 \le x \le 8`

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In all honesty, it's probably easier to integrate with respect to y since the given functions are in terms of y initially. Also, there isn't a junction point in which the curves swap places in terms of which one is larger. However, it doesn't hurt to have practice in integrating with respect to x.

If you're curious about what the y integral looks like, then it would be

`\displaystyle \int_(-2)^(4) \bigg[ (2y) - (y^2-8)\bigg] dy`

You can use a tool like WolframAlpha to check that both integral expressions result in 36 to help confirm that they represent the same overall area (just in different ways of course).

Answer 2

Yes - see attached sketch and explanation

Explanation:

As you have a sideways parabola, you would need to split the integral into 3 parts.

Find the x coordinate where the parabola first intersects the line: x = - 4.

Now you should be able to see that the parabola between where it intersects the x-axis (x = -8) and the line x = -4 can be split into 2 equal parts (above and below the x-axis). I have shaded these parts as blue and red on my attached sketch.

As the areas above and below the x-axis and the parabola are equal, this can be written as 2 x the area ABOVE the x-axis (the positive integral). So yes, this is why the 2 has been added before the integral, as without the 2 the integral is finding the area ABOVE the x-axis and the parabola curve ONLY.

Then you would find the area between the parabola and the line between x = -4 and x = 8 (shaded grey on my sketch), and as it's only the positive part of the parabola, a normal "integral between 2 curves" can be applied, i.e.

`\int\limits^a_b {(top-bottom)} \, dx`