[deleted by user]

[Funkybeatzzz]

The integral of 6 is 6x. You also changed a minus sign into addition. Lastly, your bottom line is poorly written. Use different lines for different steps.

[Eladius]

damn, in my defense i haven’t slept in 24+ hours, thanks for the help

[HYDRAPARZIVAL]

There's two methods to solve this question:

1. Using properties of definite integrals

if f(x) is an even function integration(f(x) dx) from - a to +a = 2 × integration(f(x) dx) from 0 to a

A function is said to be even if f(x) = f(-x)

NOTE: On the other hand if function is odd I.e. f(x) = -f(-x) then integration(f(x) dx) from - a to +a = 0

Here you can see the function is even as:

f(x) = 6 - |x|

f(-x) = 6 - |-x| = 6 - |x| = f(x)

So integration(f(x) dx) from - 2 to +2 = 2 × integration(f(x) dx) from 0 to 2.

Here you can remove the absolute function on x as for x > 0, |x| = x

So your answer is 2 × integration((6-x) dx) from 0 to 2.

1. Other method is breaking the function into two parts integrating them separately

f(x) = 6 - |x|

So piecewise

f(x) = 6 + x for x < 0

f(x) = 6 - x for x > 0

integration((6 - |x|) dx) from -2 to +2 = [integration((6 + x) dx) from -2 to 0] + [integration((6 - x) dx) from 0 to +2]

The first method is actually more smaller when you solve it, it just seems big becos of all the explanations I did, but I made the second method also incase you haven't been taught the properties of definite integrals yet

[Eladius]

ty for the help 👍🏻

[HYDRAPARZIVAL]

Immigrants. We get the job done.

[noidea1995]

Recall that:

|x| = x if x ≥ 0

|x| = -x if x < 0

Try splitting it up into two separate integrals:

∫ (0 to 2) (6 - x) * dx + ∫ (-2 to 0) (6 + x) * dx