Switching bounds of an integral doesn't change sign?

Question

I know that when switching the upper and lower bounds of an integral, you also switch signs. But what about this:

Integral (-2 to -1) ((16-x^3) / x^3) = Integral (-1 to -2) ((16-x^3) / x^3)

-8x^-2 - x + C = -8x^-2 + x + C

(-8(-2)^-2 - (-2)) - (-8(-1)^-2 - (-1)) = (-8(-1)^-2 + (-1)) - (-8(-2)^-2 + (-2))

0--7 = 7-0

0+7 = 7-0

7 = 7

Even though the bounds changed, the answer didn't change signs. Why is this?

Answer 1

When you wrote done –8x^–2 – x + C = –8x^–2 + x + C

you introduced a sign error in the second term.

It shouldn't be – x on the left and + x on the right.

It shouldn be – x on both sides.

This is why you are incorrectly getting the same sign of the integral on the left and right when you switch the order of the upper and lower bounds of the integral.

Without the sign error (and plugging the upper and lower bounds in the correct order), this is what you get:

–8x^–2 – x evaluated from –2 to –1 is

–8(–1)^–2 – (–1) – [–8(–2)^–2 – (–2)] = –8 + 1 – [–2 + 2] = –7 – 0 = –7

–8x^–2 – x evaluated from –1 to –2 is

–8(–2)^–2 – (–2) – [–8(–1)^–2 – (–1)] = –2 + 2 – [–8 + 1] = 0 – (–7) = 7

Switching the order of the upper and lower bounds of the integral does change the sign of the integral (as it must).