y = cos(x), y = 1, −π ≤ x ≤ π
If i try to solve it finding the integral I get
$$\int_{-pi}^{pi} cos(x)-1 dx$$
$$\int_{-pi}^{pi} sin(pi)-(pi) - (sin(-pi) - (-pi)) dx$$
= -6.283185 which is just 2 x pi and getting rid of the - becomes 6.283185
y = cos(x), y = 1, −π ≤ x ≤ π
If i try to solve it finding the integral I get
$$\int_{-pi}^{pi} cos(x)-1 dx$$
$$\int_{-pi}^{pi} sin(pi)-(pi) - (sin(-pi) - (-pi)) dx$$
= -6.283185 which is just 2 x pi and getting rid of the - becomes 6.283185
Sorry, but I can't read your mind. Is this another area problem? If so ...
first of all, $1 \ge \cos{x}$
second, you can take advantage of symmetry ...
$\displaystyle A = 2\int_0^\pi 1-\cos{x} \, dx$
If the problem was to integrate $$\int_{-\pi}^\pi cos(x) -1 dx$$ then your second line is written incorrectly- you have already done the integration, you don't write the integration symbols again:
$$\int_{-\pi}^\pi cos(x)- 1 dx= \left[sin(x)- x\right]_{-\pi}^\pi= (sin(\pi)- \pi)- (sin(-\pi)- (-\pi))= (0- \pi)- (0+ \pi)= -2\pi$$
Why "get rid of the -"?= -6.283185 which is just 2 x pi and getting rid of the - becomes 6.283185
absolute value is required when the curve in question changes sign across the axis ... for the area between curves $f$ and $g$, abs value is required when the curves cross each other and change sign of $f -g$. For a pencil/paper solution, you still have to break it into 2 or more integrals if that happens. If using a calculator with an integrate function, enter the integrand using absolute value over the whole interval.