Thread: Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

1. Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

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

2. Re: Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

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 ...

$\displaystyle A = 2\int_0^\pi 1-\cos{x} \, dx$

3. Re: Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

Originally Posted by bee77
y = cos(x), y = 1, −π ≤ x ≤ π

If i try to solve it finding the integral I get

$$\int_{-pi}^{pi} cos(x)-1 dx$$

$$\int_{-pi}^{pi} sin(pi)-(pi) - (sin(-pi) - (-pi)) dx$$

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:
$$\int_{-\pi}^\pi cos(x)- 1 dx= \left[sin(x)- x\right]_{-\pi}^\pi= (sin(\pi)- \pi)- (sin(-\pi)- (-\pi))= (0- \pi)- (0+ \pi)= -2\pi$$

= -6.283185 which is just 2 x pi and getting rid of the - becomes 6.283185
Why "get rid of the -"?

4. Re: Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

I thought with areas we take the absolute value ...is that right ?
Thanks

5. Re: Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

Yes another area problem ,sorry .

6. Re: Should I just try and remember this conceptually ? with cos (x) = 1 or solve?

Originally Posted by bee77
I thought with areas we take the absolute value ...is that right ?
Thanks
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.

Thank you!