So my assignment states the following:
$$\text{Let}\;\;f(x) = -\sin(x),\,\;\text{where}\;\,x \in \left[-\dfrac{\pi}{4}, \dfrac{\pi}{6}\right]$$
Ok, so I know the forum isn't here to do my homework, but I do need some help in the area I'm getting stuck in. Now I know this somehow relates to the Unit circle, but I'm not sure how to use it.
My best guess is that I somehow have to figure out the $x$ values to use so I can calculate the definite integral. Now how do I use the $-\sin(x)$ function to find those values on the unit circle? Or do I instead use the cosine function (anti-derivate) to find the $x$ values?
It's a multiple choice question and I've come close to the answer, but not the actual answer.
What am I missing?
This is what I have so far. I get the $F$ of $-\sin x$ which I believe is $cos x.$ Then I do the following
$$F\left(\frac{\pi}{6}\right) - F\left(-\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{6}\right) - \cos\left(-\frac{\pi}{4}\right) =\frac{\sqrt 3}{2} - \left(-\frac{\sqrt 2}{2} \right) $$
Is that correct?
2 Answers
You are looking for the area under the curve $f(x) = - \sin x$ between the values $x = -\pi/4$ and $x = \pi/6$.
$$\int_{-\pi/4}^{\pi/6} -\sin x \,dx$$
But if you're looking for area between the curve $f(x) = -\sin x$ and the $x$-axis, then you need to divide the integral into two integrals if you are to find "absolute area" between the curve and the x-axis, which is the curve/line $y = 0$, since on the interval $(0, \pi/6)$, $-\sin x < 0$.
Close up of positive region (above x-axis) and negative region (below x-axis). So if you want area between the curve and the x-axis (i.e., the line $y = 0$), you need to integrate the following two integrals, and in each, we take the top curve and subtract the bottom curve. For the first integral, $-\sin x > 0$ so the integrand is $-\sin x - 0$, and in the second integral, $0 > -\sin x$, so we need for the integrand to be $0 - (-\sin x)$:
$$\begin{align}\text{Area}\;& =\;\int_{-\pi/4}^0 (-\sin x - 0)\,dx + \int_0^{\pi/6}(0 - (-\sin x))\,dx \\ \\ & = \int_{-\pi/4}^0 -\sin x\,dx \quad +\quad \int_0^{\pi/6}\sin x\,dx \\ \\ & = \cos x\Big|_{-\pi/4}^0 \quad + \quad -\cos x\Big|_0^{\pi/6}\\ \\ & = \left(1 - \frac{\sqrt 2}2\right) - \left(\frac{\sqrt 3}2 - 1\right)\\ \\ & = 2 - \frac 12(\sqrt 3 + \sqrt 2) \end{align}$$
Area under any function $f(x)$: $$A=\int_{a}^{b}|f(x)|dx$$ $\sin x$ is negative between $-\pi/4$ and $0$, so $|-\sin x|=-\sin x$ for this interval. For $x$ between $0$ and $\pi/6$ you have $|-\sin x|=\sin x$.
Thus, your area is: $$ A=\int_{-\pi/4}^{0}(-\sin{x})dx +\int_{0}^{\pi/6}\sin{x} {dx}$$
And i think you can solve from here on.
