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1) Find the area between $y=(x+4)^{1/2}$ and $y=(4-x)^{1/2}$ and the X-axis, from $x=-4$ to $x=4$.

2) Find the area between $y=\sin(x)$ and $y=\cos(x)$ from $x=\dfrac{\pi}{4}$ to $x=\dfrac{5\pi}{4}$.

I got both questions wrong on my quiz yesterday, and I really want to see the approach and steps to the answers. Much appreciation if someone could help.

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up vote 1 down vote accepted

$(1)\;$ First question: $$y = (4 + x)^{1/2}, \;\; y = (4 - x)^{1/2}, \;\; y = 0 \;\text{(x-axis)}\;\;-4\leq x \leq 4$$

Graphing always helps. See the picture below, compliments of Wolfram Alpha. By setting the equations equal to one another, we can find (and see) that the point of intersection of the first two equations is at $(0, 2)$.

Note that for $x \in [-4, 0]$, we need the area bounded by $y = \sqrt{x + 4}$ and $y = 0$. For $x \in [0, 4]$, we want to compute the area bounded by $y = \sqrt{4 - x}\,$ and $\,y = 0.\;$.

So to compute the area bounded by the respective curves and the $x$-axis $\,(y = 0),\;$ we will compute the sum of two integrals:

$$\textrm{AREA }\; = \int_{-4}^0 \sqrt{x + 4} \, dx + \int_0^4 \sqrt{4 - x} \; dx\tag{1}$$

$(2)\;$ Second problem: $$y = \sin x,\quad y = \cos x, \quad \frac{\pi}{4} \leq x \leq \frac{5 \pi}4$$

Try to find the points of intersection for your second problem by equating $\sin x = \cos x$. Graph the curves so you can see which function is the upper bound of the area to be computed, on the given interval, and which function is the lower bound.

We see $\sin x > \cos x$ over the given interval, and that the $x$-coordinates of the two points of intersection of the curves are precisely the bounds for integrating: from $\;\frac{\pi}{4}\;$ to $\;\frac{5 \pi}4.\;$ In the integrand, we subtract $\cos x\,$ (lower bound) from $\,\sin x\,$ (upper bound).

$$\textrm{AREA } \; = \int_{\pi/4}^{5\pi/4}\left( \sin x - \cos x \right)\; dx\tag{2}$$

I'll let you take the evaluation of the integrals from here!

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Case 1 is basically just realizing how two functions look like, then using the symmetry $$ 2\int_{0}^4 \sqrt{4-x}dx=\left. 2\left(\frac{-2}{3} (4-x)^{3/2}\right) \right|_{0}^4= \frac{32}{3} $$

Case 2 is more of the same as I see.

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Check you can prove the following...drawing a sketch may help a lot:

$$\begin{align*}(1)&\;\;\;\;A=\int\limits_{-4}^0\sqrt{4+x}\,dx+\int\limits_0^4\sqrt{4-x}\,dx\\{}\\(2)&\;\;\;\;A=\int\limits_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx\end{align*}$$

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