calculusintegration
I have been at this for over an hour, can someone please help.
Let $F(a)$ be the area between the $x$-axis and the graph of $y=x^2\sin(x/4)$ between $x=0$ and $x=a$, for $a>0$ (consider the area to be negative if the graph lies below the axis).
$F(a) =$ ?
I thought $F(a) = 375.655$ but the webwork I am doing says that I am wrong.
I took the integral of x^2 sin(x/4) from 0 to 12.566371 and I got 375.655
$(1)\;$ 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!
I hope you understand from this figure why you need two integrals:
Best Answer
It is very nice you be told that your answer is not correct.
$\sin(x/4)$ is zero if $x = 4 \pi$ (this is your 12.566371 which is just an approximate value). Then $a = 4 \pi$. If you plug this number in your integral (we are still waiting for it), the exact result is $64 (\pi^2 - 4)$, the approximate value of which being
$$375.65468166971895160540742399207367266\ldots$$
So, your value of $375.655$ is wrong.