What is the net area between #f(x) = cscx -xsinx# and the x-axis over #x in [pi/6, (5pi)/8 ]#?

Answer 1

#F(x) = F_1(x) - F_2 =[lntan(frac(x)(2)) -sinx+xcosx]_(pi/6)^((5pi)/8)#
# F(x) ~~ .09143 #

The area under #f(x) = cscx-xsinx# and the #x# axis over the interval #[pi/6, (5pi)/8]# is obtained by compute the integral of #f(x)#,
#F(x) = int_(pi/6)^((5pi)/8) f(x) dx = int_(pi/6)^((5pi)/8) (cscx - xsinx) dx#
Apply the Sum Rule:
#F(x) = int_(pi/6)^((5pi)/8) (cscx) dx - int_(pi/6)^((5pi)/8)(xsinx) dx#
Let
#F_1(x) = int_(pi/6)^((5pi)/8) (cscx) dx# ============> (1)
#F_2 = int_(pi/6)^((5pi)/8)(xsinx) dx# =============> (2)

Let's integrate #color(red)((1))# applying integral substitution:
Let #u= tan(x/2);# then #csc(x) = \frac{1+u^2}{2u} #
and #dx=\frac{2}{1+u^2}du #
#I_1(u)=\int \frac{1+u^2}{2u}\frac{2}{1+u^2}du = \int \frac{1}{u}du#
#I_1(u)= |ln(u)|;# substituting #u= tan(x/2)#
#F_1 = [|ln(tan(x/2))|]#

Now let integrate #color(blue)((2))# applying integration by parts. Recall from the product rule of derivative of 2 function f and g
#(f*g)'= f'g+g'f# integrate both side and you end with
#int(f*g)' dx= int(f'g+g'f) dx #
#(f*g) = int(f'g+g'f) dx# we can rewrite this as:
#color(purple)(int(g'f )dx = (f*g) - int(f'g) dx)# ======>(3)
From what we derived in (3) and letting:
#u = x; (du)/dx = u' = 1 #
#v' = sinx; v = int sinx dx = -cosx #
#color(purple)(int uv'dx = uv - int u'v dx) #
#intxsinx dx = x(-cosx) - int-cosx dx#
#F_2(x) = -xcosx + sinx = [sinx - xcosx]#

#F(x) = F_1(x) - F_2 =[lntan(frac(x)(2)) -sinx+xcosx]_(pi/6)^((5pi)/8)#

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Answer 2

To find the net area between the curve (f(x) = \csc(x) - x \sin(x)) and the x-axis over the interval (x \in [\frac{\pi}{6}, \frac{5\pi}{8}]), we need to calculate the definite integral of (f(x)) with respect to (x) over that interval and take the absolute value of the result. The net area is the absolute value of the integral because the function crosses the x-axis multiple times within the given interval, resulting in both positive and negative areas that cancel each other out when summed.

The integral can be calculated as follows:

[ \text{Net area} = \left| \int_{\frac{\pi}{6}}^{\frac{5\pi}{8}} (\csc(x) - x \sin(x)) , dx \right| ]

After evaluating this integral, you'll get the net area between the curve and the x-axis over the given interval.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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