Respuesta :
[tex]y(t)=8\sin(bt)[/tex] has a period of [tex]\dfrac{2\pi}b[/tex], which is to say one "arch" of the curve occurs over the interval [tex]0\le t\le\dfrac\pi b[/tex].
a. Then the area under one such arch is
[tex]\displaystyle\int_0^{\pi/b}8\sin(bt)\,\mathrm dt[/tex]
b. Substitute [tex]u=bt[/tex], so that [tex]\dfrac{\mathrm du}b=\mathrm dt[/tex]. When [tex]t=0[/tex], [tex]u=0[/tex]; when [tex]t=\dfrac\pi b[/tex], [tex]u=\pi[/tex].
Then the integral is
[tex]\displaystyle\frac1b\int_0^\pi8\sin u\,\mathrm du[/tex]
The required area is [tex]\int^{\frac{\pi }{b}}_0 8sin(bt).dt\\[/tex]
The appropriate Substitution is [tex]\dfrac{1}{b} \int^\pi _08sinu.du[/tex]
Given that,
The area under one arch of the curve y(t) = 8sin(bt) for t ≥ 0 where b is a positive constant.
We have to find,
Set up the definite integral needed to find the area.
Make an appropriate substitution.
According to the question,
The area under one arch of the curve y(t) = 8sin(bt) for t ≥ 0 where b is a positive constant.
- The curve y(t) = 8sin(bt) has a period of 2π\b, which is one arch of the curve occur over the interval [tex]0\leq t\leq \frac{\pi }{b}[/tex].
The area under one arch is given by,
[tex]Area = \int^{\frac{\pi }{b}}_0 8sin(bt).dt\\[/tex]
The required area is [tex]Area = \int^{\frac{\pi }{b}}_0 8sin(bt).dt\\[/tex]
- Appropriate Substitute u= bt ,
Then,
[tex]\dfrac{du}{b} = dt \\\\when \ t=0, \ and \ u=0\\\\when\ t = \dfrac{\pi }{b}, u = \pi[/tex]
Then,
The required integral is ,
[tex]\dfrac{1}{b} \int^\pi _08sinu.du[/tex]
The appropriate Substitution is [tex]\dfrac{1}{b} \int^\pi _08sinu.du[/tex].
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