Answer:
[tex]-7.3[/tex]
Step-by-step explanation:
We want to evaluate the line integral:
[tex]\int\limits^{(4,6)}_{(0,3)} {x\sin y} \, ds[/tex]
where [tex]ds=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2 }dt[/tex]
The parametric equation of the straight line joining (0,3) and (4,6) is
[tex]x=4t[/tex] and [tex]y=3t+3[/tex]
This implies that;
[tex]ds=\sqrt{(4)^2+(3)^2 }dt[/tex]
[tex]ds=\sqrt{25}dt[/tex]
[tex]ds=5dt[/tex]
Our line integral then becomes;
[tex]\int\limits^{1}_{0} {4t\sin (3t+3)} \, 5dt[/tex]
Using, using integration by parts, we obtain;
[tex]20\int\limits^{1}_{0} {t\sin (3t+3)} \, dt=-7.3[/tex] to the nearest tenth.
The solution to the line integral for the equation [tex]\mathbf{\int ^{(4,6)}_{(0,3)} \ x \ sin y \ ds}[/tex] where c is the line segment from (0,3) to (4,6) is -7.3
A line integral is a type of integral in mathematics in which the variable function to be integrated is measured across a curve.
From the given information, we are to evaluate the line integral:
[tex]\mathbf{\int ^{(4,6)}_{(0,3)} \ x \ sin y \ ds}[/tex]
where;
[tex]\mathbf{ds = \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2}}[/tex]
Using the parametric function to determine the straight line joining (0,3) and (4,6), we have:
Then, we can now have ds to be:
[tex]\mathbf{ds = \sqrt{(\dfrac{4}{1})^2 + (\dfrac{3}{1})^2} \ dt}[/tex]
[tex]\mathbf{ds = \sqrt{(16 + 9} \ dt}[/tex]
[tex]\mathbf{ds = \sqrt{25} \ dt}[/tex]
ds = 5dt
Now, the line integral can be written as:
[tex]\mathbf{=\int^1_0 4t sin (3t + 3) \ 5 dt}[/tex]
By applying integration by parts, we have:
[tex]\mathbf{= 20 \int^1_0 t sin (3t + 3) \ dt}[/tex]
= -7.3
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