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Evaluate the line integral, where c is the given curve. ∫c x sin(y)ds, c is the line segment from (0, 1) to (4, 4)

Sagot :

From calculations, the given integral ∫c x sin(y)ds is equal to [tex]20\left(-\frac{1}{3}\cos \left(4\right)+\frac{1}{9}\sin \left(4\right)-\frac{1}{9}\sin \left(1\right)\right)=0.806[/tex].

Integration

The integrals are the opposite of derivatives. They are used in several applications, like: calculations of areas, volumes and others.

For solving an integration, you should know its rules. For this question will be necessary to apply the following integration rules:

  • For constant function -  ∫b dx = b ∫ dx= bx+C
  • For sin function -  ∫sin(x) dx = cos(x) + C
  • For integration by parts - ∫u v dx = uv -∫v du

First, you should calculate the segment from the points (0, 1) and (4, 4).

segment=(4-0,4-1)=(4,3).

After that you should parametrize the segment:

r(t)=(0,1)+(4t,3t)= (4t,3t+1), where 0≤t≤1

Now, you can find dr/dt.

r'(t)=(4,3)

Consequently, the magnitude of |r'(t)| will be:

|r'(t)|  =[tex]\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25} =5[/tex]

Finally you can evaluate the integral: ∫c x sin(y)ds. From r(t), you know that x=4t and y=3t+1.

[tex]\int _0^1\:xsin\left(y\right)\:ds=\int _0^1\:4t\cdot sin\left(3t+1\right)\:\cdot 5ds=\int _0^1\:20t\cdot sin\left(3t+1\right)\:\cdot ds[/tex]

Applying the Rule Integration for a Constant.

[tex]\int _0^1\:20t\cdot sin\left(3t+1\right)\:\cdot dt\\ \\ 20\cdot \int _0^1t\sin \left(3t+1\right)dt\\ \\[/tex]

Applying the Rule Integration by Parts.

∫u v dx = uv -∫v du

u=t

dv= sin(3t +1 )dt, then v=

[tex]=20\left[-\frac{1}{3}t\cos \left(3t+1\right)-\int \:-\frac{1}{3}\cos \left(3t+1\right)dt\right]^1_0\\ \\=20\left[-\frac{1}{3}t\cos \left(3t+1\right)+\frac{1}{9}\sin \left(3t+1\right)\right]^1_0\\ \\ =20\left(-\frac{1}{3}\cos \left(4\right)+\frac{1}{9}\sin \left(4\right)-\frac{1}{9}\sin \left(1\right)\right)\\ \\ =0.806[/tex]

Read more about integration rules here:

https://brainly.com/question/14405394

The value of the integral  ∫c x sin(y)ds where c is the curve is 0.806 if the line segment is from (0,1) to (4,4).

What is integration?

It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.

First, we have to calculate line segment (0,1 ) to (4, 4)

= (4-0, 4-1) = (4, 3)

Parametric form of the segment:

P(t) = (0+4t, 3t)   where 0 ≤ t ≤ 1

Now differentiate the segment:

P'(t) = (4, 3)

The magnitude of the P'(t)

[tex]\rm P'(t) = \sqrt{4^2+3^2}[/tex]

P'(t)  = 5

Now the integration can be evaluated from the P(t)

[tex]\rm \int\limits^1_0 {xsin(y)} \, ds = \int\limits^1_0 {4tsin(3t+1)} \, 5ds[/tex]    ( x= 4t, y = 3t+1)

[tex]\rm \int\limits^1_0 {xsin(y)} \, ds = 20\int\limits^1_0 {tsin(3t+1)} \, ds[/tex]

The value of the integration:

[tex]\rm \int \limits^1_0 {tsin(3t+1)} \, ds = 0.040[/tex]

[tex]\rm \int\limits^1_0 {xsin(y)} \, ds = 20(0.04)[/tex]

[tex]\rm \int\limits^1_0 {xsin(y)} \, ds =0.806[/tex]

Thus, the value of the integral  ∫c x sin(y)ds where c is the curve is 0.806 if the line segment is from (0,1) to (4,4).

Learn more about integration here:

brainly.com/question/18125359

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