Question

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)

Answer 1

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

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)|  =$\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25} =5$

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

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

Applying the Rule Integration for a Constant.

$\int _0^1\:20t\cdot sin\left(3t+1\right)\:\cdot dt\\ \\ 20\cdot \int _0^1t\sin \left(3t+1\right)dt$

Applying the Rule Integration by Parts.

∫u v dx = uv -∫v du

u=t

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

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

Read more about integration rules here:

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