Question

Find the line integral of f(x,y)= ye^x^2 along the curve r(t)= 4ti-3tj, -1<= t <= 1The integral of f is ??

Answer 1

$\displaystyle\int_Cf(x,y)\,\mathrm dS=\int_{-1}^1f(x(t),y(t))\|\vec r'(t)\|\,\mathrm dt$

$=\displaystyle\int_{-1}^1(-3t)e^{(4t)^2}\sqrt{4^2+(-3)^2}\,\mathrm dt$

$=\displaystyle-15\int_{-1}^1 te^{16t^2}\,\mathrm dt$

Let $u=16t^2$, so that $\mathrm du=32t\,\mathrm dt$:

$=\displaystyle-\frac{15}{32}\int_{16}^{16} e^u\,\mathrm du=\boxed0$

Answer 2

Integrating function f implies that the parts of the function are added up to find the whole

The integral of f is 0.

The given parameters are:

$\mathbf{f(x,y) = ye^{x^2}}$

$\mathbf{r(t) = 4ti - 3tj}$

$\mathbf{ -1 \le t \le 1}$

$\mathbf{r(t) = 4ti - 3tj}$ means that:

$\mathbf{x = 4t}$

$\mathbf{y = -3t}$

Start by calculating: r->

$\mathbf{r\to(t) = \sqrt{4^2 + (-3)^2}}$

$\mathbf{r\to(t) = \sqrt{25}}$

$\mathbf{r\to(t) = 5}$

So, the integrand is:

$\mathbf{\int_Cf(x,y) dS = \int\limits^1_{-1}f(x,y)||r\to(t)||dt}$

This gives

$\mathbf{\int_Cf(x,y) dS = \int\limits^1_{-1}ye^{x^2} \times 5\ dt}$

Substitute values for x and y

$\mathbf{\int_Cf(x,y) dS = \int\limits^1_{-1}(-3t)e^{(4t)^2} \times 5\ dt}$

$\mathbf{\int_Cf(x,y) dS = \int\limits^1_{-1}(-15t)e^{16t^2}} \ dt}$

Remove the constant

$\mathbf{\int_Cf(x,y) dS = -15\int\limits^1_{-1}te^{16t^2}} \ dt}$

Represent

$\mathbf{16t^2}}$ with u.

So, we have:

$\mathbf{u = 16t^2}$

When t = 1 and -1, we have:

$\mathbf{u = 16(1)^2 = 16}$

$\mathbf{u = 16(-1)^2 = 16}$

Differentiate

$\mathbf{du =32t\ dt}$

Make t dt the subject

$\mathbf{t\ dt =\frac{du}{32t}}$

So, we have:

$\mathbf{\int_Cf(x,y) dS = -15\int\limits^1_{-1}te^{16t^2}} \ dt}$

$\mathbf{\int_Cf(x,y) dS = -15\int\limits^{16}_{-16}e^{u}\ du}$

Integrate

$\mathbf{\int_Cf(x,y) dS = -15 \times [e^u] \limits^{16}_{16} }$

Expand

$\mathbf{\int_Cf(x,y) dS = -15 \times [e^{16} - e^{16}]}$

$\mathbf{\int_Cf(x,y) dS = -15 \times0}$

$\mathbf{\int_Cf(x,y) dS = 0}$

Hence, the integral of f is 0.

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