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I've went up to tutors at my school and they don't know how to solve this.

Evaluate the line integral of f(x,y) along the curve C

$$f(x,y) = \frac{x^4}{\sqrt{(1+4y)}} , C: y=x^2, 0\le{x}\le{2}$$

I'm as far as setting it as a definite integral between 0 and 2 but not sure how to proceed.

$$\int^2_0$$

The other line integrals I've had were vectors that I could paramaterize and use those to plug into the integral using the formula

$$\int^b_a f[g(t)i,h(t)j,k(t)k]|v(t)|dt$$

This one doesn't look like those

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  • $\begingroup$ Please show what progress you've made so far. $\endgroup$ – rogerl Dec 11 '17 at 21:27
  • $\begingroup$ I've updated it with my progress. I'm truly stuck $\endgroup$ – Samuel Uribe Dec 11 '17 at 21:30
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    $\begingroup$ Notice that your curve is $\left(x,y\right)=\left(t,t^{2}\right)$, a vector! Can you plug this into your formula? $\endgroup$ – eranreches Dec 11 '17 at 21:33
  • $\begingroup$ How am I able to noticed that my curve can be parameterized into $(t,t^2)$? $\endgroup$ – Samuel Uribe Dec 11 '17 at 21:44
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Parametrizise $C$ as $r(t)=(t,t^2),$ $t\in[0,2]$ and note that $r'(t)=(1,2t)$. $$\int_C f(r(t))ds=\int_0^2f(r(t))|r'(t)|dt=\int_0^2 \frac{t^4}{\sqrt{1+4t^2}}\sqrt{1^2+(2t)^2}dt=\int_0^2t^4dt$$

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  • $\begingroup$ I don't understand how you went from $f(r(t))$ to the fraction afterwards or even how you parameterized C. How did you parameterize it if it was $f(x)=x^2$ $\endgroup$ – Samuel Uribe Dec 11 '17 at 21:43
  • $\begingroup$ @Samuel Uribe Note that the curve $y=f(x)$ where $x\in[a,b]$ consists of points $(t,f(t)), $ $t\in[a,b]$ in the plane. See: en.wikipedia.org/wiki/…. $\endgroup$ – dromastyx Dec 11 '17 at 21:48
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There is a formula for the line integral in cartesian coordinates:

$$\int_\gamma g\ ds=\int_{a}^{b}g\left(x, f(x)\right)|s'(x)|dx$$

Where, in cartesian coordinates, $|s'(x)|=\sqrt{1+\left[f'(x)\right]^2},$ because

$$s(x)=(x, f(x))=(x, x^2)$$

So

$$|s'(x)|=|\left(1, 2x)\right|=\sqrt{1+4x^2}$$

Hence the integral is given by

$$\int_\gamma g\ ds=\int_{0}^{2}\frac{x^4}{\sqrt{1+4x^2}}\sqrt{1+4x^2}\ dx=\int_{0}^{2}x^4\ dx=\left[\frac{x^5}{5}\right]_{0}^{2}=\frac{32}{5}.$$

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