I need to calculate the line integral for $f(x,y)=x+2y$ in the curve $y=x^2+1$ from point $(1,2)$ to $(3,10)$. So I parameterized the parabola as $C(t)=\langle t;t^2+1\rangle$ with $1 \le t \le 3$.
I defined $ds=\sqrt {1+4t^2}. dt$
So I got the line integral as: $\int_1^3 (2t^2+t+2). \sqrt {1+4t^2} .dt$
Now this is where I got stuck. What can I do to get an easier integral? That root is driving me crazy. Thanks!
With $u=1+4t^2$, $du=8t\;dt$, therefore: $$ \int_1^3 2t^2 \sqrt{1+4t^2}\; dt =\int_5^{14} 2 \frac{u-1}{4} \sqrt{u}\; du =\frac{1}{2}\int_5^{14} u^{3/2}-\sqrt{u}\; d{u} $$ The last integral is now easy to compute! The same change of variables work for the rest too.