I'm trying to understand how to solve this integral, but unsuccessfully. Could you help me to get the answer $H$, or at least $Z$?
$$Z = \int_L^{}(x-y) dx + x \, dy$$
$$\text{when:} ~~~ L: y = 4 - x^{2}, ~A(1;3), ~B(2;0);$$
$$H = \sqrt[5]{90876845839099} \cdot (Z+0,5) \cdot 3 + 1$$
We're trying to integrate the vector field $X(x,y)=(x-y,x)$ over the line $\gamma (t)=(t,4-t^2)$ with $t\in \left[ 1,2\right] $. We have:
$$ \int_\gamma X(r)dr =\int_1^2X(\gamma(t))\cdot\gamma '(t)dt $$ as per definition (the point "$\cdot$" stands for inner product of vectors). Substituting the expressions we already know, $$ \int_\gamma X(r)dr =\int_1^2(t-4+t^2,t)\cdot(1,-2t)dt=\int_1^2t-4+t^2-2t^2= $$ $$ =\int_1^2(-t^2+t-4)dt=\left(-\dfrac{t^3}{3}+\dfrac{t^2}{2}+t\right)\mid^2_1=-\dfrac{8}{3}+2+2+\dfrac{1}{3}-\dfrac{1}{2}-1= $$ $$ -\dfrac{7}{3}+\dfrac{5}{2}=\dfrac{1}{6} $$
You check it tough; I'm very sleepy.