Question

Evaluate each of the following integrals in the easiest way you can.

$\mathbf{\int }\left(y^2-x^2\right)\mathit{d}\mathit{x}\mathbf{+}\left(2xy+3\right)\mathit{d}\mathit{y}$,along the xaxis from (0,0) and localid="1659182150932" $\left(\surd 5,0\right)$then along

a circular arc from $\mathbf{\left(}\mathbf{\surd }\mathbf{5}\mathbf{,}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{to}\mathbf{}\left(1,2\right)\mathbf{.}$

Short answer

The solution to this problem is$/=\frac{29}{3}$

Step-by-step solution

Given Information.

The given information is$\int \left(y^2-x^2\right)dx+\left(2xy+3\right)dy$

Definition of Green’s Theorem.

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Find the solution.

Write the values of P and Q.

$$\begin{gathered} P=y^2-x^2 \\ Q=2xy+3 \end{gathered}$$

Find the difference of their differentiation.

$$\begin{gathered} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2y-2y \\ =0 \end{gathered}$$

The integral is in the form mentioned below.

$$\begin{gathered} I=-{\int }_1^0\left(3x^2+8x^2+6\right)dx \\ ={\int }_1^0\left(11x^2+6\right)dx \\ ={\left[\frac{11}{3}{x}^3+6x\right]}_0^1 \\ =\frac{29}{3} \end{gathered}$$

Hence, the solution to this problem is$I=\frac{29}{3}$