$$\begin{gathered} \mathbf{Evaluate}\mathbf{}\mathbf{the}\mathbf{}\mathbf{line}\mathbf{}\mathbf{integral}\mathbf{}\mathbf{\int }\left(x^2-y^2\right)\mathbf{dx}\mathbf{-}\mathbf{2}\mathbf{xydy}\mathbf{}\mathbf{along}\mathbf{}\mathbf{each}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{following}\mathbf{}\mathbf{paths}\mathbf{}\mathbf{from}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{)}\mathbf{}\mathbf{to}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{)}\mathbf{.}\mathbf{} \\ \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{.} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{x}\mathbf{=}{\mathbf{t}}^{\mathbf{2}} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathbf{from}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{;}\mathbf{}\mathbf{then}\mathbf{}\mathbf{along}\mathbf{}\mathbf{the}\mathbf{}\mathbf{straight}\mathbf{}\mathbf{line}\mathbf{}\mathbf{joining}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{)}\mathbf{}\mathbf{to}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{2}\mathbf{)}\mathbf{.} \end{gathered}$$

$$\begin{gathered} \mathrm{The}\mathrm{following}\mathrm{is}\mathrm{the}\mathrm{given}\mathrm{information}: \\ \mathrm{y}=2{\mathrm{x}}^2 \\ \mathrm{y}=2\sqrt{\mathrm{x}} \\ \mathrm{y}=4-2\mathrm{x}. \end{gathered}$$

A line integral is integral in which the function to be integrated is determined along a curve in thecoordinate system.In mathematics and physics, scalar field or scalar-valued function referred to a scalarvalueto everypointin aspace– possiblyphysical space. The scalar may either be a (dimensionless) mathematical number or aphysical quantity.

a) Substitute in the integral with $y=2x^2$.

Its derivative is dy=4xdx .

Find the integral.

$$\begin{gathered} I=\int \left(x^2-4x^2\right)dx-2x\left(2x^2\right)4xdx \\ ={\int }_0^1\left(-20x^4+x^2\right)dx \\ ={\left[-20\frac{x^5}{5}+\frac{x^3}{3}\right]}_0^1 \\ =\frac{-11}{3} \end{gathered}$$

b)Substitute in the integral with $y=2\sqrt{x}$.

Its derivative is$dy=\frac{1}{2\sqrt{x}}dx$.

Find the integral.

$$\begin{gathered} I=\int \left(X^2-4x\right)dx-2x\left(2\sqrt{x}\right)\frac{1}{\sqrt{x}}dx \\ ={\int }_0^1\left(x^2+8x\right)dx \\ ={\left[\frac{x^3}{3}+8\frac{x^2}{2}\right]}_0^1 \\ =\frac{-11}{3} \end{gathered}$$

c) In this part the integral is summed over two paths.

The first path is Y=0 .

Find the integral.

$$\begin{gathered} I_1={\int }_0^2{x}^{2}dx \\ ={\overline{)\frac{x^3}{3}}}_0^2 \\ =\frac{8}{3} \end{gathered}$$

The Second path is from (2, 0) to (1, 2).

Find the equation of line from point (2, 0) to (1, 2).

$$\begin{gathered} \frac{y-2}{x-1}=\frac{0-2}{2-1} \\ y=4-2x \end{gathered}$$

Substitute in the integral with y=4-2x .

Its derivative is dy=-2dx .

$$\begin{gathered} I_2=\int \left(x^2-\left(16-4x^2-16x\right)\right)dx-2x\left(4-2x\right)-2dx \\ ={\int }_2^1\left(-11x^2+32x-16\right)dx \\ ={\left[-\frac{11x^3}{3}+\frac{32x^2}{2}-16x\right]}_2^1 \\ =\frac{-19}{3} \end{gathered}$$

The following is the total integral.

$$\begin{gathered} I=I_1+I_2 \\ =\frac{8}{3}+\left(\frac{-19}{3}\right) \\ =-\frac{11}{3} \end{gathered}$$

Hence,the solution to this problem is,

$$\begin{gathered} a)I=\frac{-11}{3} \\ b)I=\frac{-11}{3} \\ c)I=\frac{-11}{3} \end{gathered}$$