### Select your language

Suggested languages for you:

Americas

Europe

Q7E

Expert-verified
Found in: Page 771

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the line integral, where C is the given curve. $$\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}$$, $${\rm{C}}$$consists of line segments from $${\rm{(0,0)}}$$to$${\rm{(2,1)}}$$and from $${\rm{(2,1)}}$$to$${\rm{(3,0)}}$$.

The line integral $$\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\frac{{\rm{5}}}{{\rm{2}}}$$.

See the step by step solution

## Step 1: Explanation of solution.

Find the line integral

$$\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}$$,

Where $${\rm{C}}$$ is the line segment from $${\rm{(0,0)}}$$ to $${\rm{(2,1)}}$$and from $${\rm{(2,1)}}$$ to $${\rm{(3,0)}}$$. Note$${\rm{C = }}{{\rm{C}}_1}{\rm{ + }}{{\rm{C}}_2}$$, where $${{\rm{C}}_1}$$is the line segment joining $$(0,0)$$ to $$(2,1)$$ and $${{\rm{C}}_2}$$is the line segment joining $${\rm{(2,1)}}$$ to $${\rm{(3,0)}}$$.

$$\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\int_{{{\rm{C}}_{\rm{1}}}{\rm{ + }}{{\rm{C}}_{\rm{2}}}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}$$.

## Step 2: Graph and Calculation.

On $${C_1},y = \frac{x}{2}$$ which implies $$dy = \frac{{dx}}{2}$$, and $${{\rm{C}}_{{\rm{2,y}}}}{\rm{ = - x + 3}} \Rightarrow {\rm{dy = - dx}}$$.

$$\begin{array}{l}{{\rm{C}}_{\rm{1}}}{\rm{:y = }}\frac{{\rm{x}}}{{\rm{2}}}{\rm{,0}} \le x \le {\rm{2}}\\{{\rm{C}}_{\rm{2}}}{\rm{:y = - x + 3,2}} \le x \le 3\end{array}$$

Thus,

$$\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\int_{{{\rm{C}}_{\rm{1}}}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy + }}\int_{{{\rm{C}}_{\rm{2}}}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}$$

$$\begin{array}{l}{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {\left( {{\rm{x + 2 \times }}\frac{{\rm{x}}}{{\rm{2}}}} \right){\rm{dx + }}{{\rm{x}}^{\rm{2}}}\frac{{{\rm{dx}}}}{{\rm{2}}}} \right)} {\rm{ + }}\int_{\rm{2}}^{\rm{3}} {\left( {{\rm{(x + 2 \times - x + 3)dx + }}{{\rm{x}}^{\rm{2}}}{\rm{ - dx}}} \right)} \\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {{\rm{2xdx + }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{dx}}} \right)} {\rm{ + }}\int_{\rm{2}}^{\rm{3}} {{\rm{(6 - x)}}} {\rm{dx - }}{{\rm{x}}^{\rm{2}}}{\rm{dx}}\\{\rm{ = }}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{6}}}} \right)_{\rm{0}}^{\rm{2}}{\rm{ + }}\left( {{\rm{6x - }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{2}}^{\rm{3}}\\{\rm{ = 4 + }}\frac{{\rm{4}}}{{\rm{3}}}{\rm{ + 6 - }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{19}}}}{{\rm{3}}}\\{\rm{ = 10 - }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{ - 5}}\\{\rm{ = }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{.}}\end{array}$$

Therefore the line integral $$\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\frac{{\rm{5}}}{{\rm{2}}}$$.