# Evaluate the line integral of __f(x,y)__ along the curve__ C. 2) f(x, y) = x 5 1 + 4y , C: y = x...

## Question:

Evaluate the line integral of f(x,y) along the curve C. 2) f(x, y) = x 5 1 + 4y , C: y = x 2, 0 K x K 3

## Techniques of Integration:

Several techniques and method can be used to evaluate integrals. Integration by parts, integration by substitution, parameterization, change of coordinate system, partial fraction decomposition, etc. are some of the prominent methods used for evaluating integrals.

The line integral is

{eq}\int\limits_{c}{f\left( x,y \right)}ds =\int\limits_{c}{\left( 51x+4y \right)}ds \\ {/eq}

The curve is {eq}y={{x}^{2}},0\le x\le 3 {/eq}

{eq}\begin{align} & ds=\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}} \\ & =\sqrt{1+{{\left( 2x \right)}^{2}}} \\ & =\sqrt{1+4{{x}^{2}}} \\ \end{align} {/eq}

Thus, the line integral is

{eq}\int\limits_{x=0}^{3}{\left( 51x+4{{x}^{2}} \right)\sqrt{1+4{{x}^{2}}}dx} {/eq}

We split the integrals into two and solve them separately.

{eq}\int\limits_{0}^{3}{51x\sqrt{1+4{{x}^{2}}}dx}+\int\limits_{0}^{3}{4{{x}^{2}}}\sqrt{1+4{{x}^{2}}}dx {/eq}

Substitute {eq}u=1+4{{x}^{2}} {/eq} in the first integral.

{eq}\begin{align} & \frac{du}{dx}=8x \\ & \Rightarrow \frac{du}{8}=xdx \\ \end{align} {/eq}

When {eq}x=0,u=1 {/eq} and when {eq}x=3,u-37 {/eq}

Thus, on substitution, the integral becomes,

{eq}\begin{align} & \int\limits_{1}^{37}{\frac{51}{8}\sqrt{u}du=\frac{51}{8}\left[ \frac{{{u}^{\frac{3}{2}}}}{\frac{3}{2}} \right]}_{1}^{37} \\ & =\frac{17}{4}\left[ {{37}^{\frac{3}{2}}}-1 \right] \\ & \approx 952.27 \\ \end{align} {/eq}

The second integral is

{eq}4\int\limits_{0}^{3}{{{x}^{2}}}\sqrt{1+4{{x}^{2}}}dx {/eq}

Substitute {eq}x=\frac{\tan \left( u \right)}{2}\to dx\frac{{{\sec }^{2}}\left( u \right)}{2}du {/eq}

{eq}\begin{align} & =4\int{\frac{{{\sec }^{2}}\left( u \right){{\tan }^{2}}\left( u \right)\sqrt{{{\tan }^{2}}\left( u \right)+1}}{8}du} \\ & =\frac{1}{2}\int{{{\sec }^{3}}\left( u \right){{\tan }^{2}}\left( u \right)}du \\ & =\frac{1}{2}\int{{{\sec }^{3}}\left( u \right)\left( {{\sec }^{2}}\left( u \right)-1 \right)}du \\ & =\frac{1}{2}\int{\left[ {{\sec }^{5}}\left( u \right)-{{\sec }^{3}}\left( u \right) \right]}du \\ \end{align} {/eq}

Using the reduction formula for sec

{eq}\int{{{\sec }^{n}}}\left( u \right)du=\frac{n-2}{n+1}{{\int{\sec }}^{n-2}}\left( u \right)du+\frac{{{\sec }^{n-2}}\left( u \right)\tan \left( u \right)}{n-1} {/eq}

On applying the reduction formula repeatedly on {eq}{{\sec }^{5}}\left( u \right) {/eq} and {eq}{{\sec }^{3}}\left( u \right) {/eq} in the above integral, we finally get

{eq}=\frac{-\ln \left( \tan \left( u \right)+\sec \left( u \right) \right)}{16}+\frac{{{\sec }^{3}}\left( u \right)\tan \left( u \right)}{8}-\frac{\sec \left( u \right)\tan \left( u \right)}{16} {/eq}

Now, we undo the substitution and plug in the limits

{eq}\begin{align} & =\left[ \frac{-ar\sinh \left( 2x \right)-4x{{\left( 4{{x}^{2}}+1 \right)}^{\frac{3}{2}}}+2x\sqrt{4{{x}^{2}}+1}}{16} \right]_{0}^{3} \\ & \approx 166.36 \\ \end{align} {/eq}

Thus, the overall integral is

{eq}\begin{align} & =452.21+166.36 \\ & =1118.63 \\ \end{align} {/eq}