# Calculate the integral \int_{0}^{\sqrt{2}}\frac{4x}{\sqrt{1 + 4x^2}}dx

## Question:

Calculate the integral {eq}\int_{0}^{\sqrt{2}}\frac{4x}{\sqrt{1 + 4x^2}}dx {/eq}

## Utilizing Substitution to Evaluate an Integral:

The method of substitution is utilized when an integrand is a composite function.

{eq}\int f(g(x))g'(x)dx=\int f(u)du {/eq}

A general strategy for utilizing the substitution method is when given an integral with a composite function, {eq}f(g(x)) {/eq} is to identify an inner function that contains a multiple of {eq}g'(x) {/eq} in the integrand. Let {eq}u=g(x) {/eq} and calculate the relation between {eq}du {/eq} and {eq}dx {/eq}, make appropriate substitutions, and integrate with respect to {eq}u {/eq}.

Step 1. Let {eq}u=4x^{2}+1 {/eq} and solve for {eq}4x\:dx {/eq}.

{eq}\begin{align} \displaystyle u\displaystyle &=4x^{2}+1\\ du\displaystyle & =8x\:dx\\ \frac{1}{2}du \displaystyle &=4x\:dx\\ \end{align} {/eq}

Step 2. Calculate the boundaries of the integral in terms of the variable {eq}u {/eq}.

When {eq}x=0: {/eq}

{eq}\begin{align} \displaystyle u\displaystyle &=4x^{2}+1\\ \displaystyle &=4(0)^{2}+1\\ \displaystyle &=1\\ \end{align} {/eq}

When {eq}x=\sqrt{2} {/eq}:

{eq}\begin{align} \displaystyle u\displaystyle &=4x^{2}+1\\ \displaystyle &=4(\sqrt{2})^{2}+1\\ \displaystyle &=4(2)+1\\ \displaystyle &=9\\ \end{align} {/eq}

Step 3. Rewrite the integral in terms of the variable {eq}u {/eq} (substitute {eq}u {/eq} for {eq}1+4x^{2} {/eq}, {eq}\frac{1}{2}du {/eq} for {eq}4x\:dx {/eq} and {eq}u=1 {/eq} and {eq}u=9 {/eq} for the integral's boundaries).

{eq}\begin{align} \displaystyle \int_{0}^{\sqrt{2}}\frac{4x}{\sqrt{1 + 4x^2}}dx\displaystyle &=\int_{1}^{9}\frac{\frac{1}{2}}{\sqrt{u}}du\\ \displaystyle &=\int_{1}^{9}\frac{1}{2\sqrt{u}}du\\ \end{align} {/eq}

Step 4. Integrate {eq}\frac{1}{2\sqrt{u}} {/eq} with respect to {eq}u {/eq} from {eq}u=1 {/eq} to {eq}u=9 {/eq}.

{eq}\begin{align} \displaystyle \int_{1}^{9}\frac{1}{2\sqrt{u}}du=\displaystyle &=\left[\sqrt{u}\right]_{u=1}^{u=9}\\ \displaystyle &=\sqrt{9}-\sqrt{1}\\ \displaystyle &=3-1\\ \displaystyle &=2\\ \end{align} {/eq}

Therefore {eq}\displaystyle \int_{0}^{\sqrt{2}}\frac{4x}{\sqrt{1 + 4x^2}}dx=2 {/eq}.