# Use an appropriate binomial series and integration of power series to estimate the integral value...

## Question:

Use an appropriate binomial series and integration of power series to estimate the integral value to 3 decimal place accuracy.

{eq}\displaystyle \int_{0}^{0.6}\sqrt{1 + x^{3}}\,dx {/eq}

## Estimating the Integral Using Binomial Series:

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. The Binomial Theorem is a quick way of expanding a binomial expression that has been raised to some (generally inconveniently large) power. And the formula for the binomial series is, {eq}\left ( 1+x \right )^n=1+nx+\frac{n\left ( n-1 \right )}{2!}x^2+\frac{n\left ( n-1 \right )\left ( n-2 \right )}{3!}x^{3}+... {/eq} By using this series we need to apply the given integral function to find the solution.

From the given information, the integral function is {eq}\int_{0}^{0.6}\sqrt{1+x^{3}}dx. {/eq}

And the general form of binomial series is, {eq}\left ( 1+x \right )^n=1+nx+\frac{n\left ( n-1 \right )}{2!}x^2+\frac{n\left ( n-1 \right )\left ( n-2 \right )}{3!}x^{3}+...... {/eq}

We can write, {eq}\sqrt{1+x^{3}} {/eq} as {eq}\left ( 1+x^{3} \right )^{\frac{1}{2}} {/eq}

So, the value of {eq}x {/eq} is {eq}x^{3} {/eq} and the value of {eq}n {/eq} is {eq}\frac{1}{2} {/eq}

Now, substitute the values in the above series,

{eq}1+\frac{1}{2}x^3+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}\left(x^3\right)^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}\left(x^3\right)^3... \\ =1+\frac{1\cdot \:x^3}{2}+\left ( -\frac{1}{4\cdot \:2!}\left(x^3\right)^2 \right )+\frac{3}{8\cdot \:3!}\left(x^3\right)^3... \\ =1+\frac{x^3}{2}-\frac{x^6}{8}+\frac{x^9}{16}... {/eq}

Now,

{eq}\int \sqrt{1+x^{3}}dx=\int \left ( 1+x^{3} \right )^{\frac{1}{2}}dx=x+\frac{x^{4}}{8}-\frac{x^{7}}{56}+\frac{x^{10}}{160}... \\ \int_{0}^{0.6}\sqrt{1+x^{3}}dx=\left [ x+\frac{x^4}{8}-\frac{x^7}{56}+\frac{x^{10}}{160} \right ]_{0}^{0.6} \\ =0.6+\frac{0.6^4}{8}-\frac{0.6^7}{56} \\ =0.6+0.0162-0.0004998857 \\ =0.6157001143 {/eq}

{eq}0.6157001143+\frac{\left ( 0.6 \right )^{10}}{160} \\ =0.6157379056 {/eq}

Hence the integral value should be three decimal place accuracy.

So,

{eq}\int_{0}^{0.6}\sqrt{1+x^{3}}dx=0.616. {/eq}