Evaluate the given integral by using three terms of the appropriate series. Round to four decimal...

Question:

Evaluate the given integral by using three terms of the appropriate series. Round to four decimal places. {eq}\int^{0.3}_0 \sqrt[3]{ 1+ 3x^2} dx {/eq}

Binomial Rule for Definite Integral:

For the simplification of the definite integral, we'll simplify the given cube root function using the binomial theorem by computing the first three non-zero terms of the function.

For that, we need to rewrite the given function as the function {eq}(1+y)^m {/eq} with positive exponent {eq}m {/eq} and replace the varaible {eq}y {/eq} and exponent of the fomrula for the first three terms of the expansion.

• {eq}\displaystyle (1+y)^m=1+my+\frac{m(m-1)}{2!}y^2+\dots {/eq}

The given definite integral with cube root function is:

{eq}I=\displaystyle \int^{0.3}_0 \sqrt[3]{ 1+ 3x^2}\ dx {/eq}

Rewriting the cube root function with fractional exponent, we get:

{eq}\displaystyle \sqrt[3]{ 1+ 3x^2}=( 1+ 3x^2)^{\frac{1}{3}} {/eq}

According to the standard form of binomial function {eq}(1+x)^m {/eq}, we have:

{eq}\displaystyle x=3x^2\\ \displaystyle m=\frac{1}{3} {/eq}

For the first three terms of binomial expansion of the function {eq}( 1+ 3x^2)^{\frac{1}{3}} {/eq}, we'll substitute the required values in the fomrula of binomial expansion.

{eq}\begin{align*} \displaystyle ( 1+ 3x^2)^{\frac{1}{3}}&=1+\frac{1}{3}(3x^2)+\frac{\frac{1}{3}\left ( \frac{1}{3}-1 \right )}{2!}(3x^2)^2+\dots\\ &=1+x^2+\frac{\frac{1}{3}\left ( -\frac{2}{3} \right )}{2}(9x^4)+\dots&\because (x^m)^n=x^{mn}\\ &=1+x^2-x^4+\dots \end{align*} {/eq}

Using the above expression in the given definite integral, we get:

{eq}I=\displaystyle \int^{0.3}_0 (1+x^2-x^4)\ dx {/eq}

Simplifying the above expression of the definite integral using the exponent rule and fundamental rule of integrals, we get:

{eq}\begin{align*} \displaystyle I&=\displaystyle \int^{0.3}_0 1\ dx+\int^{0.3}_0 x^2\ dx-\int^{0.3}_0x^4\ dx\\ &=\displaystyle \left [ x \right ]^{0.3}_0 +\left [ \frac{x^{2+1}}{2+1} \right ]^{0.3}_0-\left [ \frac{x^{4+1}}{4+1} \right ]^{0.3}_0\\ &=\displaystyle \left ( 0.3-0 \right) +\frac{1}{3}\left [ x^{3} \right ]^{0.3}_0-\frac{1}{5}\left [ x^{5}\right ]^{0.3}_0\\ &=\displaystyle 0.3+\frac{1}{3}\left ( (0.3)^{3}-0 \right )-\frac{1}{5}\left ( (0.3)^{5}-0\right )\\ &=\displaystyle 0.3+\frac{0.027}{3}-\frac{0.00243}{5}\\ &=\displaystyle 0.3+0.009-0.000486\\ &\approx 0.3085 \end{align*} {/eq}