# Calculates \int^{2}_{0} 4x^{3} dx by using the formal definition of the define integral...

## Question:

Calculates {eq}\int^{2}_{0} 4x^{3} dx{/eq} by using the formal definition of the define integral

{eq}\int^{2}{0} f(x)dx=\lim_{x \rightarrow \infty} \left[ \sum^{n}_{k=1} f(xi) \Delta x \right]{/eq}

## Integration:

Integration is used to find the area under the curve and to find the volume.

It is a reverse process of differentiation.

Derivative indicates rate of change. It is used to find the slope of the curve.

We know that {eq}\int_{a}^{b}f(x)\,dx=\lim_{n\rightarrow \infty }\sum_{k=1}^{n}f(x_k)\Delta x {/eq} where {eq}x_k=a+k\Delta x {/eq} and {eq}\Delta x=\frac{b-a}{n} {/eq}

Take {eq}a=0\,,\,b=2 {/eq}

{eq}\Delta x=\frac{2-0}{n}=\frac{2}{n} {/eq} and {eq}x_k=0+\frac{2k}{n}=\frac{2k}{n} {/eq}

Therefore,

{eq}\int_{0}^{2}f(x)\,dx=\lim_{n\rightarrow \infty }\sum_{k=1}^{n}f\left ( \frac{2k}{n} \right )\frac{2}{n}\\ =\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{2}{n}(4)\left ( \frac{2k}{n} \right )^3\\ =\lim_{n\rightarrow \infty }\sum_{k=1}^{n}64\left ( \frac{k^3}{n^4} \right )\\ =64\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{1}{n^4}k^3\,\,\left \{ \because \sum_{k=1}^{n}k^3=\left [ \frac{n(n+1)}{2} \right ]^2 \right \}\\ =64\lim_{n\rightarrow \infty }\left [ \frac{1}{n^4}\left ( \frac{n(n+1)}{2} \right )^2 \right ]\\ =16\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^2\\ =16(1+0) {/eq} 