Answer:
Therefore, area under the curve is [tex]\frac{1}{4}[/tex]
Step-by-step explanation:
We have to find the area under curve y = x³ from 0 to 1 as limit.
Since Area 'A' = [tex]\lim_{n \to \infty} \sum_{i=1}^{n}f(x_{i})\triangle x[/tex]
The given function is f(x) = x³
Since [tex]x_{i}=a+\triangle x.i[/tex]
Here a = 0 and [tex]\triangle x=\frac{1-0}{n}=\frac{1}{n}[/tex]
[tex]f(x_{i})=(\frac{i}{n})^{3}[/tex]
Now A = [tex]\lim_{n \to \infty} \sum_{i=1}^{n}f(x_{i})\triangle x= \lim_{n \to \infty}\sum_{i=1}^{n}(\frac{i}{n})^{3}(\frac{1}{n})[/tex]
[tex]= \lim_{n \to \infty}\frac{1}{n^{4}}\sum_{i=1}^{n}i^{3}[/tex]
[tex]= \lim_{n \to \infty}\frac{1}{n^{4}}( \frac{n(n+1)}{2})^{2}[/tex] Since 1³ + 2³ + 3³..............n³ = [tex][\frac{n(n+1)}{2}]^{2}[/tex]
[tex]= \lim_{n \to \infty}\frac{n^{2}(n+1)^{2}}{4n^{4}}[/tex]
[tex]= \lim_{n \to \infty}\frac{1}{4}(1+\frac{1}{n})^{2}[/tex]
[tex]=\frac{1}{4}(1+0)[/tex]
[tex]=\frac{1}{4}[/tex]
Therefore, area under the curve is [tex]\frac{1}{4}[/tex]
