anh puvid ~ chị hangibich giúp em ạ !!!!!! $\to \displaystyle\int\dfrac{x^2}{1}dx$
2 câu trả lời
Đáp án:
$\dfrac{x^3}{3} + C$
Giải thích các bước giải:
$\quad \displaystyle\int \dfrac{x^2}{1}dx$
$= \displaystyle\int x^2dx$
$= \dfrac{x^{2+1}}{2+1} + C$
$=\dfrac{x^3}{3} + C$
Đáp án:
Ta có :
$\displaystyle\int\dfrac{x^2}{1}dx$
$= \displaystyle\int x^2dx$
$ = \displaystyle\int xxdx$
`= 1/2x^3 - 1/2` $\displaystyle\int xxdx$
`= 1/2x^3 - 1/4 x^3 + 1/4` $\displaystyle\int xxdx$
`= 1/2x^3 - 1/4x^3 + 1/8x^3 - 1/8` $\displaystyle\int xxdx$
`...`
`= \sum_{n=1}^{∞} \frac{(-1)^{n+1}}{2^n}x^3+C`
`= x^3 \sum_{n=1}^{∞} \frac{(-1)^{n+1}}{2^n}+C`
`= x^3 ( \sum_{n=0}^{∞} \frac{(-1)^{2n}}{2^{2n+1}}+ \sum_{n=1}^{∞} \frac{(-1)^{2n+1}}{2^{2n}} ) + C`
`= x^3 (\sum_{n=0}^{∞} \frac{1}{2^{2n+1}}- \sum_{n=1}^{∞} \frac{1}{2^{2n}}) + C`
`= x^3 (\frac{1}{2}\sum_{n=0}^{∞} \frac{1}{4^n}- \frac{1}{4}\sum_{n=0}^{∞} \frac{1}{4^n}) + C`
`= x^3 (1/4\sum_{n=0}^{∞} \frac{1}{4^n}) + C`
`= x^3 (1/4 1/(1 - 1/4)) + C`
`= 1/3 x^3 + C`