Math, asked by wwwmadelinechetri, 11 months ago

int int ( x²-1) dx x =2 to 1

Answers

Answered by Anonymous

31

Answer:

$\large\bold\red{-\frac{4}{3}}$

Step-by-step explanation:

$\displaystyle \int _{2}^{1}( {x}^{2} - 1)dx$

Further simplifying,

We get,

$= \displaystyle \int _{2}^{1} {x}^{2} dx - \displaystyle \int _{2}^{1}dx \\ \\ = </p><p>\bigg[\dfrac{{x}^{3}}{3}-x\bigg]^{1} _2\\ \\ = (\frac{ {1}^{3} }{3} - 1) - ( \frac{ {2}^{3} }{3} - 2) \\ \\ = \frac{1}{3} - 1 - \frac{8}{3} + 2 \\ \\ = 1 + \frac{ 1- 8}{3} \\ \\ = 1 - \frac{7}{3} \\ \\ = \frac{3 - 7}{3} \\ \\ = - \frac{4}{3}$

Concepts used :-

$\bold{\displaystyle\int{x}^{n}=\frac{{x}^{n+1}}{n+1}}$

Answered by lucky997761

4

$<body bgcolor="black"></h2><h2><font color="white">$

☆AnswEr:

$\large\bold\red{-\frac{4}{3}}$

☆Step-by-step explanation:

$\displaystyle \int _{2}^{1}( {x}^{2} - 1)dx$

➡Further simplifying,

➡We get,

$\begin{lgathered}= \displaystyle \int _{2}^{1} {x}^{2} dx - \displaystyle \int _{2}^{1}dx \\ \\ = \bigg[\dfrac{{x}^{3}}{3}-x\bigg]^{1} _2\\ \\ = (\frac{ {1}^{3} }{3} - 1) - ( \frac{ {2}^{3} }{3} - 2) \\ \\ = \frac{1}{3} - 1 - \frac{8}{3} + 2 \\ \\ = 1 + \frac{ 1- 8}{3} \\ \\ = 1 - \frac{7}{3} \\ \\ = \frac{3 - 7}{3} \\ \\ = - \frac{4}{3}\end{lgathered}$

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