√limit-22 x^2019 dx is equals to

1.	√limit-22 x^2019 dx is equals to
Answer» $\implies\displaystyle\int\limits^{ \sf2}_{ \sf - 2} \sf{x}^{2019} dx$ We KNOW that :- $\sf\implies\displaystyle\int\limits^{ \sf \: n}_{ \sf \: m} \sf \: f(x)dx = g(x) {\bigg]}^{n}_{ m} = g(n) - g(m)$ $\implies\displaystyle\int\limits^{ \sf2}_{ \sf - 2} \sf{x}^{2019} dx = \dfrac{1}{2020} { \large \: x}^{2020} {\bigg]}^{2}_{ - 2}$ Now PUT the value of UPPER limit and LOWER limit. $\sf\implies \dfrac{1}{2020} \: { \large \: x}^{2020} {\bigg]}^{2}_{ - 2} = \dfrac{1}{2020} \bigg[ {2}^{2020} - { ( - 2)}^{2020} \bigg]$ $\sf\implies \dfrac{1}{2020} \bigg[ {2}^{2020} - \bigg( { {( - 1)}^{2020} \times ( 2)}^{2020} \bigg) \bigg]$ $\sf\implies \dfrac{1}{2020} \bigg[ {2}^{2020} - \bigg( { 1\times ( 2)}^{2020} \bigg) \bigg]$ $\sf\implies \dfrac{1}{2020} \bigg({2}^{2020} - {2}^{2020} \bigg)$ $\sf\implies \dfrac{1}{2020} \times 0$ $\sf\implies 0$ Answer : OPTION (a)0 is correct.

Answer»

$\implies\displaystyle\int\limits^{ \sf2}_{ \sf - 2} \sf{x}^{2019} dx$