Question

integration 0 to 2 X square + x dx as the limit of sum ​

Answer 1

Step-by-step explanation:

plzz see the attachment

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore \int\limits_{0}^{2}(x^{2}+x)dx=\frac{14}{3}}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{Given : } \\ \implies \int \limits_{0} ^{2}( {x}^{2} + x)dx \\ \\ \underline \bold{To \:Find: } \\ \implies \int \limits_{0} ^{2}( {x}^{2} + x)dx = ?$

• According to given question :

$\implies y= {x}^{2} + x \\ \\ \implies \frac{dy}{dx} = {x}^{2} + x \\ \\ \implies dy = ({x}^{2} + x)dx \\ \\ \bold{Both \: side \: integrating : } \\ \implies \int dy = \int( {x}^{2} + x)dx \\ \\ \bold{Putting \: limits }\\ \implies \int \limits_{0}^{y} dy = \int \limits_{0}^{2} ( {x}^{2} + x)dx \\ \\ \implies y - 0 = \int \limits_{0}^{2} ( \frac{ {x}^{3} }{3} + \frac{ {x}^{2} }{2} )dx \\ \\ \implies y = (\frac{ {2}^{3} }{3} + \frac{ {2}^{2} }{2}) - (0) \\ \\ \implies y = \frac{8}{3} + \frac{2}{1} \\ \\ \implies y = \frac{8 + 6}{3} \\ \\ \bold{\implies y = \frac{14}{3} }$