Question

$\displaystyle \sf\int_{0}^{2} \int_{x }^{ {x}^{2} } x \cdot {y}^{2} - xdydx$​

Answer 1

Step-by-step explanation:

= 2⁸/24 - 2⁴/4 - 2⁶/18 + 2³/3

= 2³ ( 2⁵/24- 2/4 - 2³/18 + 1/3]

= 8 ( 32(3) - 2(18) -8(4) + 1(24))/72

= ( 96-36-32+24) /9

= 42/9 = 14/3+C

Answer 2

$\: \: \: \: \: \: \: \bf\red{Answer:-}$

$\longmapsto\rm\displaystyle \sf\int_{0}^{2}( \frac{x}{3} ( {y}^{3} )^{ {x}^{3} } _{x} - x(y) ^{ {x}^{2} } _{n})dydn$

$\rm \longmapsto \displaystyle{ \rm\int\limits_{0}^{2}( \frac{x}{3} ( {x}^{6} - {x}^{3}) - x( {x}^{2} - x))dydn}$

$\longmapsto\rm \displaystyle \int\limits_{0}^{2} \frac{1}{3} ( \rm{x}^{7} - {x}^{4} ) - {x}^{3} + {x}^{2} )dydn$

$\rm \longmapsto \displaystyle\cancel{\int\limits}_{ \cancel{0}}^{ \cancel{2}} \frac{1}{3} ( \rm \: x \frac{8}{8} - \dfrac{ {x}^{5} }{5} ) ^{2} - \frac{1}{4} ( {x}^{4} ) ^{2} _{0} + \frac{1}{3} ( {x}^{3} )^{2} _{0}$

$\longmapsto \rm \frac{1}{3} (32 - \frac{32}{5} ) - \frac{1}{ \cancel4} ( \cancel{ - 16} \frac{4}{}\: ) + \frac{8}{3}$

$\longmapsto \rm \frac{1}{3} ( \frac{128}{5} ) = \frac{128}{15} + \frac{8}{3} - \frac{4}{1}$

$\longmapsto \rm \frac{128 + 40 - 60}{15}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \fbox{[460 - 160 = 300]}$

$\longmapsto \rm \frac{300}{15}$

$\longmapsto \rm20 \times 2$

$\longmapsto \rm \fbox \red{200}$

$▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬$

@Shivam

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