ExpandProcessLogUse the common integral rule: \n \( \mathrm { \displaystyle\int u ^ { n } du = \displaystyle\frac { 1 } { n + 1 } u ^ { n + 1 } + c , n \neq - 1 } \)Str\( \mathrm { \displaystyle\int\limits_{0}^{3} x^{2}\, dx = \displaystyle\frac{x^{3}}{3} \Bigg|_{ 0 } ^ { 3 } } \)ProcessLogOrganize the integrals according to upper and lower limits. \n Upper limit: \n \( \mathrm { x = 3 \Longrightarrow \displaystyle\frac{3^{3}}{3} } \) \n Lower limit: \n \( \mathrm { x = 0 \Longrightarrow \displaystyle\frac{0^{3}}{3} } \) \n The value of the upper limit is subtracted from the value of the lower limit.Str\( \mathrm { \displaystyle\frac{3^{3}}{3} - \displaystyle\frac{0^{3}}{3} } \)ExpandProcessLogSimplify the exponents: \n \( \mathrm { 0^{3} = 0 } \)Str\( \mathrm { \displaystyle\frac{3^{3}}{3} - 0 } \)ProcessLogSimplify the exponents: \n \( \mathrm { 3^{3} = 27 } \)Str\( \mathrm { \displaystyle\frac{27}{3} - 0 } \)ProcessLog\( \mathrm { \displaystyle\frac{27}{3} - 0 = 9 } \)Str\( \mathrm { 9 } \)\( \mathrm { 9 } \)Simplify\( \mathrm { \displaystyle\int\limits_{0}^{3} x^{2}\, dx } \)\( \mathrm { 9 } \)Evaluate the integralProcessLogDecimal form: \[ 9.0 \]

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Solution: Find the area under $y=x^2$ from $x=0$ to $x=3$.

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