ExpandProcessLogApply the sum rule of integration: \n \( \mathrm { \displaystyle\int ( f ( u ) + g ( u ) ) du = \displaystyle\int f ( u ) du + \displaystyle\int g ( u ) du} \)ExpandStr1. STEP:Str\( \mathrm { \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} \left(- 3 \cdot y^{2}\right)\, dy } \)ProcessLogApply the integral constant multiple rule: \n \( \mathrm { \displaystyle\int a \cdot f ( u ) du = a \displaystyle\int f ( u ) du } \)Str\( \mathrm { - 3 \cdot \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} y^{2}\, dy } \)ProcessLogUse the common integral rule: \n \( \mathrm { \displaystyle\int u ^ { n } du = \displaystyle\frac { 1 } { n + 1 } u ^ { n + 1 } + c , n \neq - 1 } \)Str\( \mathrm { - 3 \cdot \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} y^{2}\, dy = - 3 \cdot \displaystyle\frac{y^{3}}{3} \Bigg|_{ 0 } ^ { \displaystyle\frac{1}{6} } } \)ProcessLogOrganize the integrals according to upper and lower limits. \n Upper limit: \n \( \mathrm { y = \displaystyle\frac{1}{6} \Longrightarrow \left(- \displaystyle\frac{3}{3}\right) \cdot \left(\displaystyle\frac{1}{6}\right)^{3} } \) \n Lower limit: \n \( \mathrm { y = 0 \Longrightarrow \left(- \displaystyle\frac{3}{3}\right) \cdot 0^{3} } \) \n The value of the upper limit is subtracted from the value of the lower limit.Str\( \mathrm { \left(- \displaystyle\frac{3}{3}\right) \cdot \left(\displaystyle\frac{1}{6}\right)^{3} + \displaystyle\frac{3}{3} \cdot 0^{3} } \)ExpandProcessLogSimplify the exponents: \n \( \mathrm { 0^{3} = 0 } \)Str\( \mathrm { \left(- \displaystyle\frac{3}{3}\right) \cdot \left(\displaystyle\frac{1}{6}\right)^{3} + \displaystyle\frac{3 \cdot 0}{3} } \)ProcessLog\( \mathrm { \left(- \displaystyle\frac{3}{3}\right) \cdot \left(\displaystyle\frac{1}{6}\right)^{3} + 0 = - \displaystyle\frac{1}{216} } \)Str\( \mathrm { - \displaystyle\frac{1}{216} } \)\( \mathrm { - \displaystyle\frac{1}{216} } \)SimplifyProcessLogSubstitute backStr\( \mathrm { - \displaystyle\frac{1}{216} + \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} \displaystyle\frac{y}{2}\, dy } \)Str2. STEP:Str\( \mathrm { \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} \displaystyle\frac{y}{2}\, dy } \)ProcessLogApply the integral constant multiple rule: \n \( \mathrm { \displaystyle\int a \cdot f ( u ) du = a \displaystyle\int f ( u ) du } \)Str\( \mathrm { \displaystyle\frac{\displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} y\, dy}{2} } \)ProcessLogUse the common integral rule: \n \( \mathrm { \displaystyle\int u ^ { n } du = \displaystyle\frac { 1 } { n + 1 } u ^ { n + 1 } + c , n \neq - 1 } \)Str\( \mathrm { \displaystyle\frac{\displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} y\, dy}{2} = \displaystyle\frac{\displaystyle\frac{1}{2} \cdot y^{2}}{2} \Bigg|_{ 0 } ^ { \displaystyle\frac{1}{6} } } \)ProcessLogOrganize the integrals according to upper and lower limits. \n Upper limit: \n \( \mathrm { y = \displaystyle\frac{1}{6} \Longrightarrow \displaystyle\frac{1}{2 \cdot 2} \cdot 1 \cdot \left(\displaystyle\frac{1}{6}\right)^{2} } \) \n Lower limit: \n \( \mathrm { y = 0 \Longrightarrow \displaystyle\frac{1}{2 \cdot 2} \cdot 1 \cdot 0^{2} } \) \n The value of the upper limit is subtracted from the value of the lower limit.Str\( \mathrm { \displaystyle\frac{1}{2 \cdot 2} \cdot 1 \cdot \left(\displaystyle\frac{1}{6}\right)^{2} - \displaystyle\frac{0^{2}}{2 \cdot 2} } \)ExpandProcessLogSimplify the exponents: \n \( \mathrm { 0^{2} = 0 } \)Str\( \mathrm { \displaystyle\frac{1}{2 \cdot 2} \cdot 1 \cdot \left(\displaystyle\frac{1}{6}\right)^{2} + \displaystyle\frac{1}{2 \cdot 2} \cdot 0 } \)ProcessLog\( \mathrm { \displaystyle\frac{1}{144} + 0 = \displaystyle\frac{1}{144} } \)Str\( \mathrm { \displaystyle\frac{1}{144} } \)\( \mathrm { \displaystyle\frac{1}{144} } \)SimplifyProcessLogSubstitute backStr\( \mathrm { - \displaystyle\frac{1}{216} + \displaystyle\frac{1}{144} } \)\( \mathrm { \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} \left(- 3 \cdot y^{2}\right)\, dy + \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} \displaystyle\frac{y}{2}\, dy } \)\( \mathrm { - \displaystyle\frac{1}{216} + \displaystyle\frac{1}{144} } \)Evaluate the integralsExpandProcessLog\( \mathrm { - \displaystyle\frac{1}{216} + \displaystyle\frac{1}{144} = \displaystyle\frac{1}{432} } \)Str\( \mathrm { \displaystyle\frac{1}{432} } \)\( \mathrm { \displaystyle\frac{1}{432} } \)Simplify\( \mathrm { \displaystyle\int\limits_{0}^{\displaystyle\frac{1}{6}} \left(- 3 \cdot y^{2} + \displaystyle\frac{y}{2}\right)\, dy } \)\( \mathrm { \displaystyle\frac{1}{432} } \)Evaluate the integralProcessLogDecimal form: \[ 0.00231481481481481 \]

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