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\( \mathrm { \displaystyle\int\limits_{0}^{1} x\, dx } \)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\( \mathrm { \displaystyle\int\limits_{0}^{1} x\, dx = \displaystyle\frac{x^{2}}{2} \Bigg|_{ 0 } ^ { 1 } } \)ProcessLogOrganize the integrals according to upper and lower limits. \n Upper limit: \n \( \mathrm { x = 1 \Longrightarrow \displaystyle\frac{1^{2}}{2} } \) \n Lower limit: \n \( \mathrm { x = 0 \Longrightarrow \displaystyle\frac{0^{2}}{2} } \) \n The value of the upper limit is subtracted from the value of the lower limit.Str\( \mathrm { \displaystyle\frac{1^{2}}{2} - \displaystyle\frac{0^{2}}{2} } \)ExpandProcessLogSimplify the exponents: \n \( \mathrm { 0^{2} = 0 } \)Str\( \mathrm { \displaystyle\frac{1^{2}}{2} + \displaystyle\frac{0}{2} } \)ProcessLogSimplify the exponents: \n \( \mathrm { 1^{2} = 1 } \)Str\( \mathrm { \displaystyle\frac{1}{2} + \displaystyle\frac{0}{2} } \)ProcessLog\( \mathrm { \displaystyle\frac{1}{2} + \displaystyle\frac{0}{2} = \displaystyle\frac{1}{2} } \)Str\( \mathrm { \displaystyle\frac{1}{2} } \)\( \mathrm { \displaystyle\frac{1}{2} } \)SimplifyProcessLogSubstitute backStr\( \mathrm { \displaystyle\frac{1}{2} + \displaystyle\int\limits_{0}^{1} \left(- x^{2}\right)\, dx } \)Str2. STEP:Str\( \mathrm { \displaystyle\int\limits_{0}^{1} \left(- x^{2}\right)\, dx } \)ProcessLogApply the integral constant multiple rule: \n \( \mathrm { \displaystyle\int a \cdot f ( u ) du = a \displaystyle\int f ( u ) du } \)Str\( \mathrm { - \displaystyle\int\limits_{0}^{1} x^{2}\, dx } \)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\( \mathrm { - \displaystyle\int\limits_{0}^{1} x^{2}\, dx = - \displaystyle\frac{x^{3}}{3} \Bigg|_{ 0 } ^ { 1 } } \)ProcessLogOrganize the integrals according to upper and lower limits. \n Upper limit: \n \( \mathrm { x = 1 \Longrightarrow - \displaystyle\frac{1^{3}}{3} } \) \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\( \mathrm { - \displaystyle\frac{1^{3}}{3} + \displaystyle\frac{0^{3}}{3} } \)ExpandProcessLogSimplify the exponents: \n \( \mathrm { 0^{3} = 0 } \)Str\( \mathrm { - \displaystyle\frac{1^{3}}{3} - 0 } \)ProcessLogSimplify the exponents: \n \( \mathrm { 1^{3} = 1 } \)Str\( \mathrm { - \displaystyle\frac{1}{3} - 0 } \)ProcessLog\( \mathrm { - \displaystyle\frac{1}{3} - 0 = - \displaystyle\frac{1}{3} } \)Str\( \mathrm { - \displaystyle\frac{1}{3} } \)\( \mathrm { - \displaystyle\frac{1}{3} } \)SimplifyProcessLogSubstitute backStr\( \mathrm { \displaystyle\frac{1}{2} - \displaystyle\frac{1}{3} } \)\( \mathrm { \displaystyle\int\limits_{0}^{1} x\, dx + \displaystyle\int\limits_{0}^{1} \left(- x^{2}\right)\, dx } \)\( \mathrm { \displaystyle\frac{1}{2} - \displaystyle\frac{1}{3} } \)Evaluate the integralsExpandProcessLog\( \mathrm { \displaystyle\frac{1}{2} - \displaystyle\frac{1}{3} = \displaystyle\frac{1}{6} } \)Str\( \mathrm { \displaystyle\frac{1}{6} } \)\( \mathrm { \displaystyle\frac{1}{6} } \)Simplify\( \mathrm { \displaystyle\int\limits_{0}^{1} \left(x - x^{2}\right)\, dx } \)\( \mathrm { \displaystyle\frac{1}{6} } \)Evaluate the integralProcessLogDecimal form: \[ 0.166666666666667 \]

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