Question

integration of 3t2-4t3). 0_1

Answer 1

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$= \int _{0} ^{1} (3 {t}^{2} - 4 {t}^{3})dt \\ \\ = \int _{0} ^{1} 3 {t}^{2}dt - \int _{0} ^{1} 4 {t}^{3}dt \\ \\ = 3 \int _{0} ^{1} {t}^{2}dt - 4\int _{0} ^{1} {t}^{3}dt \\ \\ = 3 \times [ \frac{ {t}^{3} }{3}] _{0} ^{1} - 4 \times [\frac{ {t}^{4} }{4}] _{0} ^{1} \\ \\ = [{t}^{3}] _{0} ^{1} - [{t}^{4} ]_{0} ^{1} \\ \\ = [ {1}^{3} - {0}^{3} ] - [ {1}^{4} - {0}^{4} ] \\ \\ = 1 - 1 \\ \\ = 0$
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