Question-171253

Question Number 171253 by vonem1 last updated on 11/Jun/22

Answered by haladu last updated on 11/Jun/22

8∫_1 ^4 t^(−(1/2)) dt −12 ∫ t^(3/2) dt 8 (t^(−(1/2) +1) /(−(1/2) +1)) −12 (t^((3/2) +1) /((3/2) +1)) + C 8 (t^(1/2) /(1/2)) −12 (t^(5/2) /(5/2)) + C ∣_1 ^4 16 ( 4^(1/2) −1^(1/2) ) −((24)/5) ( 4^(3/2) −1^(3/2) ) = 16 ( 2−1 ) −((24)/5) ( 8 −1 ) = 16 −((24×7)/5) = ((16×5−24×7)/5) = ((−88)/5)

$$\:\mathrm{8}\int_{\mathrm{1}} ^{\mathrm{4}} \:\boldsymbol{\mathrm{t}}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:\:\boldsymbol{\mathrm{dt}}\:−\mathrm{12}\:\:\int\:\boldsymbol{\mathrm{t}}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\boldsymbol{\mathrm{dt}} \\ $$$$\:\:\: \\ $$$$\:\:\mathrm{8}\:\:\:\frac{\boldsymbol{\mathrm{t}}^{−\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}} }{−\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}}\:\:−\mathrm{12}\:\:\:\frac{\boldsymbol{\mathrm{t}}^{\frac{\mathrm{3}}{\mathrm{2}}\:+\mathrm{1}} }{\frac{\mathrm{3}}{\mathrm{2}}\:+\mathrm{1}}\:\:+\:\boldsymbol{\mathrm{C}} \\ $$$$\:\:\: \\ $$$$\:\:\mathrm{8}\:\:\:\frac{\boldsymbol{\mathrm{t}}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\frac{\mathrm{1}}{\mathrm{2}}}\:\:−\mathrm{12}\:\:\frac{\boldsymbol{\mathrm{t}}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\frac{\mathrm{5}}{\mathrm{2}}}\:+\:\:\boldsymbol{\mathrm{C}}\:\underset{\mathrm{1}} {\overset{\mathrm{4}} {\mid}} \\ $$$$\:\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\mathrm{16}\:\left(\:\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \:−\mathrm{1}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)\:−\frac{\mathrm{24}}{\mathrm{5}}\:\left(\:\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\mathrm{1}^{\frac{\mathrm{3}}{\mathrm{2}}} \right) \\ $$$$\:\: \\ $$$$\:=\:\mathrm{16}\:\left(\:\mathrm{2}−\mathrm{1}\:\right)\:−\frac{\mathrm{24}}{\mathrm{5}}\:\left(\:\mathrm{8}\:−\mathrm{1}\:\right) \\ $$$$\:\: \\ $$$$=\:\:\:\mathrm{16}\:\:−\frac{\mathrm{24}×\mathrm{7}}{\mathrm{5}} \\ $$$$\:\: \\ $$$$\:=\:\:\frac{\mathrm{16}×\mathrm{5}−\mathrm{24}×\mathrm{7}}{\mathrm{5}}\:=\:\:\frac{−\mathrm{88}}{\mathrm{5}} \\ $$

Answered by thfchristopher last updated on 11/Jun/22

=∫_1 ^4 (8t^(−(1/2)) −12t^(3/2) )dt =[16t^(1/2) −((24)/5)t^(5/2) ]_1 ^4 =32−((768)/5)−16+((24)/5) =−((664)/5)

$$=\int_{\mathrm{1}} ^{\mathrm{4}} \left(\mathrm{8}{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{12}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \right){dt} \\ $$$$=\left[\mathrm{16}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{24}}{\mathrm{5}}{t}^{\frac{\mathrm{5}}{\mathrm{2}}} \right]_{\mathrm{1}} ^{\mathrm{4}} \\ $$$$=\mathrm{32}−\frac{\mathrm{768}}{\mathrm{5}}−\mathrm{16}+\frac{\mathrm{24}}{\mathrm{5}} \\ $$$$=−\frac{\mathrm{664}}{\mathrm{5}} \\ $$

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