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

Answered by haladu last updated on 11/Jun/22

$$8{\int }_1^4{\mathbf{t}}^{-\frac{1}{2}}\mathbf{dt}-12\int {\mathbf{t}}^{\frac{3}{2}}\mathbf{dt}$$$$$$$$8\frac{{\mathbf{t}}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-12\frac{{\mathbf{t}}^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\mathbf{C}$$$$$$$$8\frac{{\mathbf{t}}^{\frac{1}{2}}}{\frac{1}{2}}-12\frac{{\mathbf{t}}^{\frac{5}{2}}}{\frac{5}{2}}+\mathbf{C}\underset{1}{\stackrel{4}{\mid }}$$$$$$$$$$$$16(4^{\frac{1}{2}}-1^{\frac{1}{2}})-\frac{24}{5}(4^{\frac{3}{2}}-1^{\frac{3}{2}})$$$$$$$$=16(2-1)-\frac{24}{5}(8-1)$$$$$$$$=16-\frac{24\times 7}{5}$$$$$$$$=\frac{16\times 5-24\times 7}{5}=\frac{-88}{5}$$

Answered by thfchristopher last updated on 11/Jun/22

$$={\int }_1^4(8t^{-\frac{1}{2}}-12t^{\frac{3}{2}})dt$$$$={[16t^{\frac{1}{2}}-\frac{24}{5}{t}^{\frac{5}{2}}]}_1^4$$$$=32-\frac{768}{5}-16+\frac{24}{5}$$$$=-\frac{664}{5}$$

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