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Question Number 66071 by AnjanDey last updated on 08/Aug/19

$$\mathrm{Evaluate}\mathrm{the}\mathrm{integral}\mathrm{as}\mathrm{a}\mathrm{limit}\mathrm{of}\mathrm{sums}:$$$$1.{\int }_1^3(e^{2-3x}+x^2+1)dx$$

Answered by meme last updated on 08/Aug/19

$$$$$$={\int }_1^3{e}^{2-3x}dx+{\int }_1^3{x}^{2}dx+{\int }_1^{3}dx$$$$=-3{\int }_1^3-\frac{1}{3}{e}^{2-3x}dx+{\int }_1^3{x}^{2}dx+{\int }_1^{3}dx$$$$=-3{\left[e^{2-3x}\right]}_1^3+\frac{1}{3}{\left[x^3\right]}_1^3+{\left[x\right]}_1^3$$$$=-3e^{-7}+3e^{-1}+9-\frac{1}{3}+3-1$$$$=3(e^{-1}-e^{-7})+\frac{32}{3}$$

Commented by AnjanDey last updated on 09/Aug/19

$$\mathrm{There}\mathrm{is}\mathrm{a}\mathrm{condition}\mathrm{and}\mathrm{the}\mathrm{condition}\mathrm{is}:-\mathrm{evaluate}\mathrm{the}\mathrm{integral}\mathrm{as}\mathbf{limit}\mathbf{of}\mathbf{sums}....\mathrm{So}{\mathrm{you}}^{\prime }\mathrm{ve}\mathrm{to}\mathrm{proceed}\mathrm{it}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}.$$

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