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Question Number 102313 by bemath last updated on 08/Jul/20

Answered by 1549442205 last updated on 08/Jul/20

$$\mathrm{The}\mathrm{abscissa}\mathrm{of}\mathrm{intersection}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{curve}$$$$\mathrm{y}=\frac{4}{\sqrt{\mathrm{x}}}-\frac{4}{\mathrm{x}}\mathrm{and}\mathrm{x}-\mathrm{axis}\mathrm{being}\mathrm{the}\mathrm{root}\mathrm{of}\mathrm{the}\mathrm{eqs}.\frac{4}{\sqrt{\mathrm{x}}}-\frac{4}{\mathrm{x}}=0$$$$\iff \mathrm{x}-\sqrt{\mathrm{x}}=0(\mathrm{x}\ne 0)\iff \mathrm{x}=1.\mathrm{Hence},$$$$\mathrm{S}={\int }_0^1{\mathrm{e}}^{1-\frac{1}{4}\mathrm{x}}\mathrm{dx}+{\int }_1^4[{\mathrm{e}}^{1-\frac{1}{4}\mathrm{x}}-(\frac{4}{\sqrt{\mathrm{x}}}-\frac{4}{\mathrm{x}})]\mathrm{dx}$$$${\int }_0^4{\mathrm{e}}^{1-\frac{1}{4}\mathrm{x}}\mathrm{dx}+{\int }_1^4\frac{4}{\mathrm{x}}\mathrm{dx}-{\int }_1^4\frac{4}{\sqrt{\mathrm{x}}}\mathrm{dx}$$$$=-{4\mathrm{e}}^{1-\frac{1}{4}\mathrm{x}}{\mid }_0^4+4\mathrm{lnx}{\mid }_1^4-8\sqrt{\mathrm{x}}{\stackrel{4}{\mid }}_1$$$$=-4-(-4\mathrm{e})+4(\ln 4-0)-8(2-1)$$$$=4\mathrm{e}+8\ln 2-12$$

Commented by I want to learn more last updated on 08/Jul/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by bemath last updated on 08/Jul/20

$$igot$$$$Area=\underset{0}{\stackrel{4}{\int }}{e}^{1-\frac{x}{4}}dx-\underset{1}{\stackrel{4}{\int }}(4x^{-1/2}-4x^{-1})dx$$$$=-4\underset{0}{\stackrel{4}{\int }}{e}^{1-\frac{x}{4}}d(1-\frac{x}{4})-{\{8\sqrt{x}-4\ln \left(x\right)\}}_1^4$$$$=-4{\left\{e^{1-\frac{x}{4}}\right\}}_0^4-\{(16-4\ln \left(4\right)-8\}$$$$=-4(1-\mathrm{e})+4\ln \left(4\right)-8$$$$=4\mathrm{e}+4\ln \left(4\right)-12$$$$\approx 4.4183$$

Commented by I want to learn more last updated on 08/Jul/20

$$\mathrm{Thanks}\mathrm{sir}$$

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