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0-1-2-2x-1-5-2x-1-10-x-dx-




Question Number 69327 by mhmd last updated on 22/Sep/19
 ∫_( 0) ^1   ((2^(2x+1) − 5^(2x−1) )/(10^x )) dx =
$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\frac{\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} −\:\mathrm{5}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{10}^{{x}} }\:{dx}\:=\: \\ $$
Answered by MJS last updated on 22/Sep/19
∫((2^(2x+1) −5^(2x−1) )/(10^x ))dx=2∫((2/5))^x dx−(1/5)∫((5/2))^x dx=  =(2/(ln (2/5)))((2/5))^x −(1/(5ln (5/2)))((5/2))^x =(1/(ln 2 −ln 5))(2((2/5))^x +(1/5)((5/2))^x )+C  ∫_0 ^1 ...dx=(9/(10(ln 5 −ln 2)))
$$\int\frac{\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} −\mathrm{5}^{\mathrm{2}{x}−\mathrm{1}} }{\mathrm{10}^{{x}} }{dx}=\mathrm{2}\int\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{{x}} {dx}−\frac{\mathrm{1}}{\mathrm{5}}\int\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{x}} {dx}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{ln}\:\frac{\mathrm{2}}{\mathrm{5}}}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{{x}} −\frac{\mathrm{1}}{\mathrm{5ln}\:\frac{\mathrm{5}}{\mathrm{2}}}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{x}} =\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}\:−\mathrm{ln}\:\mathrm{5}}\left(\mathrm{2}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)^{{x}} +\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{{x}} \right)+{C} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}…{dx}=\frac{\mathrm{9}}{\mathrm{10}\left(\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\mathrm{2}\right)} \\ $$

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