calculate ∫_0 ^1 ((5^(2x+1) −2^(2x−1) )/(10^x )) dx


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Question Number 53474 by maxmathsup by imad last updated on 22/Jan/19

$c a l c u l a t e \int_{0}^{1} \frac{5^{2 x + 1} - 2^{2 x - 1}}{10^{x}} d x$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jan/19
	$∫_0 ^1 ((5^x ×5^x ×5−((2^x ×2^x )/2))/(5^x ×2^x ))dx 5∫_0 ^1 ((5/2))^x dx−(1/2)∫_0 ^1 ((2/5))^x dx ∣5×((((5/2))^x )/(ln((5/2))))−(1/2)×((((2/5))^x )/(ln((2/5))))∣_0 ^1 =[{5×((((5/2)))/(ln((5/2))))−(1/2)×((((2/5)))/(ln((2/5))))}−{(5/(ln((5/2))))−(1/2)×(1/(ln((2/5))))}] =[((((25)/2)−5)/(ln((5/2))))−(1/2)×((2/5)/(ln((2/5))))+(1/2)×(1/(ln((2/5))))] =((15)/(2ln((5/2))))+(1/(2ln((2/5)))){1−(2/5)}] =((15)/(2ln((5/2))))+(3/(10ln((2/5))))$
	$\int_{0}^{1} \frac{5^{x} \times 5^{x} \times 5 - \frac{2^{x} \times 2^{x}}{2}}{5^{x} \times 2^{x}} d x$ $5 \int_{0}^{1} {(\frac{5}{2})}^{x} d x - \frac{1}{2} \int_{0}^{1} {(\frac{2}{5})}^{x} d x$ $∣ 5 \times \frac{{(\frac{5}{2})}^{x}}{l n (\frac{5}{2})} - \frac{1}{2} \times \frac{{(\frac{2}{5})}^{x}}{l n (\frac{2}{5})} ∣_{0}^{1}$ $= [{5 \times \frac{(\frac{5}{2})}{l n (\frac{5}{2})} - \frac{1}{2} \times \frac{(\frac{2}{5})}{l n (\frac{2}{5})}} - {\frac{5}{l n (\frac{5}{2})} - \frac{1}{2} \times \frac{1}{l n (\frac{2}{5})}}]$ $= [\frac{\frac{25}{2} - 5}{l n (\frac{5}{2})} - \frac{1}{2} \times \frac{\frac{2}{5}}{l n (\frac{2}{5})} + \frac{1}{2} \times \frac{1}{l n (\frac{2}{5})}]$ $= \frac{15}{2 l n (\frac{5}{2})} + \frac{1}{2 l n (\frac{2}{5})} {1 - \frac{2}{5}}]$ $= \frac{15}{2 l n (\frac{5}{2})} + \frac{3}{10 l n (\frac{2}{5})}$
	Commented by malwaan last updated on 23/Jan/19

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