solve ∫_(−1) ^1 ∣3^x −2^x ∣dx


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Question Number 112135 by mathdave last updated on 06/Sep/20

$s o l v e$ $\int_{- 1}^{1} ∣ 3^{x} - 2^{x} ∣ d x$
	Answered by MJS_new last updated on 06/Sep/20

	$0 ⩽ x ⩽ 1 : 3^{x} ⩾ 2^{x} \Rightarrow$ $\underset{0}{\int^{1}} ∣ 3^{x} - 2^{x} ∣ d x = \underset{0}{\int^{1}} 3^{x} - 2^{x} d x = {[\frac{3^{x}}{\ln 3} - \frac{2^{x}}{\ln 2}]}_{0}^{1} =$ $= \frac{2}{\ln 3} - \frac{1}{\ln 2} = \frac{2 \ln 2 - \ln 3}{\ln 2 \ln 3} = \frac{\ln \frac{4}{3}}{\ln 2 \ln 3}$
	Commented by mathdave last updated on 06/Sep/20

	$a m r e a l l y r e a l l y s o r r y t h i s i s t h e$ $q u a s t i o n n o t t h e o n e i w r o t e b e f o r e . i t$ $w a s a m i s t a k e$
	Answered by MJS_new last updated on 06/Sep/20

	$\underset{- 1}{\int^{1}} ∣ 3^{x} - 2^{x} ∣ d x = \underset{- 1}{\int^{0}} 2^{x} - 3^{x} d x + \underset{0}{\int^{1}} 3^{x} - 2^{x} d x =$ $= {[\frac{2^{x}}{\ln 2} - \frac{3^{x}}{\ln 3}]}_{- 1}^{0} + {[\frac{3^{x}}{\ln 3} - \frac{2^{x}}{\ln 2}]}_{0}^{1} =$ $= \frac{1}{2 \ln 2} - \frac{2}{3 \ln 3} + \frac{2}{\ln 3} - \frac{1}{\ln 2} =$ $= \frac{4}{3 \ln 3} - \frac{1}{2 \ln 2} = \frac{8 \ln 2 - 3 \ln 3}{6 \ln 2 \ln 3} = \frac{\ln \frac{256}{27}}{6 \ln 2 \ln 3}$
	Answered by mathmax by abdo last updated on 06/Sep/20
	$I =∫_(−1) ^1 ∣3^x −2^x ∣dx ⇒I =∫_(−1) ^1 2^x ∣((3/2))^x −1∣ dx we have ((3/2))^x −1>0 ⇔((3/2))^x >1 ⇒e^(xln((3/2))) >1 ⇔xln((3/2))>0 ⇔x>0 ⇒I =∫_(−1) ^0 2^x (1−((3/2))^x )dx +∫_0 ^1 2^x (((3/2))^x −1)dx =∫_(−1) ^0 (2^x −3^x )dx +∫_0 ^1 (3^x −2^x )dx but ∫_(−1) ^0 (2^x −3^x )dx =∫_(−1) ^0 (e^(xln2) −e^(xln3) )dx =[(1/(ln2))2^x −(1/(ln3))3^x ]_(−1) ^0 ={(1/(ln2))−(1/(ln3)) −( (2^(−1) /(ln2))−(3^(−1) /(ln3)))} =(1/(2ln2))−(2/(3ln3)) ∫_0 ^1 (3^x −2^x )dx =[(1/(ln3))3^x −(1/(ln2))2^x ]_0 ^1 =(3/(ln3))−(2/(ln2))−(1/(ln3))+(1/(ln2)) =(2/(ln3))−(1/(ln2)) ⇒ I =(1/(2ln2))−(2/(3ln3)) +(2/(ln3))−(1/(ln2)) =−(1/(2ln2)) +(4/(3ln3))$
	$I = \int_{- 1}^{1} ∣ 3^{x} - 2^{x} ∣ dx \Rightarrow I = \int_{- 1}^{1} 2^{x} ∣ {(\frac{3}{2})}^{x} - 1 ∣ dx$ $we have {(\frac{3}{2})}^{x} - 1 > 0 \Leftrightarrow {(\frac{3}{2})}^{x} > 1 \Rightarrow e^{xln (\frac{3}{2})} > 1 \Leftrightarrow xln (\frac{3}{2}) > 0 \Leftrightarrow x > 0$ $\Rightarrow I = \int_{- 1}^{0} 2^{x} (1 - {(\frac{3}{2})}^{x}) dx + \int_{0}^{1} 2^{x} ({(\frac{3}{2})}^{x} - 1) dx$ $= \int_{- 1}^{0} (2^{x} - 3^{x}) dx + \int_{0}^{1} (3^{x} - 2^{x}) dx but$ $\int_{- 1}^{0} (2^{x} - 3^{x}) dx = \int_{- 1}^{0} (e^{xln 2} - e^{xln 3}) dx = {[\frac{1}{\ln 2} 2^{x} - \frac{1}{\ln 3} 3^{x}]}_{- 1}^{0}$ $= {\frac{1}{\ln 2} - \frac{1}{\ln 3} - (\frac{2^{- 1}}{\ln 2} - \frac{3^{- 1}}{\ln 3})} = \frac{1}{2 \ln 2} - \frac{2}{3 \ln 3}$ $\int_{0}^{1} (3^{x} - 2^{x}) dx = {[\frac{1}{\ln 3} 3^{x} - \frac{1}{\ln 2} 2^{x}]}_{0}^{1} = \frac{3}{\ln 3} - \frac{2}{\ln 2} - \frac{1}{\ln 3} + \frac{1}{\ln 2}$ $= \frac{2}{\ln 3} - \frac{1}{\ln 2} \Rightarrow I = \frac{1}{2 \ln 2} - \frac{2}{3 \ln 3} + \frac{2}{\ln 3} - \frac{1}{\ln 2} = - \frac{1}{2 \ln 2} + \frac{4}{3 \ln 3}$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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