Menu Close

solve-1-1-3-x-2-x-dx-




Question Number 112135 by mathdave last updated on 06/Sep/20
solve  ∫_(−1) ^1 ∣3^x −2^x ∣dx
$${solve} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \mid\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mid{dx} \\ $$
Answered by MJS_new last updated on 06/Sep/20
0≤x≤1: 3^x ≥2^x  ⇒  ∫_0 ^1 ∣3^x −2^x ∣dx=∫_0 ^1 3^x −2^x dx=[(3^x /(ln 3))−(2^x /(ln 2))]_0 ^1 =  =(2/(ln 3))−(1/(ln 2))=((2ln 2 −ln3)/(ln 2 ln 3))=((ln (4/3))/(ln 2 ln 3))
$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}:\:\mathrm{3}^{{x}} \geqslant\mathrm{2}^{{x}} \:\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mid\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mid{dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {dx}=\left[\frac{\mathrm{3}^{{x}} }{\mathrm{ln}\:\mathrm{3}}−\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$=\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{2ln}\:\mathrm{2}\:−\mathrm{ln3}}{\mathrm{ln}\:\mathrm{2}\:\mathrm{ln}\:\mathrm{3}}=\frac{\mathrm{ln}\:\frac{\mathrm{4}}{\mathrm{3}}}{\mathrm{ln}\:\mathrm{2}\:\mathrm{ln}\:\mathrm{3}} \\ $$
Commented by mathdave last updated on 06/Sep/20
am really really sorry this is the  quastion not the one i wrote before .it  was a mistake
$${am}\:{really}\:{really}\:{sorry}\:{this}\:{is}\:{the} \\ $$$${quastion}\:{not}\:{the}\:{one}\:{i}\:{wrote}\:{before}\:.{it} \\ $$$${was}\:{a}\:{mistake} \\ $$
Answered by MJS_new last updated on 06/Sep/20
∫_(−1) ^1 ∣3^x −2^x ∣dx=∫_(−1) ^0 2^x −3^x dx+∫_0 ^1 3^x −2^x dx=  =[(2^x /(ln 2))−(3^x /(ln 3))]_(−1) ^0 +[(3^x /(ln 3))−(2^x /(ln 2))]_0 ^1 =  =(1/(2ln 2))−(2/(3ln 3))+(2/(ln 3))−(1/(ln 2))=  =(4/(3ln 3))−(1/(2ln 2))=((8ln 2 −3ln 3)/(6ln 2 ln 3))=((ln ((256)/(27)))/(6ln 2 ln 3))
$$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\mid\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mid{dx}=\underset{−\mathrm{1}} {\overset{\mathrm{0}} {\int}}\mathrm{2}^{{x}} −\mathrm{3}^{{x}} {dx}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {dx}= \\ $$$$=\left[\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}−\frac{\mathrm{3}^{{x}} }{\mathrm{ln}\:\mathrm{3}}\right]_{−\mathrm{1}} ^{\mathrm{0}} +\left[\frac{\mathrm{3}^{{x}} }{\mathrm{ln}\:\mathrm{3}}−\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3ln}\:\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}= \\ $$$$=\frac{\mathrm{4}}{\mathrm{3ln}\:\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2ln}\:\mathrm{2}}=\frac{\mathrm{8ln}\:\mathrm{2}\:−\mathrm{3ln}\:\mathrm{3}}{\mathrm{6ln}\:\mathrm{2}\:\mathrm{ln}\:\mathrm{3}}=\frac{\mathrm{ln}\:\frac{\mathrm{256}}{\mathrm{27}}}{\mathrm{6ln}\:\mathrm{2}\:\mathrm{ln}\:\mathrm{3}} \\ $$
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))
$$\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \mid\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} \mid\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \:\mathrm{2}^{\mathrm{x}} \mid\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} −\mathrm{1}\mid\:\mathrm{dx} \\ $$$$\mathrm{we}\:\mathrm{have}\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} −\mathrm{1}>\mathrm{0}\:\Leftrightarrow\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} >\mathrm{1}\:\Rightarrow\mathrm{e}^{\mathrm{xln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} >\mathrm{1}\:\Leftrightarrow\mathrm{xln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)>\mathrm{0}\:\Leftrightarrow\mathrm{x}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{0}} \:\mathrm{2}^{\mathrm{x}} \left(\mathrm{1}−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} \right)\mathrm{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{2}^{\mathrm{x}} \left(\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{x}} −\mathrm{1}\right)\mathrm{dx} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{0}} \:\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right)\mathrm{dx}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} \right)\mathrm{dx}\:\:\mathrm{but}\: \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{0}} \:\left(\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\mathrm{x}} \right)\mathrm{dx}\:=\int_{−\mathrm{1}} ^{\mathrm{0}} \:\left(\mathrm{e}^{\mathrm{xln2}} −\mathrm{e}^{\mathrm{xln3}} \right)\mathrm{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{ln2}}\mathrm{2}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{ln3}}\mathrm{3}^{\mathrm{x}} \right]_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=\left\{\frac{\mathrm{1}}{\mathrm{ln2}}−\frac{\mathrm{1}}{\mathrm{ln3}}\:−\left(\:\frac{\mathrm{2}^{−\mathrm{1}} }{\mathrm{ln2}}−\frac{\mathrm{3}^{−\mathrm{1}} }{\mathrm{ln3}}\right)\right\}\:=\frac{\mathrm{1}}{\mathrm{2ln2}}−\frac{\mathrm{2}}{\mathrm{3ln3}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} \right)\mathrm{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{ln3}}\mathrm{3}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{ln2}}\mathrm{2}^{\mathrm{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{3}}{\mathrm{ln3}}−\frac{\mathrm{2}}{\mathrm{ln2}}−\frac{\mathrm{1}}{\mathrm{ln3}}+\frac{\mathrm{1}}{\mathrm{ln2}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{ln3}}−\frac{\mathrm{1}}{\mathrm{ln2}}\:\Rightarrow\:\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2ln2}}−\frac{\mathrm{2}}{\mathrm{3ln3}}\:+\frac{\mathrm{2}}{\mathrm{ln3}}−\frac{\mathrm{1}}{\mathrm{ln2}}\:=−\frac{\mathrm{1}}{\mathrm{2ln2}}\:+\frac{\mathrm{4}}{\mathrm{3ln3}} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *