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Question Number 52953 Answers: 1 Comments: 0

Can you pls help me with that but only by using the derivative of f(1) and not using the defined defivative of f(x) itself ? Thanks f(x) = x^2 − x f ′ (1) = ?

$$ \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{pls}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{that}\:\mathrm{but}\:\mathrm{only} \\ $$$$\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathrm{derivative}\:\mathrm{of}\:{f}\left(\mathrm{1}\right)\:\mathrm{and}\:\boldsymbol{\mathrm{not}} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{defined}\:\mathrm{defivative}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{itself}\:? \\ $$$$ \\ $$$$\mathrm{Thanks} \\ $$$$ \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:−\:{x} \\ $$$$ \\ $$$${f}\:'\:\left(\mathrm{1}\right)\:=\:? \\ $$

Question Number 52950 Answers: 1 Comments: 1

∫_( 0) ^(π/4) ((sin x+cos x)/(3+sin 2x)) dx =

$$\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{3}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\:= \\ $$

Question Number 52949 Answers: 0 Comments: 4

If f(x) is an odd function, then ∫_( 0) ^π f (cos x) dx = 2∫_( 0) ^(π/2) f (cos x) dx

$$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{function},\:\mathrm{then} \\ $$$$\underset{\:\mathrm{0}} {\overset{\pi} {\int}}\:{f}\:\left(\mathrm{cos}\:{x}\right)\:{dx}\:=\:\mathrm{2}\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:{f}\:\left(\mathrm{cos}\:{x}\right)\:{dx} \\ $$

Question Number 52948 Answers: 2 Comments: 1

If f(x) =∫_( 1) ^x ((log t)/(1+t)) dt, then f(x)+f ((1/x) )=(1/2)(log x)^2

$$\mathrm{If}\:\:\:{f}\left({x}\right)\:=\underset{\:\mathrm{1}} {\overset{{x}} {\int}}\:\frac{\mathrm{log}\:{t}}{\mathrm{1}+{t}}\:{dt},\:\mathrm{then} \\ $$$$\:{f}\left({x}\right)+{f}\:\left(\frac{\mathrm{1}}{{x}}\:\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} \\ $$

Question Number 52947 Answers: 1 Comments: 0

∫_( 0) ^π ((x tan x)/(sec x+cos x)) dx =

$$\underset{\:\mathrm{0}} {\overset{\pi} {\int}}\:\:\frac{{x}\:\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:= \\ $$

Question Number 52944 Answers: 1 Comments: 0

∫_( 0) ^( 1) ((x^3 − 1)/((1 + x^2 ) ln x)) dx

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:−\:\mathrm{1}}{\left(\mathrm{1}\:+\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\:\boldsymbol{\mathrm{ln}}\:\boldsymbol{\mathrm{x}}}\:\:\boldsymbol{\mathrm{dx}} \\ $$

Question Number 52938 Answers: 0 Comments: 3

Question Number 52936 Answers: 0 Comments: 1

Question Number 52946 Answers: 1 Comments: 1

∫_( 0) ^(π/2) log sin 2x dx =

$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\:\mathrm{log}\:\mathrm{sin}\:\mathrm{2}{x}\:{dx}\:= \\ $$

Question Number 52945 Answers: 3 Comments: 2

Question Number 52911 Answers: 2 Comments: 2

Question Number 52898 Answers: 1 Comments: 0

∫arcsin x arccos x dx=?

$$\int\mathrm{arcsin}\:{x}\:\mathrm{arccos}\:{x}\:{dx}=? \\ $$

Question Number 52900 Answers: 3 Comments: 0

∫_0 ^(π/2) sin x (√(sin 2x)) dx=? ∫_(−(π/4)) ^(π/4) cos x (√(cos 2x)) dx=?

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}\:{x}\:\sqrt{\mathrm{sin}\:\mathrm{2}{x}}\:{dx}=? \\ $$$$\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:{dx}=? \\ $$

Question Number 52867 Answers: 1 Comments: 0

Question Number 52859 Answers: 0 Comments: 3

Question Number 52847 Answers: 0 Comments: 2

Question Number 52846 Answers: 1 Comments: 0

Question Number 52845 Answers: 1 Comments: 0

Question Number 52841 Answers: 1 Comments: 4

Question Number 52881 Answers: 1 Comments: 9

Question Number 52825 Answers: 1 Comments: 4

Question Number 52820 Answers: 1 Comments: 0

y = log{(√x) +(1/((√x) ))}^2 show that x(x+1)^2 y_2 + (x+1)^2 y_1 = 2

$${y}\:=\:{log}\left\{\sqrt{{x}}\:+\frac{\mathrm{1}}{\sqrt{{x}}\:\:\:}\right\}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${show}\:{that} \\ $$$${x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} {y}_{\mathrm{2}} \:+\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} {y}_{\mathrm{1}} \:=\:\mathrm{2} \\ $$

Question Number 52818 Answers: 1 Comments: 0

Question Number 52814 Answers: 0 Comments: 6

Question Number 52808 Answers: 0 Comments: 0

Question Number 52806 Answers: 1 Comments: 2

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