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Question Number 192452 by Spillover last updated on 18/May/23

Evaluate     ∫_0 ^1 2^(x ) dx

$${Evaluate}\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}^{{x}\:} {dx}\:\:\: \\ $$$$ \\ $$

Answered by senestro last updated on 18/May/23

1/ln 2

$$\mathrm{1}/\mathrm{ln}\:\mathrm{2} \\ $$

Commented by Spillover last updated on 18/May/23

show your solution

$${show}\:{your}\:{solution} \\ $$

Answered by manxsol last updated on 19/May/23

y=2^x   lny=xln2  (dy/y) =ln2dx  x=0   y=1  x=1    y=2  ∫_1 ^2 y(dy/(yln2))=(y/(ln2))∣_1 ^2 = (2/(ln2))−(1/(ln2))=(1/(ln2))

$${y}=\mathrm{2}^{{x}} \\ $$$${lny}={xln}\mathrm{2} \\ $$$$\frac{{dy}}{{y}}\:={ln}\mathrm{2}{dx} \\ $$$${x}=\mathrm{0}\:\:\:{y}=\mathrm{1} \\ $$$${x}=\mathrm{1}\:\:\:\:{y}=\mathrm{2} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \cancel{{y}}\frac{{dy}}{\cancel{{y}ln}\mathrm{2}}=\frac{{y}}{{ln}\mathrm{2}}\mid_{\mathrm{1}} ^{\mathrm{2}} =\:\frac{\mathrm{2}}{{ln}\mathrm{2}}−\frac{\mathrm{1}}{{ln}\mathrm{2}}=\frac{\mathrm{1}}{{ln}\mathrm{2}} \\ $$$$ \\ $$

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