Question Number 226880 by Spillover last updated on 17/Dec/25
Answered by TonyCWX last updated on 18/Dec/25
![∫_0 ^1 [2^x ]dx = ∫_0 ^1 [e^(xln(2)) ]dx = ∫_0 ^1 [(e^(ln(2)) )^x ]dx = [(((e^(ln(2)) )^x )/(ln(e^(ln(2)) )))]_0 ^1 =[(2^x /(ln(2)))]_0 ^1 =(2^1 /(ln(2)))−(2^0 /(ln(2))) =(1/(ln(2)))](https://www.tinkutara.com/question/Q226889.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{2}^{{x}} \right]\mathrm{d}{x} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[{e}^{{x}\mathrm{ln}\left(\mathrm{2}\right)} \right]\mathrm{d}{x} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left({e}^{\mathrm{ln}\left(\mathrm{2}\right)} \right)^{{x}} \right]\mathrm{d}{x} \\ $$$$=\:\left[\frac{\left({e}^{\mathrm{ln}\left(\mathrm{2}\right)} \right)^{{x}} }{\mathrm{ln}\left({e}^{\mathrm{ln}\left(\mathrm{2}\right)} \right)}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left[\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\left(\mathrm{2}\right)}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{1}} }{\mathrm{ln}\left(\mathrm{2}\right)}−\frac{\mathrm{2}^{\mathrm{0}} }{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{2}\right)} \\ $$