Solved: "Evaluate the integral. ∫012−0d0"

aortiH

Answered question

2021-11-07

Evaluate the integral.

\int_{0}^{1} 2^{- 0} d 0

d2saint0

Skilled2021-11-08Added 89 answers

\int_{0}^{1} 2^{- 0} d 0

Now, using the formula of integration as,

\int a^{x} d x = \frac{a^{x}}{\log a} + c

to integrate the given integral.

\int_{0}^{1} 2^{- 0} d 0 = - {[\frac{2^{- 0}}{\log (2)}]}_{0}^{1}

= - {[\frac{1}{2^{0} \log (2)}]}_{0}^{1}

= - \frac{1}{2^{1} \log (2)} + \frac{1}{2^{0} \log (2)}

= - \frac{1}{2 \log (2)} + \frac{1}{\log (2)}

= \frac{- 1 + 2}{2 \log (2)}

= \frac{1}{2 \log (2)}

Therefore, the value of the integral is

\frac{1}{2 \log (2)}

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