Question

The value of $\int_{0}^{[x]} \frac{2^x}{2^{[x]}}dx$ is

• A. $[x] \log 2$
• B. $\frac{[x]}{\log 2}$
• C. $\frac{1}{2} \cdot \frac{[x]}{\log 2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{[x]}{\log 2}$

Answer 2

Let the given integral be $I$. The upper limit of the integral is $[x]$. Let $n = [x]$. Since the lower limit is 0, we consider the case where $n$ is a non-negative integer. If $x \ge 0$, then $[x] \ge 0$. The integral is $I = \int_{0}^{n} \frac{2^x}{2^{[x]}}dx$.

We can split the integral into a sum of integrals over the intervals $[k, k+1)$ for $k = 0, 1, \dots, n-1$. In each interval $[k, k+1)$, the value of $[x]$ is $k$. So, we can write the integral as: $I = \sum_{k=0}^{n-1} \int_{k}^{k+1} \frac{2^x}{2^{[x]}}dx$

For $x \in [k, k+1)$, $[x] = k$. Thus, the integrand becomes $\frac{2^x}{2^k}$. The integral over the interval $[k, k+1)$ is: $\int_{k}^{k+1} \frac{2^x}{2^k}dx = \frac{1}{2^k} \int_{k}^{k+1} 2^x dx$

Now, we evaluate the integral $\int 2^x dx$. The formula for the integral of $a^x$ is $\int a^x dx = \frac{a^x}{\log a} + C$. So, $\int 2^x dx = \frac{2^x}{\log 2} + C$.

Now, evaluate the definite integral $\int_{k}^{k+1} 2^x dx$: $\int_{k}^{k+1} 2^x dx = \left[\frac{2^x}{\log 2}\right]_{k}^{k+1} = \frac{2^{k+1}}{\log 2} - \frac{2^k}{\log 2} = \frac{2^{k+1} - 2^k}{\log 2}$ $= \frac{2 \cdot 2^k - 2^k}{\log 2} = \frac{2^k(2-1)}{\log 2} = \frac{2^k}{\log 2}$

Substitute this result back into the integral over the interval $[k, k+1)$: $\int_{k}^{k+1} \frac{2^x}{2^k}dx = \frac{1}{2^k} \cdot \frac{2^k}{\log 2} = \frac{1}{\log 2}$

The total integral is the sum of these integrals over the intervals from $k=0$ to $n-1$. There are $n$ such intervals. $I = \sum_{k=0}^{n-1} \frac{1}{\log 2}$

This is a sum of $n$ terms, each equal to $\frac{1}{\log 2}$. $I = n \cdot \frac{1}{\log 2} = \frac{n}{\log 2}$

Since $n = [x]$, the value of the integral is $\frac{[x]}{\log 2}$.