First slide

Evaluation of definite integrals

Question

The value of $\int_{0}^{[x]} \frac{2^{t}}{2^{[t]}} d t$ (where [ ] denotes greatest integer function)

Moderate

A

$[x] \log 2$
B

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

$\frac{1}{2} \frac{[x]}{\log 2}$
D

$\frac{1}{4} \frac{[x]}{\log 2}$

Solution

$I = \int_{0}^{[x]} \frac{2^{t}}{2^{[t]}} d t = \int_{0}^{[x]} 2^{t - [t]} d t$
The function $g (t) = 2^{t - [t]}$ is periodic with period 1,
Therefore
$\begin{array}{l} I = [x] \int_{0}^{1} 2^{t - [t]} d t = [x] \int_{0}^{1} 2^{t} d t = \frac{[x]}{\log 2} [2 t]_{0}^{1} \\ = \frac{[x]}{\log 2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App