Integral Calculus> can u answer this quickly...

can u answer this quickly

Aditya Kartikeya , 11 Years ago

Grade 10

1 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

$I = \int_{0}^{x}\frac{2^{t}}{2^{[t]}}dt$
$I = \int_{0}^{x}2^{(t-[t])}dt$
$(t) = t - [t]$
where () is an fractional part of t.
$I = \int_{0}^{x}2^{(t)}dt$
The time period of fractional part is ‘1’.
If x is an integer, then
$I = x\int_{0}^{1}2^{t}dt$
$I = x[\frac{2^{t}}{ln2}]_{0}^{1}$
$I = \frac{x}{ln2}$

If x is not an integer, then break the limit to lower closer integer (let say I). Then
$I = \int_{0}^{I}2^{(t)}dt + \int_{I}^{x}2^{(t)}dt$
$I = I\int_{0}^{1}2^{t}dt + \int_{I}^{x}2^{t-[t]}dt$
$I = \frac{I}{ln2}+ \int_{I}^{x}\frac{2^{t}}{2^{[t]}}dt$
$I = \frac{I}{ln2}+ \int_{I}^{x}\frac{2^{t}}{2^{I}}dt$
$I = \frac{I}{ln2}+ \frac{1}{2^{I}}[\frac{2^{t}}{ln2}]_{I}^{x}$
$I = \frac{I}{ln2}+ \frac{1}{2^{I}}.\frac{2^{x}-2^{I}}{ln2}$
$I = \frac{I}{ln2}+ \frac{2^{x-I}-1}{ln2}$