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can u answer this quickly

can u answer this quickly

Question Image
Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
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}

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