int_{0}^{100} e^{x-[x]} d x is equal to | vedclass

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(C) Let $I = \int_{0}^{100} e^{x-[x]} d x$. Since $f(x) = x - [x]$ is a periodic function with period $T = 1$, we can use the property $\int_{0}^{nT}

Vedclass · Accepted Answer

$100(e-1)$

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(C) Let $I = \int_{0}^{100} e^{x-[x]} d x$. Since $f(x) = x - [x]$ is a periodic function with period $T = 1$, we can use the property $\int_{0}^{nT} f(x) d x = n \int_{0}^{T} f(x) d x$. Here, $n = 100$ and $T = 1$, so $I = 100 \int_{0}^{1} e^{x-[x]} d x$. For $0 Thus, $I = 100 \int_{0}^{1} e^{x} d x$. Evaluating the integral, we get $I = 100 [e^{x}]_{0}^{1}$. $I = 100 (e^{1} - e^{0}) = 100 (e - 1)$.

$\int_{0}^{100} e^{x-[x]} d x$ is equal to

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