If $\int_0^{10} f(x) d x=5$, then $\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x=$

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Aicharya · Accepted Answer

Let $I=\int_0^1 f(k-1+x) d x$ Put $k-1+x=t \Rightarrow d x=d t$ Limits: $ \begin{aligned} & x=0 \Rightarrow t=k-1 \\ & x=1 \Rightarrow t=k \end{aligned} $ $ \therefore \quad I=\int_{k-1}^k f(t) d t $ It can also be written as, $ \begin{aligned} & I=\int_{k-1}^k f(x) d x \\ & \text { Now, } \sum_{k=1}^{10} \int_0^1 f(k-1+x) d x \\ & =\sum_{k=1}^{10} \int_{k-1}^k f(x) d x \\ & =\int_0^1 f(x) d x+\int_1^2 f(x) d x+\ldots .+\int_9^{10} f(x) d x \\ & =\int_0^{10} f(x) d x=5 \quad \text { (from Eq. (i)) } \end{aligned} $

If $\int_0^{10} f(x) d x=5$, then $\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x=$

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