If int_0^{10} f(x) d x=5,then sum_{k=1}^{10} | vedclass

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(C) Let $I = \int_0^1 f(k-1+x) d x$. Substitute $k-1+x = t$,then $d x = d t$. When $x=0$,$t=k-1$. When $x=1$,$t=k$. Therefore,$I = \int_{k-1}^k f(t) d

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$5$

Vedclass · Answer

(C) Let $I = \int_0^1 f(k-1+x) d x$. Substitute $k-1+x = t$,then $d x = d t$. When $x=0$,$t=k-1$. When $x=1$,$t=k$. Therefore,$I = \int_{k-1}^k f(t) d t = \int_{k-1}^k f(x) d x$. Now,we need to evaluate the summation: $\sum_{k=1}^{10} \int_0^1 f(k-1+x) d x = \sum_{k=1}^{10} \int_{k-1}^k f(x) d x$. Expanding the summation: $= \int_0^1 f(x) d x + \int_1^2 f(x) d x + \dots + \int_9^{10} f(x) d x$. Using the property of definite integrals $\int_a^b f(x) d x + \int_b^c f(x) d x = \int_a^c f(x) d x$,we get: $= \int_0^{10} f(x) d x$. Given that $\int_0^{10} f(x) d x = 5$,the final value is $5$.

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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