Question

$\small \displaystyle \sf\int_{0}^{1} \bigg[ \prod \limits_{k = 1}^{100}(x + k) \bigg] \bigg[ \sum \limits_{k = 1}^{100} \bigg( \frac{1}{x + 1} \bigg) \bigg] \: dx$​

Answer 1

100(100!)

Step-by-step explanation:

$\small \displaystyle \sf\int_{0}^{1} \bigg[ \prod \limits_{k = 1}^{100}(x + k) \bigg] \bigg[ \sum \limits_{k = 1}^{100} \bigg( \frac{1}{x + 1} \bigg) \bigg] \: dx \\ \\ = \small \displaystyle \sf\int_{0}^{1} d\bigg[ \prod \limits_{k = 1}^{100}(x + k) \bigg]$

To see how just differentiate backwards using chain rule and take (x+1)(x+2) . . . .(x+100) common .

$\bigg[ \prod \limits_{k = 1}^{100}(x + k) \bigg] _{0} ^{1} \\ \\ = \bigg[ \prod \limits_{k = 1}^{100}(1+ k) \bigg] - \bigg[ \prod \limits_{k = 1}^{100}(0 + k) \bigg] \\ \\ = 101! - 100! \\ \\ = 100!(101 - 1) = 100(100!)$

Answer 2

Step-by-step explanation:

Answer:

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Step-by-step explanation:

hope it help you.

thanks