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Prove that ∫〖∑_(i=1)K_i f_i (x)dx〗=∑_(i=1)K_i ∫∑_(i=0)〖f_i (x)dx〗
Prove that ∫〖∑_(i=1)K_i f_i (x)dx〗=∑_(i=1)K_i ∫∑_(i=0)〖f_i (x)dx〗
Rohan Byanjankar
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September 21, 2020
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