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Let g:R→R be a function satisfying g(x)=x2+x21∫−1tg(t)dt+x31∫−1g(t)dt. Then the value of 111∫−1(g(x)+g(−x))dx is

Answer» Let g:RR be a function satisfying g(x)=x2+x211tg(t)dt+x311g(t)dt. Then the value of 1111(g(x)+g(x))dx is


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