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from (1/32)S(1/u).du * = (1/32)ln(u) then sub u in? = (1/32)ln(25+16(x^2)) = [ln(25+16(x^2))]/32 Is it this? *the S is the integral sign
Originally Posted by likearollingstone from (1/32)S(1/u).du * = (1/32)ln(u) then sub u in? = (1/32)ln(25+16(x^2)) = [ln(25+16(x^2))]/32 Is it this? *the S is the integral sign I think i see my mistake right?
You have to substitute back for u. So it would be (ln(25+16x^2))/32
Originally Posted by VitaX I think i see my mistake right? Yep thats right
Originally Posted by emeryj You have to substitute back for u. So it would be (ln(25+16x^2))/32 Yah I know you have to sub back but why are you writing parenthesis isnt it absolute value?
Originally Posted by VitaX Yah I know you have to sub back but why are you writing parenthesis isnt it absolute value? 25+16(x^2) is always greater than zero so you dont really need the absolute sign but i give you permision to put it in anyway.
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