integral of 1/(2+x^3)dx bounds:0 to .5
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Originally Posted by cm3pyro integral of 1/(2+x^3)dx bounds:0 to .5 integrate ... using the first three terms should get you within 0.00001 of the actual value
Originally Posted by skeeter integrate ... using the first three terms should get you within 0.00001 of the actual value i need a maclaurin series
Originally Posted by cm3pyro i need a maclaurin series what do you think I posted?
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