so we just learned integrals, and i'm having troubles solving this one: dy/dx = (2e^x + sec(x)tan(x) - x^(1/2)) I've gotten to y= (2e^x)dx + sec(x) - (2/3)x^(3/2), but I can't figure out how to integrate 2e^x. please help?
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Originally Posted by jahichuanna so we just learned integrals, and i'm having troubles solving this one: dy/dx = (2e^x + sec(x)tan(x) - x^(1/2)) I've gotten to y= (2e^x)dx + sec(x) - (2/3)x^(3/2), but I can't figure out how to integrate 2e^x. please help? $\displaystyle \int e^x\,dx=e^x+C$...
I get that part, but where did the 2 come from?
Originally Posted by jahichuanna I get that part, but where did the 2 come from? The 2 appears in the original problem. When you evaluate $\displaystyle \int 2e^x\,dx$, you can rewrite it by applying one of the integral properties. Its equivalent to $\displaystyle 2\int e^x\,dx=2e^x+C$ Does this make sense?
ah, ok. thanks. so the answer is 2e^x + sec x + (2/3)x^(3/2) + c?
Originally Posted by jahichuanna ah, ok. thanks. so the answer is 2e^x + sec x + (2/3)x^(3/2) + c?
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