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how to integrate e^2x ?
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\[\int\limits_{}^{}e^xdx = e^x + c\]
\[\int\limits_{}^{}e^{2x}.dx = e^{2x}/2 + c\]
yes, but how?
explain.
differentiate rhs and see u will get the lhs
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rhs?? lhs??
you could substitute to be more explicit \[\alpha x=u\qquad\qquad\alpha \text d x=\text du\qquad\qquad \text dx=\frac{\text du}{\alpha}\] \[\int e^{\alpha x}\text dx=\int e^{u}\frac{\text du}{\alpha}=\frac 1\alpha \int e^u\text du=\frac 1 \alpha e^u+c=\frac{e^{\alpha x}}{\alpha}+c\]
UnkleRhaukus is right btw rhs and lhs are right hand side & left hand side
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