Here's the question you clicked on:
virtus
integrate y=2^x from x=0 to x=3
\[\frac d{dx}a^x=a^x\ln a\]so then what is the integral?
actually i don't know hahaha
The indefinite integral of a^x is a^x/lna + C
\[\int a^x\ln adx=a^x\]so\[\int a^xdx=\frac{a^x}{\ln a}\]+C if it is indefinite, but yours is definite anyway...
turning test can you please show me how this was derived. Thank you very much!
do you need me to prove that\[\frac d{dx}a^x=a^x\ln a\]? that would be the most thorough way to start
we can prove it with logarithmic differentiation\[y=a^x\]taking the natural of of both sides\[\ln y=\ln(a^x)\]\[\ln y=x\ln a\]now differentiate implicitly\[\frac{y'}y=\ln a\]\[y'=y\ln a=a^x\ln a\]so what does this tell us about the antiderivative?
\[y'=a^x\ln a\iff y=a^x\]so\[\int a^x\ln a=a^x\]now we can use this info and do your integral with a u-substitution
watch the sub carefully:\[\int a^xdx\]\[u=a^x\]\[du=a^x\ln a dx\to\frac{du}{\ln a}=a^xdx\]subbing in our expresison for a^xdx we get\[\frac1{\ln a}\int du=\frac u{\ln a}+C=\frac{a^x}{\ln a}+C\]
any questions about that?
THANK YOU A MILLION TIMES OVER turning test! greatly appreciated