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Calculus question: Use Part 1 of FTC to find the derivative of the function
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\[g(x)=\int\limits_{4}^{x}e ^{4t ^{2}-4} dt\] gprime(x)=
Just replace the x with the variable t
It's that easy
wow...i feel dumb >_<
\[\huge g'(x) = e^{4x^2-4}\]
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Thanks!
Np :)
The fundamental theorem of calc part 1 states this, If f is continuous on [a,b] then \[g(x) = \int\limits_{a}^{x} f(t) dt~~~~a \le x \le b\] is continuous on [a,b] differentiable on (a,b) and \[g'(x) = f(x)\]
Basically saying plug in x for t haha.
Or what ever the variable is :P
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