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Find F'(x) a) F(x)= integral (top bound x, lower bound -4) (t-1)dt
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I'm sorry I have no idea how to write integrals in a normal text form
\[\int\limits_{x}^{-4} -12x^2(4x^3 -1)dx\]
You want this: \[\frac{ d }{ dx } \int\limits_{-4}^{x}(t-1)dt\] This is one of those fundamental theorem of calculus things. In general: \[\frac{ d }{ dx }\int\limits_{g(x)}^{f(x)}h(t)dt = h(f(x))\cdot f'(x) - h(g(x)) \cdot g'(x)\] Applying this idea gives: \[\frac{ d }{ dx }\int\limits_{-4}^{x}(t-1)dt = (x-1)(1) - (-4-1)(0) = x-1\]
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