Asked by Anonymous
find the derivative using fundamental theorem of calculus integral of sin^3tdt from e^x to 0
Answers
Answered by
Steve
d/dx ∫[e^x,0] sin^3(t) dt
= d/dx -∫[0,e^x] sin^3(t) dt
= -sin^3(e^x) * d/dx (e^x)
= -e^x sin^3(e^x)
= d/dx -∫[0,e^x] sin^3(t) dt
= -sin^3(e^x) * d/dx (e^x)
= -e^x sin^3(e^x)
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