Question
evaluate d/dx ∫x,5] (t^5)tan(t)dt
Answers
Answered by
oobleck
per the usual rules, with ∫[lower .. upper],
d/dx ∫[5 .. x] (t^5) tan(t) dt = x^5 tanx
d/dx ∫[5 .. x] (t^5) tan(t) dt = x^5 tanx
Answered by
oobleck
this is just the Chain Rule in reverse. For, if
F(t) = ∫f(t) dt then if u and v are functions of x,
∫[u..v] f(t) dt = F(v) - F(u)
so
d/dx ∫[u..v] f(t) dt = F'(v) dv/dx - F'(u) du/dx = f(v) dv/dx - f(u) du/dx
F(t) = ∫f(t) dt then if u and v are functions of x,
∫[u..v] f(t) dt = F(v) - F(u)
so
d/dx ∫[u..v] f(t) dt = F'(v) dv/dx - F'(u) du/dx = f(v) dv/dx - f(u) du/dx
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