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\int _{2}^{x}\frac{\left(t^{2}-2\right)^{6}}{64}\mathrm{d}t
Diferenciar w.r.t. x
\frac{\left(x^{2}-2\right)^{6}}{64}
Cuestionario
Integration
5 problemas similares a:
f ( x ) = \int _ { 2 } ^ { x } ( \frac { 1 } { 2 } t ^ { 2 } - 1 ) ^ { 6 } d t
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A useful approach for these sorts of estimates is to use the mean value theorem. (a) Let g(x) = {1 \over 1+x^2}, then g'(x) = -{2x \over (1+x^2)^2 } and g''(x) = 2 { 3 x^2 -1 \over (1+x^2)^3 }. ...
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With polar coordinates, the first integral becomes \int_0^a\int_0^{2\pi}(r\cos (t))^2 (r)drdt =\int_0^ar^3dr \int_0^{2\pi}\cos^2 (t)dt =\frac {a^4}{4}\Bigl [\frac {t}{2}+\frac {1}{4}\sin(2t)\Bigr]_0^{2\pi} ...
Let f be a continuous function and f(x)=\frac{2}{t^2}\int^t_0 f(x+s)sds, \forall t>0. Show that f is a constant
https://math.stackexchange.com/q/2471603
While your point is taken, I think some rephrasing will do this the world of good. Fix x. Define h(t) = \frac 2{t^2} \int_{0}^t f(x+s)sds. By definition, we know that h(t) is a constant, for all ...
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You are on the correct path. You applied the fundamental theorem of calculus correctly and deduced (almost) correctly that F'(x) = \frac{9x^8}{5+(x^9)^2} = \frac{9x^8}{5+x^{18}}. All you have to ...
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\begin{align} y & = \int_0^x (1+t^3)^{-1/2} \, dt, \\[10pt] \frac{dy}{dx} & = (1+x^3)^{-1/2}, \\[10pt] \frac{dx}{dy} & = \frac{1}{(1+x^3)^{-1/2}} = (1+x^3)^{1/2} = \text{some function of }y. \\[10pt] ...
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