Find $$\frac{d}{dx}\int_{-x^2}^{0} sin(8t^2)dt$$
Is this correct? $$\frac{d}{dx}\int_{-x^2}^{0} sin(8t^2)dt = -2x(-sin(8x^4))$$
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No. ${d\over dt}\int_u^v f(x) = v’f(v)-u’f(u)$ – Certainly not a dog Nov 04 '19 at 11:36
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Will the answer be $$2x sin(-8x^4)$$? – Jisbon Nov 04 '19 at 11:38
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$8(-x^2)^2=8x^4 \neq -8x^4$ – Certainly not a dog Nov 04 '19 at 11:39
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Oh yes, thanks for the help – Jisbon Nov 04 '19 at 11:40
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glad to help – Certainly not a dog Nov 04 '19 at 11:41
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Slightly more general than the "Fundamental Theorem of Calculus" is "Leigniz's rule", $\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta(x)}{dx}f(x,x)- \frac{d\alpha(x)}{dx}f(x,x)+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f(x,t)}{\partial t} dt$.
In $\frac{d}{dx}\int_{-x^2}^0 sin(8t^2)dt$, we have $\alpha(x)= -x^2$, so $\frac{d\alpha}{dx}= -2x$, $\beta(x)= 0$ so $\frac{d\beta}{dx}= 0$, and $f(x,t)= sin(8t^2)$. f(x,t) is actually a function t only so $\frac{\partial f}{\partial x}= 0$.
$\frac{d}{dx}\int_{-x^2}^0 sin(8t^2)dt= 0(sin(8x^2))- (-2x)sin(8x^2)+ 0= 2x sin(8x^2)$
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