Derivative of $\int_0^tf(x,t)dx$? Fundamental theorem of calculus works for $\int_0^tf(x)dx$
$$\frac{d}{dt}\int_0^tf(x)dx=f(t)$$
but how about this case?
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$\begingroup$ Suppose $F(x,t)$ is an antiderivative of a nice function $f(x,t)$ with respect to $x$. (It is defined up to addition by a function $C(t)$ constant with respect to $x$.) Then $\int_{a(t)}^{b(t)}f(x,t)\mathrm{d}x=F(a(t),t)-F(b(t),t)$, which can be differentiated with the chain rule. $\endgroup$– anonAug 16, 2019 at 2:32
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$\begingroup$ @runway44 can you give something only about $f$? $\endgroup$– monotonicAug 16, 2019 at 2:36
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1$\begingroup$ No, in general, you must also differentiate $f$. Try the special case where $f$ doesn't depend on $x$. What's the derivative of $t f(t)$? $\endgroup$– Calvin KhorAug 16, 2019 at 2:53
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1 Answer
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$$\frac{d}{dt}\int_0^tf(x,t)dx=\int_0^t\frac{\partial}{\partial t}f(x,t)dx~+~f(t,t)$$
Leibniz Integral Rule (Differentiation under the integral sign):
Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$, $$\frac{d}{dx}\left(\int_{a(x)}^{b(x)} f(x,t) dt\right)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$
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$\begingroup$ Can I have something only in terms of $f$? But not derivative of $f$? $\endgroup$ Aug 16, 2019 at 2:33
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1$\begingroup$ No, because for this case $~f~$ is a function of two variable $~x,~t~$. If you take $~f~$ is function of one variable, then Leibniz Integral Rule becomes Fundamental theorem of calculus. $\endgroup$– nmasantaAug 16, 2019 at 2:39