Integration of Parametric Functions

[darkchild]

Is there an analog or a more general form of the rule
$$\frac{d}{dx} \int_{a}^{t} F(t) dt = F(x)$$

that covers the case of F(t) being a composite function?

 

[mathman]

I am not exactly sure of what you have in mind. If F(x)=G(H(x)) for example, the theorem still applies.

 

[darkchild]

Well, for example, I have $$g(x) = \int_{0}^{x^{2}} e^{-t^{2}} dt$$, and $$g'(x)$$ does not equal $$e^{-x^{4}}$$. There's a factor missing; I can't just plug x (or x squared) into the integrand.

 

[lurflurf]

http://en.wikipedia.org/wiki/Leibniz_integral_rule]Leibniz rule


$$\frac{d}{dt}\int_{a(t)}^{b(t)}f(x,t)dx=\int_{a(t)}^{b(t)} \partial_tf(x,t)dx+f(x,t)\partial_tx|_{x=a(t)}^{x=b(t)}$$

 

[darkchild]

There's an error in my original post. The upper limit of integration should be x.

 

[darkchild]

http://en.wikipedia.org/wiki/Leibniz_integral_rule]Leibniz rule


$$\frac{d}{dt}\int_{a(t)}^{b(t)}f(x,t)dx=\int_{a(t)}^{b(t)} \partial_tf(x,t)dx+f(x,t)\partial_tx|_{x=a(t)}^{x=b(t)}$$

I don't see how that will help since the integrand is a function of t only and the integration is with respect to t.

 

[HallsofIvy]

I think what you want is "Leibniz's formula":

$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= f(x, \beta(x))\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x}dt$$

 

[darkchild]

I think what you want is "Leibniz's formula":

$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= f(x, \beta(x))\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x}dt$$

Thanks! Exactly what I needed.

 

[homology]

Thanks! Exactly what I needed.

That's exactly what lurflurf gave you. The integration variable is called a 'dummy variable' so:

$$\int_a^{g(t)} f(x)dx$$

is the same function as

$$\int_a^{g(x)} f(t) dt$$

 

[HallsofIvy]

http://en.wikipedia.org/wiki/Leibniz_integral_rule]Leibniz rule


$$\frac{d}{dt}\int_{a(t)}^{b(t)}f(x,t)dx=\int_{a(t)}^{b(t)} \partial_tf(x,t)dx+f(x,t)\partial_tx|_{x=a(t)}^{x=b(t)}$$

I don't see how that will help since the integrand is a function of t only and the integration is with respect to t.

No, the integration is with respect to x.

The only difference between what lurflurf posted and what I posted is that the "t" and "x" are switched.