# Partial derivative of integral with variable limit

1. Jun 12, 2010

### Gregg

1. The problem statement, all variables and given/known data

$$G(\theta, k ) = \int^{\theta}_0 g(x,k) dx$$

$$\frac{\partial G}{\partial \theta} = ?$$

$$\frac{\partial G}{\partial k} = ?$$

3. The attempt at a solution

If I say that $$\int g(x,k) dx = H(x,k)$$

$$\int^{\theta}_0 g(x,k) dx = H(\theta,k) - H(0,k)$$

Then is $$\frac{\partial G}{\partial \theta} = \frac{\partial H(\theta,k)}{\partial \theta}=g(\theta,k)$$
?

2. Jun 12, 2010

### vela

Staff Emeritus
Yup.

3. Jun 13, 2010

### HallsofIvy

That is "correct" but your derivation is more complicated than necessary! By the fundamental theorem of calculus, the derivative with respect to $\theta$ is just $g(\theta, k)$.

Lagrange's formula is an extension of the fundamental theorem of calculus:
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t) dt= f(x, \beta(x))}\frac{d\beta}{dx}- f(x, \alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}dt$$.

Since $\theta$ only appears in the upper limit of the integral, the derivative with respect to $\theta$ is just
$$f(\theta, k)\frac{d\theta}{d\theta}= f(\theta, k)$$

and the derivative with respect to k is just
$$\int_0^\theta \frac{\partial f}{\partial k} dx$$

4. Jun 13, 2010

### Gregg

That's clearer now, thank you.