# Thread: limits of integration are variables

1. ## limits of integration are variables

How would I find the derivatives of these functions?

1. $f(x) = \int_{x}^{1} \sqrt{t+sin(t)} dt$

2. $y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta$

2. ## Re: limits of integration are variables

In general:
$D\left(\int f(x)dx\right)=D[F(x)+C]=f(x)$
But because they're definite integrals it can be useful to calculate them.

3. ## Re: limits of integration are variables

So is this true then?

1. $f(x) = \int_{x}^{1} \sqrt{t+sin(t)} dt$
Let $g(t)=\sqrt{t+sin(t)$ so that $f(x) = \int_{x}^{1} g(t) dt$
Then $D\left( f(x) \right) = D\left( \int_{x}^{1} g(t) dt \right) = D\left( G(1) - G(x) \right) = D\left( G(1) \right) - D\left( G(x) \right) = D\left( -G(x) \right) = -g(x)$

2. $y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta$
Let $y = f(x)$ and $g(\theta)=\sqrt{\theta+sin(\theta)$ so that $f(x) = \int_{x}^{1} g(\theta) d\theta$
Then $D\left( f(x) \right) = D\left( \int_{x}^{1} g(\theta) d\theta \right) = D\left( G(x) - G(\sqrt{x}) \right) = g(x) - g(\sqrt{x})$

Seems correct, but it seems too simple. I suppose these problems could be just to illustrate how integration and differentiation are inverse processes.

4. ## Re: limits of integration are variables

Do not make this hard.
$D_x \left( {\int_0^x {f(t)dt} } \right)=f(x)$

Because ${\int_x^0 {f(t)dt}=-{\int_0^x {f(t)dt}$ we have $D_x \left( {\int_x^0 {f(t)dt} } \right)=-f(x)$

$D_x \left( {\int_{g(x)}^{h(x)} {f(t)dt} } \right) = f \circ h(x)h'(x) - f \circ g(x)g'(x)$

5. ## Re: limits of integration are variables

Right. Thanks. "KISS = Keep It Simple, Stupid" I'll try to remember that.

I'm going with these answers then:

1. $D(f(x)) = -\sqrt{x+sin(x)}$

2. $D(y) = \frac{cos(x)-\sqrt{x} \, cos\sqrt{x}}{x}$

(unless someone says "no, don't" - anyone?)

6. ## Re: limits of integration are variables

Originally Posted by Plato
$D_x \left( {\int_{g(x)}^{h(x)} {f(t)dt} } \right) = f \circ h(x)h'(x) - f \circ g(x)g'(x)$
Since Plato edited his response to include the above, it appears at least one of my answers is wrong.

2. $y = \int_{\sqrt{x}}^{x} \frac{cos\theta}{\theta} d\theta$
Let $y = f(x)$ and $t(\theta)=\frac{cos\theta}{\theta}$ so that $f(x) = \int_{\sqrt{x}}^{x} t(\theta) d\theta$

Also, if $g(x) = \sqrt{x}$ and $h(x) = x$ then this would conform with Plato's formula and the derivatives would be $h'(x) = 1, g'(x)=\frac{1}{2\sqrt{x}}$.

The integral would NOT be what I posted earlier: $D(y) = \frac{cos(x)-\sqrt{x} \, cos\sqrt{x}}{x}$.

Instead, this is what I'm thinking:

$D\left( f(x) \right) = D\left( \int_{g(x)}^{h(x)} t(\theta) d\theta \right) = t(x)*1 - t(\sqrt{x})*\frac{1}{2\sqrt{x}}$
$= \frac{cos(x)}{x}*1 - \frac{cos(\sqrt{x})}{\sqrt{x}}*\frac{1}{2\sqrt{x}}$
$= \frac{2cos(x)-cos(\sqrt{x})}{2x}$

Because the derivative of $x$ is 1, I'm sticking with my original results for the first problem, which was $D(f(x)) = -\sqrt{x+sin(x)}$

Once again, can anyone confirm or deny these answers? Thanks.

7. ## Re: limits of integration are variables

More generally, "Liebniz's rule":
$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta(x)}{dx}f(x,\beta(x))- \frac{d\alpha(x)}{dx}f(x, \alpha(x)) + \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x} dt$