Why must the upper limit match when integrating?

[MathewsMD]

For example, if you have the function f(x) = x₂ then find:

d/dx _{any number}^3x∫ t²dt

Why must the dx in d/dx ∫f(t)dt always match the upper limit in order to compute the integral? Why is the lower limit of no concern? I know that you must take chain rule into consideration and change 3x to u, and then do du/dx but why does the upper limit only matter?

Any help would be great!

 

[Student100]

Huh I don't understand your question?

 

[economicsnerd]

When applying the fundamental theorem of calculus, think about everything in terms of differentiation.

Say f:\mathbb R\to\mathbb R is some continuous, bounded function. For any a,b\in \mathbb R, let F(a,b)=\int_a^bf(x)\text{d}x.

The FTC tells us that differentiating F(a,b) by b gives f(b) (no matter what a is), and differentiating F(a,b) by a gives -f(a) (no matter what b is).

As an example, think about the case where f(x)>0 is your speed at time x, in which case F(a,b) (for a<b) is just how far you've traveled between time a and time b.

Talking about how fast you're going at time b doesn't require any information about when you started moving (i.e. time a).

 

[HallsofIvy]

For example, if you have the function f(x) = x₂ then find:

d/dx _{any number}^3x∫ t²dt

Why must the dx in d/dx ∫f(t)dt always match the upper limit in order to compute the integral? Why is the lower limit of no concern? I know that you must take chain rule into consideration and change 3x to u, and then do du/dx but why does the upper limit only matter?

Any help would be great!

You are making some assumptions that are unjustified- in particular that it is the upper
limit that is important!

In order to be able to differentiate with respect to x, you must have a function of x. One way to do that is to have one or the other limit of integration a function of x.
Both \int_0^x t^2 dt= (1/3)x^3 and \int_x^1 e^t dt= e- e^x are differentiable with respect to x.

Yet another is to have constant limits of integration while the integrand is a function of both x and the "variable of integration:
\int_0^1 e^{x+ t}dt= e^x\int_0^1 e^t dt= e^x[e- 1] is differentiable with respect to x.

(Strictly speaking, of course, a the integral does NOT have to have an "x" anywhere in order to be differentiable with respect to x! Of course, then, the derivative is 0.)

 

[lurflurf]

To elaborate on what Halls said what is important is the places where the variable appears. The general rule is called Leibniz integral rule and states that
$$\dfrac{d}{dt} \int_{\mathrm{a}(t)}^{\mathrm{b}(t)} \mathrm{f}(x,t) \, \mathrm{d}x = \int_{\mathrm{a}(t)}^{\mathrm{b}(t)} \mathrm{f} ^{(0,1)}(x,t) \, \mathrm{d}x + \mathrm{f}(\mathrm{b}(t),t)\mathrm{b}^\prime (t)-\mathrm{f}(\mathrm{a}(t),t)\mathrm{a}^\prime (t) \\ \text{where } \mathrm{f} ^{(0,1)}(x,t) \text{ is the derivative of f with respect to t with x treated as constant.}$$

 

[economicsnerd]

^ This.

And Leibniz's rule is really just what you get by combining FTC and the chain rule.