Understanding definite integrals with functions as bounds

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I can't quite grasp the meaning behind definite integrals defined on two bounds, which appear as functions. For instance,

$$\int_{x^2}^{\cos x}t^2dt$$

What is this notation telling me? What does it mean that the lower bound is $x^2$, and the upper bound is $\cos x$? Where does the definite integral "stop" and "end", if $x^2$ and $\cos x$ are not single values, but a collection of values? Wouldn't these $x$ values then overlap...?

Moreover, when I wish to take the derivative of such an integral, how do I know that $0$ (or any specified constant of integration for that matter) exists "between" $x^2$ and $\cos x$?

Very confused, my apologies. Thanks for any clarification you can provide.

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So, first we ask what an integral calculates? The answer is, the integral of a function $\int f(x) dx$ is the equal to the area between the curve and the $x$-axis. (This is a signed area. So area above the axis is positve and the area below is counted negatively.)

So, if we want to know about the definite integral on the interval $[a,b]$, then we are asking what the area under the curve is between the points $a$ and $b$. Or, another way to think about it is between the vertical lines running through $a$ and $b$, under the curve and above the $x$-axis. Pictorally we have

https://www.google.com/url?sa=i&source=images&cd=&ved=2ahUKEwibo7KTsaDeAhVFxoMKHXOGDCcQjRx6BAgBEAU&url=https%3A%2F%2Fsocratic.org%2Fquestions%2Fwhat-is-the-difference-between-a-definite-and-indefinite-integral&psig=AOvVaw2n1iATiVXl_xxtypNe1KNy&ust=1540515313234350

So $a$ is the lower bound and $b$ the upper. When using functions as bounds it can seem a bit funny because they are not single values as you said, but really all they're saying is given an input this is what the bounds are. So when we do the evaluation for something like

$$\int_{x^2}^{\cos(x)} t^2 dt=\frac{1}{3}(t^3)\bigg|_{x^2}^{\cos(X)}=\frac{1}{3}(\cos^3(x)-x^6)$$

for any given value of $x$ we can determine the area below the curve between those two bounds.