91Ó°ÊÓ

The indefinite integral is essentially the reverse process of differentiation. Think of it like un-doing a mathematical operation to find out what function was originally differentiated. Mathematicians call these anti-derivatives. When you see the notation \( \int f(x) \,dx \), it's asking you to find all functions \( F(x) \) whose derivative is \( f(x) \).

An important point is that there are infinitely many anti-derivatives for a given function, differing only by a constant. This is because differentiating a constant gives zero. So, if you're asked to find the indefinite integral of \( \sqrt{t} \), you're looking for a function that, when differentiated, gives \( \sqrt{t} \). Using the integration power rule, \( \int t^{n} dt = \frac{t^{n+1}}{n+1} + C \), where \( C \) is the integration constant and \( n eq -1 \), you'll arrive at the family of functions \( \frac{2}{3}t^{3/2} + C \). The \( C \) here accommodates all the possible functions with derivatives equal to \( \sqrt{t} \).

Differentiation is like applying a mathematical microscope to understand the rate at which a quantity changes. In calculus, the power rule is a shortcut that lets you find the derivative of a function with a variable raised to a power. The rule states that if you have \( x^n \), its derivative is \( nx^{n-1} \). So, the power rule essentially reduces the exponent by one and multiplies the entire expression by the original exponent.

In the context of our exercise, the function we're differentiating is \( F(x) = \frac{2}{3}(x^{3/2} - 4^{3/2}) \). The derivative of \( x^{3/2} \) using the power rule is \( \frac{3}{2}x^{3/2 - 1} = \frac{3}{2}x^{1/2} \), but since we have \( \frac{2}{3} \) as a coefficient, it simplifies to \( x^{1/2} \) – and the constant term \( 4^{3/2} \) vanishes upon differentiation.

We say that a function is continuous at a point if there's no interruption in the graph of the function at that point. In essence, you could draw the function at this point without lifting your pencil from the paper. For real-valued functions, this means that as the input values approach a certain point, the output of the function approaches the value of the function at that point.

In the exercise, the function \( f(t) = \sqrt{t} \) is continuous for all \( t > 0 \) because for all positive values of \( t \), you can draw \( \sqrt{t} \) without any jumps or breaks. Continuity is key for the Second Fundamental Theorem of Calculus, which links the concept of indefinite integration with differentiation. It guarantees that if the original function (like \( \sqrt{t} \) here) is continuous, then the function we get from integrating it (and then taking its derivative as in the solution) will indeed take us back to our original function.