~THE 2ND FUNDAMENTAL THEOREM~

~The 2^nd Fundamental Theorem of Calculus states the relationship between the two major operations of Calculus, Differentiation & Integration. The variable in the integrand is a “dummy” variable (i.e., it doesn’t matter if it’s a t, p, s, x, or any other letter, the outcome is unaffected)

In general, it would look like this: If F(x) = ∫_a^{^{^u}}f(x)dx, Then F'(x) = f(u) du/dx

u is a function of x, a is constant and u must be in the upper limit. If the u is in the lower limit, switch the limits of integration (introduces a minus sign in front of the integral).

Let's look at this problem: F(x) = ∫₂^{^{^x²}}t²dt, find F'(x). Since this integral fits the simple power rule, we would get, F(x)= t³/3]₂^{^{^x²} = (x²)³/3 - (2)³/3 = x⁶/3 - 8/3 = (1/3)x⁶ - 8/3

Since the equation for F(x) is known, we can take the derivative directly and get, F'(x) = 2x⁵. Notice that the constant term (no matter what it is) will be zero in the derivative, so, that means that it doesn't matter what the constant is in the lower limit of integration.

What the 2nd Fundamental theorem enables us to do is to find this derivative without having to go through the integration & evaluation process. This would be necessary, if the integral does not fit any of the rules or forms covered. We can still find it's derivative.

So, doing this problem using the 2^nd Fun Thm, we would have, F'(x) = (x²)² (2x) = x⁴ (2x) = 2x⁵, the same answer but a whole lot easier and quicker. The equation for F(x) does not have to be known. There lies the property of this theorem that is most useful.

The following example is a very important integral which actually defines the natural log.

F(x) = ∫₁^{^{^x}} (1/t)dt , F'(x) = 1/x This is the actual definition of ln(x) and it's derivative 1/x . In short, ln(x) is the area under the curve f(t)=1/t from 1 to x. (i.e., if x=2, that area will equal ln2, if x=3, the area will equal ln3, & so on). We actually get the derivative of ln(x) using the 2nd Fun Thm. Note that the power rule can not be used for 1/t since the power on t would be -1 in the numerator (the only number for a power that is not allowed).

More general problems dealing with the 2nd Fun Thm

Example 1): If F(x) = ∫₃^{^{^secx}} tan²x dx, find F'(x)

Solution: Sub in sec(x) for x in the integrand wherever there is an x (except in the dx), then multiply by it's derivative at the end.

we get: F'(x) =[tan²(secx)] (secx tanx). If we desired, we could use the trig identity

, tan²x = sec²x -1 and actually evaluate the integral, then take it's derivative directly. However, this option is not available in many problems, like the next example.

Example 2): If F(x) = ∫_{_x²+1}^⁵ e^3x cot(7x²-3x+7)dx , find F'(x)

Solution: Note that the variable quantity is in the lower limit. So, we need to switch the limits. This will introduce a negative sign in front of the integral.

So, F(x) = -∫₅^{^{^x²+1}}e^3x cot(7x²-3x+7)dx . Then F'(x) = - e^3(x²+1) cot[7(x² +1)² -3(x² +1) +7] (2x)}