Fundamental Theorem of Calculus

Before proceeding to the fundamental theorem, know its connection with calculus. There are 2 primary subdivisions of calculus i.e. – differential calculus and integral calculus. Both are inter-related to each other, even though the former evokes the tangent problem while the latter from the area problem. However, what creates a link between the two of them is the fundamental theorem of calculus (FTC). The Fundamental Theorem of Calculus denotes that differentiation and integration makes for inverse processes.

\[\int_{a}^{b} f(x) dx = F(x)|_{a}^{b} = F(b) - F(a)\]

Importance of Fundamental Theorem of Calculus in Mathematics

Until the inception of the fundamental theorem of calculus, it was not discovered that the operations of differentiation and integration were interlinked. And as discussed above, this mighty Fundamental Theorem of Calculus setting a relationship between differentiation and integration provides a simple technique to assess definite integrals without having to use calculating areas or Riemann sums.            

Fundamental Theorem of Calculus: Integrals & Anti Derivatives

The Fundamental Theorem of Calculus theorem that shows the relationship between the concept of derivation and integration, also between the definite integral and the indefinite integral— consists of 2 parts, the first of which, the Fundamental Theorem of Calculus, Part 1, and second is the Fundamental Theorem of Calculus, Part 2.

1. Fundamental Theorem of Calculus Part 1

Part 1 of Fundamental theorem creates a link between differentiation and integration.

By that, the first fundamental theorem of calculus depicts that, if “f” is continuous on the closed interval [a,b] and F is the unknown integral of “f” on [a,b], then

\[\int_{a}^{b} f(x) dx = F(x)|_{a}^{b} = F(b) - F(a)\].

This typically states the definite integral over an interval [a,b] is equivalent to the antiderivative calculated at ‘b’ minus the antiderivative assessed at ‘a’. This provides the link between the definite integral and the indefinite integral (antiderivative).

This outcome, while taught initially in primary calculus courses, is literally an intense outcome linking the purely algebraic indefinite integral and the purely evaluative geometric definite integral.

Given f is

• continuous on interval [a, b]

• F is any function that satisfies F’(x) = f(x)

Then

\[\int_{a}^{b} f(X)dx = F(b) - F(a)\]

2. Fundamental Theorem of Calculus Part 2

Within the theorem the second fundamental theorem of calculus, depicts the connection between the derivative and the integral— the two main concepts in calculus. The part 2 theorem is quite helpful in identifying the derivative of a curve and even assesses it at definite values of the variable when developing an anti-derivative explicitly which might not be easy otherwise.

The theorem bears ‘f’ as a continuous function on an open interval I and ‘a’ any point in I, and states that if “F” is demonstrated by

 F(x) = \[\int_a^{x} f(t)dt\],

then

 F'(x)=f(x) at each point in I.

The above expression represents that The fundamental theorem of calculus by the sides of curves shows that if f(z) has a continuous indefinite integral F(z) in an area R comprising of parameterized curve gamma:z=z(t) for alpha < = t < = beta, then

\[\int_\gamma f(z)dz = F(z(\beta))-F(z(\alpha))\].

\[\frac{d}{dx} \int_{a}^{x} f(t)dt = f(x)\]

• Derivative of an integral.

• Derivative matches the upper limit of integration.

• Lower limit of integration is a constant.

Practice Problem

Using First Fundamental Theorem of Calculus Part 1 Example

Problem

Two jockeys—Jessica and Anie are horse riding on a racing circuit. They are riding the horses through a long, straight track, and whoever reaches the farthest after 5 sec wins a prize. If Jessica can ride at a pace of f(t)=5+2t  ft/sec and Anie can ride at a pace of  g(t)=10+cos(π²t)  ft/sec. Find out who is going to win the horse race?

Solution

First, you need to combine both functions over the interval (0,5) and notice which value is bigger. For Jessica, we want to evaluate;-

∫50 (5+2t)dt.

Executing the power rule, we have

∫50[5+2t]dt=[5t+t2]∣∣50=[25+25]=50.

Thus, Jessica has ridden 50 ft after 5 sec.

 Now moving on to Anie, you want to evaluate

∫50(10) + cos(π²t)dt.

You recognize that sin ‘t’  is an antiderivative of cos, so it is rational to anticipate that an antiderivative of  cos(π²t)  would include  sin(π²t).  That said, when we know what’s what by differentiating sin(π²t),  we get  π²cos(π²t)  as an outcome of the chain theory, so we need to take into consideration this additional coefficient when we combine them. We have:

∫50 (10) + cos[π²t]dt=[10t+2πsin(π²t)]∣∣50=[50+2π]−[0−2πsin0]≈50.6.

Anie has ridden in an estimate 50.6 ft after 5 sec. Anie wins the race, but narrowly.

Second Fundamental Theorem of Calculus

Using First Fundamental Theorem of Calculus Part 1 Example

Problem

A ball is thrown straight up from the 5th floor of the building with a velocity v(t)=−32t+20ft/s, where t is calculated in seconds. identify, and interpret, ∫10v(t)dt.

Solution

Executing the Second Fundamental Theorem of Calculus, we see

∫10v[t]dt=∫10 [−32t+20]dt=[]10=4.

Therefore, if a ball is thrown upright into the air with velocity

v(t)= 4

The height of the ball, 1 second later, will be 4 feet high above the original height. Bear in mind that the ball went much farther. It traveled as high up to its peak and is falling down, still the difference between its height at t=0 and t=1 is 4ft.

Fun Facts

ü  And if you think Greeks invented calculus? No, they did not.

ü  Greeks created spectacular concepts with geometry, but not arithmetic or algebra very well.