Do I need a constant +C for definite integrals?

No. When you calculate $F(b) - F(a)$, the $+C$ cancels out: $(F(b) + C) - (F(a) + C) = F(b) - F(a)$. It is safe to just leave it out.

What if the lower limit is a variable instead of a constant?

You can flip the limits of integration. This changes the sign of the integral, so $\int_x^a f(t) dt = -\int_a^x f(t) dt$. Then you can apply the first part of the FTC normally.

Why are there two different parts to the theorem?

Part 1 establishes the theoretical connection that derivatives and integrals are opposites. Part 2 provides the practical formula we use to calculate exact areas and accumulated totals.

What is the Fundamental Theorem of Calculus saying?

The Fundamental Theorem of Calculus (FTC) says that differentiation and integration are reverse processes. If you have a function representing a rate of change, integrating it gives you the total accumulated change. Imagine you are driving a car. If you know your exact speed at every moment (the derivative), you can figure out exactly how far you have traveled over a certain time (the integral).<br><br>The theorem bridges the gap between calculating instantaneous rates (derivatives) and finding areas under curves (integrals). Before this theorem, mathematicians had to use tedious, complex formulas to find areas. The FTC showed that if you already know how to take a derivative, you already have the key to finding the area.

The First Part (Accumulation)

The first part of the FTC states that if you define a new function as the integral of $f(t)$ from a constant $a$ to a variable $x$ , the derivative of this new function is just $f(x)$ . In math terms, if $F(x) = \int_a^x f(t) dt$ , then $F'(x) = f(x)$ . This proves that integrating a function and then differentiating it gets you right back where you started.

The Second Part (Evaluation)

The second part is the one you will use most often to solve homework problems. It gives us a shortcut to evaluate definite integrals without calculating infinite sums. It says that $\int_a^b f(x) dx = F(b) - F(a)$ , where $F(x)$ is any antiderivative of $f(x)$ . You just find the antiderivative, plug in the top limit, plug in the bottom limit, and subtract.

Where Students Slip Up

A common mistake is forgetting the Chain Rule when applying the first part of the FTC. If the upper limit of your integral is a function like $x^2$ instead of just $x$ , you have to multiply by the derivative of that upper limit. Another frequent error is mixing up the order of subtraction in the second part; always subtract the value at the lower limit from the value at the upper limit, not the other way around.

Worked through

Evaluate the definite integral $\int_1^3 (2x) dx$ using the Fundamental Theorem of Calculus.

First, find the antiderivative of $f(x) = 2x$ . Using the power rule backwards, the antiderivative is $F(x) = x^2$ . Next, apply the second part of the FTC: $\int_1^3 2x dx = F(3) - F(1)$ . Plug in the limits: $F(3) = 3^2 = 9$ and $F(1) = 1^2 = 1$ . Finally, subtract: $9 - 1 = 8$ . The area under the curve $y = 2x$ from $x=1$ to $x=3$ is exactly 8.

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Where this comes from: OpenStax Calculus Volume 1, Chapter 5 · Khan Academy: Fundamental Theorem of Calculus · Stewart Calculus, Chapter 5