What are the two parts of the Fundamental Theorem of Calculus?

Part 1: the derivative of int_a^xf(t)dt is f(x). Part 2: int_a^bf(x)dx=F(b)-F(a) for any antiderivative F of f.

Why don't you need +C in a definite integral?

Because it cancels: (F(b)+C)-(F(a)+C)=F(b)-F(a). Any antiderivative gives the same answer.

How do you take the derivative of an integral with x in the upper limit?

Put the upper limit into the integrand and multiply by the upper limit's derivative. If the upper limit is plain x, that factor is 1.

Why is the Fundamental Theorem of Calculus important?

It shows that differentiation and integration undo each other, and it lets you evaluate integrals exactly with antiderivatives instead of limits of sums.

Can the result of a definite integral be negative?

Yes, when more of the area lies below the x-axis than above it. The integral is a net signed area.

The Fundamental Theorem of Calculus Explained Simply

Why is it called the fundamental theorem?

Calculus has two big ideas that look unrelated. The derivative measures a rate of change, like a slope. The definite integral measures an accumulated total, like an area, defined as a limit of sums of thin rectangles. The Fundamental Theorem of Calculus shows that these are two sides of one idea, each undoing the other. That's what makes integrals practical: instead of adding up rectangles, you can evaluate them with antiderivatives.

What is an area function?

Fix a starting point $a$ and let the right edge $x$ move. The area under the curve $y = f (t)$ from $a$ to $x$ is then a function of $x$ , called an area function:

g (x) = \int_{a}^{x} f (t) d t

The letter inside changes to $t$ only so that $x$ can name the moving edge; $t$ is just the variable along the axis.

The area function

g (x)

is the shaded area under

y = f (t)

from a fixed start to the moving edge at

x

What does Part 1 of the Fundamental Theorem say?

The Fundamental Theorem of Calculus, Part 1

f

is continuous, then

\frac{d}{d x} \int_{a}^{x} f (t) d t =

f (x)

Why it works

Move the edge from $x$ to $x + h$ . The extra area $g (x + h) - g (x)$ is a thin strip of width $h$ .
Over that thin strip, $f$ barely changes, so the strip is almost a rectangle of height $f (x)$ : its area is about $f (x) \cdot h$ .
So $\frac{g (x + h) - g (x)}{h} \approx \frac{f (x) \cdot h}{h} =$ $f (x)$ , and the approximation becomes exact as $h \to 0$ .
The left side is the difference quotient, so its limit is $g^{'} (x)$ . Therefore $g^{'} (x) = f (x)$ .

In words: the rate at which area accumulates is the height of the curve at the moving edge. So the area function is an antiderivative of $f$ , a function whose derivative is $f$ . No integrating is needed to use Part 1.

Worked example: Part 1 directly

Find $\frac{d}{d x} \int_{2}^{x} \frac{t}{1 + t^{2}} d t$ .

The lower limit is a constant and the upper limit is plain $x$ , so Part 1 applies directly.
Replace $t$ by $x$ in the integrand.

Answer: $\frac{x}{1 + x^{2}}$

If the upper limit is a function of $x$ , like $x^{3}$ , the area function is a composite, so the chain rule adds a factor: put the upper limit into the integrand, then multiply by the derivative of the upper limit.

Worked example: Part 1 with the chain rule

Find $\frac{d}{d x} \int_{1}^{x^{3}} \ln t d t$ for $x > 0$ .

Put the upper limit $x^{3}$ into the integrand: $\ln (x^{3})$ .
Multiply by the derivative of the upper limit, $3 x^{2}$ : $\ln (x^{3}) \cdot 3 x^{2}$ .
Simplify with $\ln (x^{3}) = 3 \ln x$ : $3 \ln x \cdot 3 x^{2} =$ $9 x^{2} \ln x$ .

Answer: $9 x^{2} \ln x$

Common mistake

Integrating first, or forgetting the chain rule factor. Part 1 needs no antiderivative at all, and when the upper limit isn't plain $x$ the answer must include its derivative. If $x$ is the lower limit, flip the limits first, which flips the sign.

What does Part 2 of the Fundamental Theorem say?

The Fundamental Theorem of Calculus, Part 2

f

is continuous on

[a, b]

and

F

is any antiderivative of

f

(so

F^{'} = f

), then

\int_{a}^{b} f (x) d x =

F (b) - F (a)

Why it works

By Part 1, the area function $g (x) =$ $\int_{a}^{x} f (t) d t$ is an antiderivative of $f$ .
Two antiderivatives of the same function differ by a constant (their difference has derivative $0$ ), so $F (x) = g (x) + C$ .
Then $F (b) - F (a) =$ $(g (b) + C) - (g (a) + C) =$ $g (b) - g (a)$ .
And $g (b) =$ $\int_{a}^{b} f (t) d t$ while $g (a) =$ $\int_{a}^{a} f (t) d t =$ $0$ . So $F (b) - F (a) =$ $\int_{a}^{b} f (t) d t$ .

The difference $F (b) - F (a)$ is often written $[F (x)]_{a}^{b}$ . Any antiderivative works, so the $+ C$ can be left out: it would cancel in the subtraction anyway.

Worked example: A polynomial

Evaluate $\int_{1}^{3} (3 x^{2} - 2 x +$ $1) d x$ .

Antiderivative, term by term with the power rule in reverse ( $x^{n} \to \frac{x^{n + 1}}{n + 1}$ ): $F (x) = x^{3} - x^{2} + x$ .
$F (3) = 27 - 9 + 3 = 21$ .
$F (1) = 1 - 1 + 1 = 1$ .
$F (3) - F (1) = 21 - 1 = 20$ .

Answer: $20$

Worked example: A trig function

Evaluate $\int_{0}^{π} \sin x d x$ .

An antiderivative of $\sin x$ is $- \cos x$ , because $\frac{d}{d x} (- \cos x) =$ $\sin x$ .
$[- \cos x]_{0}^{π} =$ $- \cos π - (- \cos 0)$ .
$= - (- 1) + 1 = 2$ .

Answer: $2$ : the area under one arch of the sine curve.

Worked example: A logarithm

Evaluate $\int_{1}^{e} \frac{1}{x} d x$ .

An antiderivative of $\frac{1}{x}$ is $\ln | x |$ .
$\ln e - \ln 1 = 1 - 0 = 1$ .

Answer: $1$

Common mistake

Subtracting in the wrong order. It's always the top limit minus the bottom limit: $F (b) - F (a)$ . Also, the theorem needs $f$ continuous on the whole interval: $\int_{- 1}^{1} \frac{1}{x^{2}} d x$ can't be done this way, because $\frac{1}{x^{2}}$ blows up at $x = 0$ .

How do you find total area instead of net area?

The definite integral counts area below the $x$ -axis as negative. To find the total area between a graph and the axis, find where the function changes sign, integrate over each piece separately, and add the sizes, ignoring signs.

Worked example: Net area and total area

For $f (x) = x^{2} - 2 x - 3$ on $[0, 4]$ , find $\int_{0}^{4} f (x) d x$ and the total area between the graph and the $x$ -axis.

Factor: $f (x) = (x - 3) (x + 1)$ , so on $[0, 4]$ it is $0$ at $x = 3$ : negative on $[0, 3]$ and positive on $[3, 4]$ .
Antiderivative: $F (x) =$ $\frac{x^{3}}{3} - x^{2} - 3 x$ .
$F (0) = 0$ , $F (3) = 9 - 9 - 9 = - 9$ , and $F (4) =$ $\frac{64}{3} - 16 - 12 =$ $\frac{64}{3} - 28 =$ $- \frac{20}{3}$ .
From $0$ to $3$ : $F (3) - F (0) = - 9$ , an area of $9$ below the axis.
From $3$ to $4$ : $F (4) - F (3) =$ $- \frac{20}{3} + 9 =$ $\frac{7}{3}$ , above the axis.
Net: $- 9 + \frac{7}{3} =$ $- \frac{20}{3}$ . Total: $9 + \frac{7}{3} =$ $\frac{34}{3}$ .

Answer: The integral is $- \frac{20}{3}$ ; the total area is $\frac{34}{3}$ .

y = x^{2} - 2 x - 3

[0, 4]

: the region below the axis (area

9

) outweighs the region above it (area

\frac{7}{3}

), so the integral is negative.

Tip

For velocity, the same distinction separates displacement (the integral of velocity: where you end up relative to the start) from total distance traveled (the integral of speed, found by splitting where the velocity changes sign).

When no antiderivative is obvious, the most common technique in a first course is u-substitution, which reverses the chain rule.

The Fundamental Theorem of Calculus, explained