Math Core

Riemann sums and geometry can only take you so far: most regions aren't made of triangles and circles. The Fundamental Theorem of Calculus connects integrals to derivatives, and that connection turns the evaluation of a definite integral into a short calculation with an antiderivative.

Part 1: differentiating an accumulation function

In the last few lessons you saw that the accumulation function $g(x) = \displaystyle\int_a^x f(t)\,dt$ grows at a rate equal to the height of $f$ . Stated formally:

Fundamental Theorem of Calculus, Part 1

If $f$ is continuous on an interval containing $a$ , then for every $x$ in that interval,

\frac{d}{dx} \int_a^x f(t)\,dt = f(x).

In words: integrating $f$ and then differentiating brings you back to $f$ .

This also tells you something remarkable: every continuous function has an antiderivative, namely its own accumulation function. That's true even for functions like $e^{t^2}$ , whose antiderivative has no formula built from familiar functions.

Worked example: Part 1 directly

Find $\dfrac{d}{dx} \displaystyle\int_2^x \sqrt{1 + t^3}\,dt$ .

The lower limit is a constant and the upper limit is exactly $x$ , so replace $t$ with $x$ :

\frac{d}{dx} \int_2^x \sqrt{1 + t^3}\,dt = \sqrt{1 + x^3}.

The lower limit 2 doesn't matter: changing it would only add a constant to the accumulation function, and constants have derivative 0.

When the upper limit is a function of x

If the upper limit is $u(x)$ instead of $x$ , the accumulation function is a composition, so the chain rule applies:

\frac{d}{dx} \int_a^{u(x)} f(t)\,dt = f\big(u(x)\big) \cdot u'(x).

If $x$ appears in the lower limit, reverse the limits first to get a minus sign.

Worked example: Chain rule with an integral

Find $\dfrac{d}{dx} \displaystyle\int_1^{x^2} \cos t\,dt$ and $\dfrac{d}{dx} \displaystyle\int_x^{5} (t^2 + 1)\,dt$ .

For the first, plug the upper limit into the integrand and multiply by its derivative:

\frac{d}{dx} \int_1^{x^2} \cos t\,dt = \cos(x^2) \cdot 2x.

For the second, flip the limits: $\displaystyle\int_x^5 (t^2 + 1)\,dt = -\int_5^x (t^2 + 1)\,dt$ . So

\frac{d}{dx} \int_x^{5} (t^2 + 1)\,dt = -(x^2 + 1).

Part 2: evaluating definite integrals

Now suppose $F$ is any antiderivative of $f$ , meaning $F'(x) = f(x)$ . The accumulation function $g(x) = \displaystyle\int_a^x f(t)\,dt$ is also an antiderivative, and two antiderivatives of the same function differ by a constant: $g(x) = F(x) + C$ . Plug in $x = a$ : $0 = F(a) + C$ , so $C = -F(a)$ . Then plug in $x = b$ :

\int_a^b f(t)\,dt = g(b) = F(b) - F(a).

Fundamental Theorem of Calculus, Part 2

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ , then

\int_a^b f(x)\,dx = F(b) - F(a).

The shorthand $\Big[F(x)\Big]_a^b$ means $F(b) - F(a)$ .

To use Part 2, you need antiderivatives. Each derivative rule you know, read backward, gives one:

Function $f(x)$	An antiderivative $F(x)$
$x^n$ with $n \ne -1$	$\dfrac{x^{n+1}}{n+1}$
$\dfrac{1}{x}$	$\ln\lvert x\rvert$
$e^x$	$e^x$
$\cos x$	$\sin x$
$\sin x$	$-\cos x$
$\sec^2 x$	$\tan x$

You can check any row by differentiating the right column. The next lesson develops these rules fully.

The graph of y = x² − 2x from x = 1 to x = 3. The area below the axis on [1, 2] is 2/3; the area above on [2, 3] is 4/3. The net area is 2/3.Open in grapher →

Worked example: Using Part 2

Evaluate $\displaystyle\int_1^3 (x^2 - 2x)\,dx$ .

An antiderivative is $F(x) = \dfrac{x^3}{3} - x^2$ (check: $F'(x) = x^2 - 2x$ ).

\int_1^3 (x^2 - 2x)\,dx = \left[\frac{x^3}{3} - x^2\right]_1^3 = (9 - 9) - \left(\frac{1}{3} - 1\right) = 0 + \frac{2}{3} = \frac{2}{3}.

This is the net area. The graph shows why it's smaller than the total area: the region below the axis on $[1, 2]$ cancels part of the region above it on $[2, 3]$ .

Worked example: A trigonometric integral

Evaluate $\displaystyle\int_0^{\pi/2} \cos x\,dx$ .

\int_0^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1.

Common mistake

Subtract the whole value $F(a)$ , not just its first term. Put $F(a)$ in parentheses: $(9 - 9) - \left(\frac{1}{3} - 1\right)$ . Dropping the parentheses gives $0 - \frac{1}{3} - 1$ , which is a sign error.

The net change theorem

Part 2 has a powerful reading when you apply it to a derivative. Since $f$ is an antiderivative of $f'$ ,

\int_a^b f'(x)\,dx = f(b) - f(a), \qquad \text{so} \qquad f(b) = f(a) + \int_a^b f'(x)\,dx.

Future value = present value + accumulated change. This is the backbone of countless AP free-response questions: you're given a starting value and a rate, and you integrate the rate to find a later value.

Worked example: Position from velocity

A particle moves along a line with velocity $v(t) = 3t^2 - 4$ . Its position at $t = 2$ is $s(2) = 7$ . Find $s(4)$ .

s(4) = s(2) + \int_2^4 (3t^2 - 4)\,dt = 7 + \Big[t^3 - 4t\Big]_2^4 = 7 + \big((64 - 16) - (8 - 8)\big) = 7 + 48 = 55.

Tip

On the calculator-active part of the exam, you can evaluate $\displaystyle\int_a^b f'(x)\,dx$ numerically even when you can't find an antiderivative. Write the integral expression first, then the value; the setup earns the points.

Practice

Practice 1

Find $\dfrac{d}{dx} \displaystyle\int_0^x e^{t^2}\,dt$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Find $\dfrac{d}{dx} \displaystyle\int_0^{3x} \sin(t^2)\,dt$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

Find $\dfrac{d}{dx} \displaystyle\int_x^{2} \sqrt{t^4 + 1}\,dt$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Evaluate $\displaystyle\int_1^4 3\sqrt{x}\,dx$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Evaluate $\displaystyle\int_1^{e^2} \frac{2}{x}\,dx$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

A particle moves along a line with velocity $v(t) = 2t + 1$ . Its position at $t = 1$ is 10. Find its position at $t = 6$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Let $g(x) = \displaystyle\int_0^{x^2} f(t)\,dt$ , where $f$ is continuous with $f(2) = 5$ and $f(4) = 3$ . What is $g'(2)$ ?