Blog · Calculus guides · 28 October 2026 · 4 min read

Fundamental Theorem of Calculus: Examples & Practice Problems

Q: Do you need + C in a definite integral?

No. Any constant you add to F shows up in both F(b) and F(a) and cancels when you subtract.

Q: Why is t used instead of x inside the integral?

Because x is already busy being the upper limit. The t is a dummy variable: it runs from a to x and disappears once the integral is done.

An illustration of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus says differentiation and integration undo each other. Part 1: if $F (x) = \int_{a}^{x} f (t) d t$ , then $F^{'} (x) = f (x)$ . Part 2: if $F$ is any antiderivative of $f$ , then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , which is how you evaluate almost every definite integral.

The formula

Part 1: \frac{d}{d x} \int_{a}^{x} f (t) d t = f (x) Part 2: \int_{a}^{b} f (x) d x = F (b) - F (a)

Two lines that connect the slope problem to the area problem. Newton and Leibniz each got there first, depending on who you asked in 1700.

When to use it

Use Part 2 whenever you need the exact value of a definite integral and you can find an antiderivative: it replaces a limit of Riemann sums with one subtraction. Use Part 1 when the question asks for the derivative of an integral, especially one whose limits contain $x$ . If the upper limit is something other than plain $x$ , Part 1 needs the chain rule as well, and if both limits move, you split the integral and handle each end.

How to do it, step by step

Decide which part you need. A number between two limits: Part 2. A derivative of an integral with $x$ in a limit: Part 1.
For Part 2, find any antiderivative $F$ . No $+ C$ needed: it would appear in $F (b)$ and $F (a)$ and cancel.
Evaluate at the top, then subtract the bottom. $F (b) - F (a)$ , in that order, with brackets around $F (a)$ so its minus signs survive.
For Part 1, substitute the upper limit into the integrand. Replace $t$ with the upper limit, then multiply by the upper limit's derivative.
If the lower limit moves too, subtract its version. The lower limit contributes $- f (lower) \cdot (lower)^{'}$ .

Worked examples

Example 1: Part 2 with a polynomial

Easy

Evaluate:

\int_{1}^{3} (3 x^{2} - 2 x) d x

An antiderivative of $3 x^{2} - 2 x$ is $F (x) = x^{3} - x^{2}$ . Check by differentiating: $3 x^{2} - 2 x$ . Good.

\int_{1}^{3} (3 x^{2} - 2 x) d x = [x^{3} - x^{2}]_{1}^{3} = (27 - 9) - (1 - 1)

The bottom limit happens to give zero here. It will not always, so write the bracket out every time anyway.

Geometrically, $18$ is the area between the curve $y = 3 x^{2} - 2 x$ and the $x$ -axis from $1$ to $3$ , since the curve stays above the axis there. A Riemann sum could only have estimated it; the theorem gets it exactly with one subtraction.

Answer

\int_{1}^{3} (3 x^{2} - 2 x) d x = 18

Example 2: Part 1 with the chain rule

Medium

Find the derivative:

\frac{d}{d x} \int_{1}^{x^{2}} 1 + t^{3} d t

You could try to integrate $1 + t^{3}$ first, but it has no elementary antiderivative, so good luck with that. Part 1 lets you skip integrating entirely.

Put the upper limit $x^{2}$ in place of $t$ , then multiply by the derivative of $x^{2}$ , which is $2 x$ . The lower limit is a constant, so it contributes nothing.

\frac{d}{d x} \int_{1}^{x^{2}} 1 + t^{3} d t = 1 + (x^{2})^{3} \cdot 2 x

Tidy $(x^{2})^{3}$ to $x^{6}$ and move the $2 x$ to the front.

Answer

2 x 1 + x^{6}

Example 3: Both limits moving

Harder

Find the derivative, for $x > 0$ :

\frac{d}{d x} \int_{x}^{x^{3}} ln t d t

Split the integral at a fixed point, say $1$ , so each piece has one moving limit:

\int_{x}^{x^{3}} ln t d t = \int_{1}^{x^{3}} ln t d t - \int_{1}^{x} ln t d t

Apply Part 1 to each piece. The first gives $ln (x^{3}) \cdot 3 x^{2}$ ; the second gives $ln x \cdot 1$ .

\frac{d}{d x} \int_{x}^{x^{3}} ln t d t = 3 x^{2} ln (x^{3}) - ln x = 3 x^{2} \cdot 3 ln x - ln x

The general pattern: top limit in, times its derivative, minus bottom limit in, times its derivative. As a formula, $\frac{d}{d x} \int_{a (x)}^{b (x)} f (t) d t = f (b (x)) b^{'} (x) - f (a (x)) a^{'} (x)$ .

Here you could also integrate $ln t$ directly (by parts it is $t ln t - t$ ) and differentiate the result. You would get the same answer, after considerably more work.

Answer

9 x^{2} ln x - ln x = (9 x^{2} - 1) ln x

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Evaluate:

\int_{0}^{2} (x^{3} + 4 x - 1) d x

Problem 2

Medium

Evaluate:

\int_{0}^{1} 6 e^{2 x} d x

Problem 3

Harder

Find the derivative:

\frac{d}{d x} \int_{1}^{x^{3}} \frac{1}{1 + t ^{2}} d t

Common mistakes

Subtracting the wrong way round. It is $F (b) - F (a)$ : top minus bottom. Doing it backwards flips the sign of the answer and nothing else, which makes it easy to miss.
Dropping brackets on $F (a)$ . $(27 - 9) - (1 - 1)$ is not $27 - 9 - 1 - 1$ . Put the bottom value in brackets before subtracting it.
Forgetting the chain rule in Part 1. $\frac{d}{d x} \int_{1}^{x^{2}} f (t) d t$ is $f (x^{2}) \cdot 2 x$ , not $f (x^{2})$ . The plain version only works when the upper limit is exactly $x$ .
Integrating when you only needed to differentiate. For $\frac{d}{d x} \int_{a}^{x} f (t) d t$ there is nothing to integrate: the answer is $f (x)$ . Working out the integral first is not wrong, just slow, and often impossible.
Leaving $t$ in the answer. After Part 1, the answer is a function of $x$ . If a $t$ survives, you have not substituted the limit in.

Practice it now

Five minutes, no account

Part 2 is what every definite integral in a Sprint runs on: single powers from about the third problem, then polynomials, sines, cosines and exponentials between small limits once the difficulty reaches 4. There are no Part 1 (derivative of an integral) problems in the game yet.

Play a Sprint See it on the formula sheet

FAQ

What is the difference between Part 1 and Part 2 of the Fundamental Theorem of Calculus?: Part 1 says the derivative of an accumulated area is the original function. Part 2 says you can compute an area with any antiderivative: $F (b) - F (a)$ . Part 1 is the reason Part 2 works.
Do you need $+ C$ in a definite integral?: No. Any constant you add to $F$ shows up in both $F (b)$ and $F (a)$ and cancels when you subtract.
Why is $t$ used instead of $x$ inside the integral?: Because $x$ is already busy being the upper limit. The $t$ is a dummy variable: it runs from $a$ to $x$ and disappears once the integral is done.
Does the Fundamental Theorem of Calculus need $f$ to be continuous?: Yes, on the interval in question. A function with a jump or a vertical asymptote between $a$ and $b$ breaks it, which is how people "prove" that $\int_{- 1}^{1} \frac{1}{x ^{2}} d x = - 2$ .
Where can I find Fundamental Theorem of Calculus practice problems?: Right on this page: "Your turn" has three Fundamental Theorem of Calculus practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

When to use it

x

. If the upper limit is something other than plain

x

, Part 1 needs the chain rule as well, and if both limits move, you split the integral and handle each end.

How to do it, step by step

Decide which part you need. A number between two limits: Part 2. A derivative of an integral with

x

in a limit: Part 1.

For Part 2, find any antiderivative $F$ . No

+ C

needed: it would appear in

F (b)

and

F (a)

and cancel.

Evaluate at the top, then subtract the bottom.

F (b) - F (a)

, in that order, with brackets around

F (a)

so its minus signs survive.

For Part 1, substitute the upper limit into the integrand. Replace

t

with the upper limit, then multiply by the upper limit's derivative.

If the lower limit moves too, subtract its version. The lower limit contributes

- f (lower) \cdot (lower)^{'}

Worked examples

Example 1: Part 2 with a polynomial

Easy

Evaluate:

\int_{1}^{3} (3 x^{2} - 2 x) d x

An antiderivative of $3 x^{2} - 2 x$ is $F (x) = x^{3} - x^{2}$ . Check by differentiating: $3 x^{2} - 2 x$ . Good.

\int_{1}^{3} (3 x^{2} - 2 x) d x = [x^{3} - x^{2}]_{1}^{3} = (27 - 9) - (1 - 1)

The bottom limit happens to give zero here. It will not always, so write the bracket out every time anyway.

Answer

\int_{1}^{3} (3 x^{2} - 2 x) d x = 18

Example 2: Part 1 with the chain rule

Medium

Find the derivative:

\frac{d}{d x} \int_{1}^{x^{2}} 1 + t^{3} d t

You could try to integrate $1 + t^{3}$ first, but it has no elementary antiderivative, so good luck with that. Part 1 lets you skip integrating entirely.

Put the upper limit $x^{2}$ in place of $t$ , then multiply by the derivative of $x^{2}$ , which is $2 x$ . The lower limit is a constant, so it contributes nothing.

\frac{d}{d x} \int_{1}^{x^{2}} 1 + t^{3} d t = 1 + (x^{2})^{3} \cdot 2 x

Tidy $(x^{2})^{3}$ to $x^{6}$ and move the $2 x$ to the front.

Answer

2 x 1 + x^{6}

Example 3: Both limits moving

Harder

Find the derivative, for $x > 0$ :

\frac{d}{d x} \int_{x}^{x^{3}} ln t d t

Split the integral at a fixed point, say $1$ , so each piece has one moving limit:

\int_{x}^{x^{3}} ln t d t = \int_{1}^{x^{3}} ln t d t - \int_{1}^{x} ln t d t

Apply Part 1 to each piece. The first gives $ln (x^{3}) \cdot 3 x^{2}$ ; the second gives $ln x \cdot 1$ .

\frac{d}{d x} \int_{x}^{x^{3}} ln t d t = 3 x^{2} ln (x^{3}) - ln x = 3 x^{2} \cdot 3 ln x - ln x

Here you could also integrate $ln t$ directly (by parts it is $t ln t - t$ ) and differentiate the result. You would get the same answer, after considerably more work.

Answer

9 x^{2} ln x - ln x = (9 x^{2} - 1) ln x

Your turn

Three to try, checked by the same machine that grades the game. Type the answer however you like; any correct form counts.

Problem 1

Easy

Evaluate:

\int_{0}^{2} (x^{3} + 4 x - 1) d x

Problem 2

Medium

Evaluate:

\int_{0}^{1} 6 e^{2 x} d x

Problem 3

Harder

Find the derivative:

\frac{d}{d x} \int_{1}^{x^{3}} \frac{1}{1 + t ^{2}} d t

Common mistakes

Subtracting the wrong way round. It is

F (b) - F (a)

: top minus bottom. Doing it backwards flips the sign of the answer and nothing else, which makes it easy to miss.

Dropping brackets on $F (a)$ .

(27 - 9) - (1 - 1)

is not

27 - 9 - 1 - 1

. Put the bottom value in brackets before subtracting it.

Forgetting the chain rule in Part 1.

\frac{d}{d x} \int_{1}^{x^{2}} f (t) d t

f (x^{2}) \cdot 2 x

, not

f (x^{2})

. The plain version only works when the upper limit is exactly

x

Integrating when you only needed to differentiate. For

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

there is nothing to integrate: the answer is

f (x)

. Working out the integral first is not wrong, just slow, and often impossible.

Leaving $t$ in the answer. After Part 1, the answer is a function of

x

. If a

t

survives, you have not substituted the limit in.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

What is the difference between Part 1 and Part 2 of the Fundamental Theorem of Calculus?

Part 1 says the derivative of an accumulated area is the original function. Part 2 says you can compute an area with any antiderivative:

F (b) - F (a)

. Part 1 is the reason Part 2 works.

Do you need

+ C

in a definite integral?

No. Any constant you add to

F

shows up in both

F (b)

and

F (a)

and cancels when you subtract.

Why is

t

used instead of

x

inside the integral?

Because

x

is already busy being the upper limit. The

t

is a dummy variable: it runs from

a

x

and disappears once the integral is done.

Does the Fundamental Theorem of Calculus need

f

to be continuous?

Yes, on the interval in question. A function with a jump or a vertical asymptote between

a

and

b

breaks it, which is how people "prove" that

\int_{- 1}^{1} \frac{1}{x ^{2}} d x = - 2

Where can I find Fundamental Theorem of Calculus practice problems?

Right on this page: "Your turn" has three Fundamental Theorem of Calculus practice problems, easy, medium and harder, checked as you type by the same checker that grades Asymptote, with the answer and the first step if you miss. For more practice against the clock, a five-minute Sprint is free and needs no account.

Fundamental Theorem of Calculus: Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Part 2 with a polynomial

Example 2: Part 1 with the chain rule

Example 3: Both limits moving

Your turn

Common mistakes

FAQ

Keep going

Fundamental Theorem of Calculus: Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: Part 2 with a polynomial

Example 2: Part 1 with the chain rule

Example 3: Both limits moving

Your turn

Common mistakes

FAQ

Keep going