AP Calculus BC · Study guide

The Fundamental Theorem of Calculus and Definite Integrals

AP Calculus · Integration and Accumulation of Change

The Fundamental Theorem and Definite Integrals

Infinitely many rectangles collapse to two evaluations and a subtraction — provided the function is continuous.

key idea$\int_a^b f(x)\,dx=F(b)-F(a)\quad\text{where }F'=f$

The theorem

The Fundamental Theorem of Calculus (evaluation form)

Hypotheses — all must hold

$f$ is continuous on $[a,b]$.
$F$ is any antiderivative of $f$ — that is, $F'=f$ on the interval.

What you may then conclude

$\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$. A limit of infinitely many rectangles collapses to two evaluations and a subtraction.

Continuity is not optional

$\displaystyle\int_{-1}^{1}\dfrac{1}{x^{2}}\,dx$ looks like it should be $\left[-\tfrac1x\right]_{-1}^{1}=-1-1=-2$. But the integrand is positive everywhere it is defined, so a negative answer is impossible. The function is discontinuous at $x=0$, inside the interval, so the theorem never applied.

$\int_1^3 2x\,dx$ is this shaded area

continuous → FTC applies

The definite integral is the shaded area. The FTC says you can get that number without ever summing rectangles — provided $f$ is continuous across the whole interval.

Checking the hypotheses

Evaluating a definite integral

Check continuity on $[a,b]$. Look for zero denominators, negative radicands and asymptotes inside the interval — not just at the endpoints.
Find any antiderivative $F$. The $+C$ cancels in the subtraction, so pick the simplest one.
Evaluate $F(b)-F(a)$, in that order. Reversing the order flips the sign.
Sanity-check the sign. A positive integrand cannot produce a negative answer.

Property	Statement	Why it holds
reversing limits	$\displaystyle\int_b^a f=-\int_a^b f$	$F(a)-F(b)=-\left[F(b)-F(a)\right]$
equal limits	$\displaystyle\int_a^a f=0$	$F(a)-F(a)=0$
the $+C$	cancels	$\left[F(b)+C\right]-\left[F(a)+C\right]$
net change	$\displaystyle\int_a^b F'(x)\,dx=F(b)-F(a)$	the same statement read backwards

The last row is how the FTC appears in applied questions: the integral of a rate gives the net change in the amount.

Worked examples

EX 1

Straight evaluation

$\int_1^3 2x\,dx$

Evaluate $\displaystyle\int_{1}^{3}2x\,dx$.

Check continuity

$2x$ is a polynomial, continuous everywhere. The FTC applies.

Find an antiderivative

$F(x)=x^{2}$, since $F'=2x$. Any $x^{2}+C$ would do.

Evaluate at the limits

$F(3)=9$ and $F(1)=1$.

Subtract in order

$9-1=8$.

Check geometrically

The region is a trapezoid with parallel sides $2$ and $6$ and width $2$: area $=\tfrac{2+6}{2}\cdot2=8$. ✓

EX 2

Why $+C$ does not matter

any antiderivative

Evaluate $\displaystyle\int_{0}^{2}3x^{2}\,dx$ using $F(x)=x^{3}$ and again using $F(x)=x^{3}+7$.

First antiderivative

$F(2)-F(0)=8-0=8$.

Second antiderivative

$F(2)-F(0)=(8+7)-(0+7)=15-7=8$.

Compare

Identical, as the theorem promises.

Explain

The constant appears in both terms of the subtraction and cancels.

Practical consequence

Never carry $+C$ through a definite integral. Choose the simplest antiderivative and move on.

EX 3

Net change in context

the applied form

A tank's inflow rate is $r(t)=6-t$ litres per minute for $0\le t\le 8$. How much does the volume change over that time?

Recognise the form

$r$ is a rate, so $\displaystyle\int_{0}^{8}r(t)\,dt$ gives the net change in volume.

Find an antiderivative

$R(t)=6t-\dfrac{t^{2}}{2}$.

Evaluate

$R(8)=48-32=16$; $R(0)=0$.

Subtract

$16-0=16$ litres.

Interpret carefully

The volume increased by 16 litres. Note $r$ turns negative after $t=6$, so water flowed out at the end — yet the net change is still positive.

EX 4

The discontinuity trap

when the FTC fails

A student evaluates $\displaystyle\int_{-1}^{1}\dfrac{1}{x^{2}}\,dx$ as $\left[-\tfrac1x\right]_{-1}^{1}=-1-1=-2$. Diagnose.

Check the arithmetic

$-\tfrac11=-1$ and $-\tfrac{1}{-1}=1$, so $-1-1=-2$. The computation is correct.

Sanity-check the sign

$\dfrac{1}{x^{2}}\gt 0$ wherever it is defined, so the area cannot be negative. Something is wrong.

Check the hypothesis

$\dfrac{1}{x^{2}}$ has a vertical asymptote at $x=0$, which lies inside $[-1,1]$. It is not continuous there.

State the conclusion

The FTC does not apply. The integral is improper and in fact diverges.

The habit that catches it

Scan the interval for discontinuities before antidifferentiating, and always sanity-check the sign afterwards.

When it fails

Scan the interval for discontinuities first

An asymptote strictly inside $[a,b]$ voids the theorem. The arithmetic will still produce a number — a wrong one.

A positive integrand cannot give a negative answer

This one check catches the discontinuity trap almost every time. Make it a habit.

$F(b)-F(a)$, in that order

Reversing gives the negative of the correct value. Write the upper limit first, every time.

Drop the $+C$

It cancels in the subtraction. Carrying it adds clutter and invites arithmetic slips.

Course map. Topic 6.7 · LO FUN-6.B — evaluate definite integrals analytically using the Fundamental Theorem of Calculus (EK FUN-6.B.1).