Fundamental Theorem of Calculus Part II

From Derivatives Back to Definite Integrals

The Fundamental Theorem of Calculus, Part I, connects integration to differentiation: if $f$ is continuous and $A(x)=\int_a^x f(t)\,dt$, then $A'(x)=f(x)$ at every interior point. Part II uses that connection in the opposite direction. If we already know an antiderivative of $f$, we can evaluate the definite integral by measuring the change in that antiderivative between the endpoints.

The key point is that the accumulation function $A$ is itself an antiderivative of $f$. Any other antiderivative must differ from $A$ by a constant, so both antiderivatives have the same change from $a$ to $b$. The mean value theorem supplies the reason that a differentiable function with derivative zero cannot change across an interval.

Antiderivatives Differ by a Constant

First isolate the fact that allows an antiderivative to be substituted for the accumulation function. The endpoint continuity assumption lets us use the mean value theorem on the entire closed interval, even though differentiability is only required in its interior.

Theorem (Antiderivatives Differ by a Constant): Let $a<b$, and let $F,G:[a,b]\to\mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $F'(x)=G'(x)$ for every $x\in(a,b)$, then $F-G$ is constant on $[a,b]$.

Proof. Define $H(x)=F(x)-G(x)$. The continuity of $F$ and $G$ on $[a,b]$ implies that $H$ is continuous there, and their differentiability on $(a,b)$ implies that $H$ is differentiable there. For each $x\in(a,b)$,

$$ H'(x)=F'(x)-G'(x)=0. $$

Take any $u,v\in[a,b]$ with $u<v$. The mean value theorem applies to $H$ on $[u,v]$: $H$ is continuous on that closed interval and differentiable on its interior, which lies in $(a,b)$. Thus there is $c\in(u,v)$ such that

$$ H(v)-H(u)=H'(c)(v-u)=0. $$

Therefore $H(v)=H(u)$ whenever $u<v$. Reversing the order of any two distinct points gives the same equality, so $H$ has the same value everywhere on $[a,b]$. Hence $F-G$ is constant. $\square$

This result does not say that two antiderivatives have identical values. For instance, adding a fixed number to an antiderivative does not change its derivative. It says precisely that this fixed difference is the only freedom: two antiderivatives of the same function can differ in their values, but not in how much they change across an interval.

The Evaluation Formula

Let $f$ be continuous on $[a,b]$, and form its accumulation function $A(x)=\int_a^x f(t)\,dt$. By the Fundamental Theorem of Calculus, Part I, $A$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $A'(x)=f(x)$ throughout the interior. If $F$ is another antiderivative of $f$, the theorem just proved shows that $F-A$ is constant. In particular, its values at $a$ and $b$ agree. Since $A(a)=0$ and $A(b)=\int_a^b f(t)\,dt$, this gives the promised formula.

Theorem (Fundamental Theorem of Calculus, Part II): Let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be continuous. Suppose $F:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$, differentiable on $(a,b)$, and satisfies $F'(x)=f(x)$ for every $x\in(a,b)$. Then $$ \int_a^b f(x)\,dx=F(b)-F(a). $$

Proof. Define the accumulation function

A(x)=\int_a^x f(t)\,dt,\qquad x\in[a,b].

The Fundamental Theorem of Calculus, Part I, gives that $A$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $A'(x)=f(x)$ for every $x\in(a,b)$. By hypothesis, $F'(x)=f(x)$ on the same interval, so $F'(x)=A'(x)$ there. The theorem on antiderivatives differing by a constant now implies that $F-A$ is constant on $[a,b]$. Consequently,

$$ F(b)-A(b)=F(a)-A(a). $$

The definition of $A$ gives $A(a)=\int_a^a f(t)\,dt=0$ and $A(b)=\int_a^b f(t)\,dt$. Substituting these values into the preceding equality yields

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

Rearranging proves $\int_a^b f(t)\,dt=F(b)-F(a)$, as claimed. $\square$

The notation $F(b)-F(a)$ is often abbreviated as $[F(x)]_a^b$. The order of the endpoint values matters: the value at the upper endpoint is written first, and the value at the lower endpoint is subtracted.

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

Worked Applications

Worked Example: A Polynomial Integrand

Evaluate $\int_1^3(3x^2-4x+2)\,dx$. An antiderivative is

$$ F(x)=x^3-2x^2+2x, $$

because differentiating each term gives $F'(x)=3x^2-4x+2$. The hypotheses of FTC Part II hold: the integrand is continuous on $[1,3]$, and $F$ is an antiderivative there. Evaluate at the endpoints:

\begin{aligned} F(3)&=3^3-2(3^2)+2(3)=27-18+6=15,\\ F(1)&=1^3-2(1^2)+2(1)=1-2+2=1. \end{aligned}

Therefore,

\int_1^3(3x^2-4x+2)\,dx=F(3)-F(1)=15-1=14.

The endpoint values are values of the antiderivative, not values of the integrand. The integral records their difference.

Worked Example: A Reciprocal Integrand

Evaluate $\int_1^{e^2}\frac{1}{x}\,dx$. The function $1/x$ is continuous on $[1,e^2]$, since this interval does not contain zero. On this interval, $F(x)=\ln x$ is an antiderivative because $F'(x)=1/x$. FTC Part II gives

\int_1^{e^2}\frac{1}{x}\,dx =\ln(e^2)-\ln(1) =2-0 =2.

The endpoint restrictions matter here: the formula applies on an interval where the integrand is continuous and the logarithm is defined. It would not be valid to use this same argument across an interval containing zero.

Worked Example: A Trigonometric Integrand

Evaluate $\int_{\pi/6}^{\pi/2}\sin x\,dx$. Since $-\cos x$ is an antiderivative of $\sin x$, FTC Part II yields

\begin{aligned} \int_{\pi/6}^{\pi/2}\sin x\,dx &=[-\cos x]_{\pi/6}^{\pi/2}\\ &=-\cos\left(\frac{\pi}{2}\right) -\left(-\cos\left(\frac{\pi}{6}\right)\right)\\ &=0+\frac{\sqrt{3}}{2} =\frac{\sqrt{3}}{2}. \end{aligned}

The subtraction in the second line applies to the entire lower-endpoint value. Keeping the parentheses makes the sign unambiguous.

Worked Example: Reversing the Limits

Evaluate $\int_3^1 2x\,dx$, interpreting the integral with oriented limits. The function $F(x)=x^2$ satisfies $F'(x)=2x$. Thus the endpoint formula gives

\int_3^1 2x\,dx=F(1)-F(3)=1^2-3^2=1-9=-8.

Reversing the limits reverses the sign: $\int_3^1 2x\,dx=-\int_1^3 2x\,dx$. The antiderivative formula has this property automatically because reversing the endpoints changes $F(b)-F(a)$ into $F(a)-F(b)$.

Why the Choice of Antiderivative Does Not Matter

An indefinite integral is commonly written with a constant of integration. That constant has no effect on a definite integral because it appears in both endpoint values and cancels. Indeed, if $G(x)=F(x)+C$ for a fixed constant $C$, then

$$ G(b)-G(a)=(F(b)+C)-(F(a)+C)=F(b)-F(a). $$

The Antiderivatives Differ by a Constant theorem gives more than this check for one chosen $C$: it proves that every antiderivative of the same function has this form. Thus any antiderivative can be used in the evaluation formula.

There is also a useful contrast between the two parts of the Fundamental Theorem. Part I starts with a continuous function $f$ and constructs an antiderivative by accumulating its integral. Part II starts with an antiderivative and uses it to evaluate the integral. Together they explain why differentiation and definite integration undo one another under the stated continuity and differentiability conditions.

A common pitfall is to apply the endpoint formula without checking that an antiderivative exists on the whole interval, or that the integrand is continuous there. For example, $1/x$ has an antiderivative $\ln x$ on positive intervals, but it is not continuous on any interval containing zero. The theorem cannot be applied across zero merely by substituting endpoints into a formula. The hypotheses ensure that the Riemann integral and the comparison with an antiderivative are both justified.

Check Your Understanding

Use the theorem and examples to answer the following questions.

Why does the proof of FTC Part II compare an antiderivative $F$ with the accumulation function $A(x)=\int_a^x f(t)\,dt$?
Which theorem shows that two antiderivatives with the same derivative differ by a constant?
If $F$ is an antiderivative of $f$, what is the correct order of endpoint values in the formula for $\int_a^b f(x)\,dx$?
Why does adding a constant to an antiderivative leave the value of a definite integral unchanged?
What sign change should occur when the limits of a definite integral are reversed, and how does the endpoint formula show it?