Fundamental Theorems of Integral Calculus and their Applications

Evaluating $\int_a^b f(x)\,dx$ as a limit of Riemann sums (§9.2) is correct but tedious, even for a very simple $f(x)$. Newton and Leibniz, working independently at around the same time, devised a vastly easier method based on two celebrated theorems connecting a function to its anti-derivative. Together these are called the Fundamental Theorems of Integral Calculus, and they provide the bridge between differential calculus and integral calculus.

Since $F(b)-F(a)$ is the value of the definite (Riemann) integral, adding any arbitrary constant $C$ to the anti-derivative $F(x)$ cancels out of the difference — so, unlike an indefinite integral, no $+C$ is needed when evaluating a definite integral. As shorthand, we write $F(b)-F(a)=\big[F(x)\big]_a^b$. The value of a definite integral is a unique real number.

Properties of definite integrals (all following from Theorem 9.2, stated here without proof except where a quick derivation illuminates the trick):

Property 1. $\displaystyle\int_a^b f(x)\,dx=\int_a^b f(u)\,du$ — the definite integral does not depend on the name of the integration variable.

Property 2. $\displaystyle\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$ — reversing the limits flips the sign.

Property 3. $\displaystyle\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$, $a<c<b$ — additivity over subintervals.

Property 4. $\displaystyle\int_a^b[\alpha f(x)+\beta g(x)]\,dx=\alpha\int_a^b f(x)\,dx+\beta\int_a^b g(x)\,dx$, $\alpha,\beta$ constants — linearity.

Property 5. If $x=g(u)$, then $\displaystyle\int_a^b f(x)\,dx=\int_c^d f(g(u))\,\dfrac{dg(u)}{du}\,du$, where $g(c)=a,\ g(d)=b$ — this is what justifies evaluating a definite integral by substitution (Examples 9.8-9.19 all use this).

Property 6. $\displaystyle\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$.

Proof. Substitute $u=a+b-x$, so $du=-dx$; when $x=a,u=b$ and when $x=b,u=a$. Then $\int_a^b f(x)\,dx=\int_b^a f(a+b-u)(-du)=\int_a^b f(a+b-u)\,du=\int_a^b f(a+b-x)\,dx$ (renaming $u\to x$, Property 1).

Property 7. $\displaystyle\int_0^{2a}f(x)\,dx=\int_0^a\big[f(x)+f(2a-x)\big]\,dx$.

Proof. By Property 3, $\int_0^{2a}f=\int_0^a f+\int_a^{2a}f$. In $\int_a^{2a}f(x)\,dx$ substitute $x=2a-u$: when $x=a,u=a$; when $x=2a,u=0$; $dx=-du$. So $\int_a^{2a}f(x)\,dx=\int_a^0 f(2a-u)(-du)=\int_0^a f(2a-u)\,du=\int_0^a f(2a-x)\,dx$. Adding gives the result.

Property 8 (even functions). If $f(-x)=f(x)$, then $\displaystyle\int_{-a}^a f(x)\,dx=2\int_0^a f(x)\,dx$.

Property 9 (odd functions). If $f(-x)=-f(x)$, then $\displaystyle\int_{-a}^a f(x)\,dx=0$.

Both Properties 8-9 proof idea: by Property 3, $\int_{-a}^a f=\int_{-a}^0 f+\int_0^a f$; substitute $x=-u$ in $\int_{-a}^0 f(x)\,dx$ to get $\int_0^a f(-u)\,du$, which equals $+\int_0^a f(u)\,du$ (even case, giving Property 8) or $-\int_0^a f(u)\,du$ (odd case, cancelling the other piece to give Property 9).

Property 10. If $f(2a-x)=f(x)$, then $\displaystyle\int_0^{2a}f(x)\,dx=2\int_0^a f(x)\,dx$ — immediate from Property 7.

Property 11. If $f(2a-x)=-f(x)$, then $\displaystyle\int_0^{2a}f(x)\,dx=0$ — also immediate from Property 7.

Property 12. If $f(a-x)=f(x)$, then $\displaystyle\int_0^a x\,f(x)\,dx=\dfrac a2\int_0^a f(x)\,dx$. …