Fundamental Theorem of Calculus

Theorem I

If $g$ is continuous on $[a,b]$ that is differentiable on $(a,b)$ and if $g'$ is integrable on $[a,b]$ then

$$\int_a^b g' = g(b) - g(a)$$

Proof Hint

Consider a general partition and use Mean Value Theorem on each parition.

Integration by parts

Suppose $u,v$ are continuous functions on $[a,b]$ that are differentiable on $(a,b)$. If $u'$ and $v'$ are Riemann integrable on $[a,b]$:

$$\int_a^b u(x)v'(x)\,\text{d}x + \int_a^b u'(x)v(x)\,\text{d}x = u(b)v(b) - u(a)v(a)$$

Proof Hint

Consider $g=uv$ and use FTC I. $$

Theorem II

Suppose $f$ is an Riemann integrable function on $[a,b]$. For $x\in(a,b)$.

$$F(x)=\int_a^x f(t)\,\text{d}t$$

• $F(x)$ is uniformly continuous on $[a,b]$
• $f$ is continuous at $x_0 \in (a,b) \implies F$ is differentiable and $F'(x_0) = f(x_0)$

Proof Hint

For the first point:

• Consider 2 points in the interval $x,y\,(>x)$ such that $|x-y|\lt\delta=\frac{\epsilon}{M}$
• Show $\lvert F(y)-F(x) \rvert \le \epsilon$

For the second point: Consider the continuity definition of $f$ and prove is quite trivial. $$

$$\bigg\lvert \frac {F(x) - F(x_0)} {x-x_0} -f(x_0) \bigg\rvert \lt \epsilon$$

Theorem

Suppose $f$ is Riemann integrable on an open interval $I$ containing the values of differentiable functions $a,b$. Then:

$$\frac {\text{d}} {\text{d}x} \int_{a(x)}^{b(x)} f(t)\,\text{d}t = f(b(x))b'(x) - f(a(x))a'(x)$$

Proof Hint

Can be done using FTC I and II. Proof is quite trivial.

Theorem - Change of Variable

Suppose $u$ is a differentiable function on an open interval $J$ such that $u'$ is continuous. Let $I$ be an open interval such that $\forall x \in J,\;u(x) \in I$.

If $f$ is continuous on $I$, then $f \circ u$ is continuous on $J$ and:

$$\int_a^b{(f\circ u)(x)\,u'(x)\,\text{d}x} = \int_{u(a)}^{u(b)} f(u)\,\text{d}u$$