Theorem 8.1.1. Let $F \colon [a,b] \to \R$ be a continuous function, differentiable on $(a,b)$. Let $f \in \mathcal R[a,b]$ be such that $f(x) = F'(x)$ for $x \in (a,b)$. Then \begin{equation*} \int_a^b f = F(b)-F(a) . \end{equation*}

Let $P$ a partition of $[a,b].$ We apply the mean value theorem to $F$ on each subinterval $[x_{k-1},x_k]$ of $P$. This yields a point $t_k\in (x_{k-1},x_k)$ where

Let $P$ a partition of $[a,b].$ We apply the mean value theorem to $F$ on each subinterval $[x_{k-1},x_k]$ of $P$. This yields a point $t_k\in (x_{k-1},x_k)$ where

Now, consider upper and lower sums $U(P, f)$ and $L(P,f)$. Since $m_k\leq f(t_k)\leq M_k,$ we have

Note that the sum in the middle is a telescoping sum so that

which is independent of the partition $P.$ Thus we have

Since $f\in \mathcal R[a,b],$ \[ L(P,f) = U(P,f) = \int_a^bf. \] Therefore \[ \int_a^bf = F(b)-F(a).\; \bs \]

Compute $\ds \int_0^1x^2~dx.$

Note that $x^2$ is the derivative of $\ds\frac{x^3}{3}.$ The FTC says that

Theorem 8.2.1. Let $f \colon [a,b] \to \R$ be a Riemann integrable function. Define \begin{equation*} F(x) := \int_a^x f . \end{equation*}

1. $F$ is continuous on $[a,b]$.
2. If $f$ is continuous at $c \in [a,b]$, then $F$ is differentiable at $c$ and $F'(c) = f(c)$.

In the following simulation (next slide), the top graph shows the function $f(t)$ and the value of the definite integral for each upper limit $x$, with lower limit $a$. The definite integral is represented with the shaded region between the graph of the function and the $x$-axis.

The graph below shows the accumulation function $$A(x)=\int_a^xf(t)dt$$ for each upper limit $x$, with lower limit $a$.

Take $x,y\in[a,b]$ and observe that

where $M\gt0$ (since $f$ is continuous/bounded). This means that $F$ is Lipschitz continuous $\Ra$ $F$ is uniformly continuous on $[a,b]$ $\Ra$ $F$ is continuous on $[a,b].$
This proves item 1. 😃

Now to prove item 2, we assume that $f$ is continuous at $c\in[a,b],$ we would like to show that $F'(c)= f(c).$ To do this we rewrite the limit of $F'(c)$ as follows

If we can show the last limit equals $f(c)$, we are done! 😃

Given $\epsilon\gt 0 $ we must prodive a $\delta\gt 0$ such that if $\abs{x-c}\lt \delta,$ then

The assumption of continuity of $f$ gives us control over the difference $\abs{f(t)-f(c)}.$ In particular, we know that there exists a $\delta\gt 0$ such that

To use this fact, we rewrite the constant $f(c)$ as

Keeping in mind that $\abs{x-c}\geq \abs{t-c}$, we have that for all $\abs{x-c}\lt \delta$,

Keeping in mind that $\abs{x-c}\geq \abs{t-c}$, we have that for all $\abs{x-c}\lt \delta$,

Therefore

Theorem 8.2.2. (Change of variables) Let $g \colon [a,b] \to \R$ be a continuously differentiable function, let $f \colon [c,d] \to \R$ be continuous, and suppose $g\bigl([a,b]\bigr) \subset [c,d]$. Then \begin{equation*} \int_a^b f\bigl(g(x)\bigr)\, g'(x)~ dx = \int_{g(a)}^{g(b)} f(s)~ ds . \end{equation*}

Since $g$, $g'$, and $f$ are continuous, $f\bigl(g(x)\bigr)\,g'(x)$ is a continuous function of $[a,b]$, therefore it is Riemann integrable. Similarly, $f$ is integrable on every subinterval of $[c,d]$.

Define $F \colon [c,d] \to \R$ by $ \ds F(y) \coloneqq \int_{g(a)}^{y} f(s)\,ds . $

$\Ra \, F$ is a differentiable function and $F'(y) = f(y)$
(by the FTC).

Define $F \colon [c,d] \to \R$ by $ \ds F(y) \coloneqq \int_{g(a)}^{y} f(s)\,ds . $

$\Ra \, F$ is a differentiable function and $F'(y) = f(y)$
(by the FCT).

Applying the chain rule we have

Also observe that $F\bigl(g(a)\bigr) = 0$.

Also observe that $F\bigl(g(a)\bigr) = 0$.

We know that the derivative of $\sin(x)$ is $\cos(x)$.

Using $g(x) \coloneqq x^2$, we solve

Consider $ \ds \int_{-1}^{1} \frac{\ln \abs{x}}{x} \,dx . $   ⚠️ Careful! ⚠️

If we consider $g(x) \coloneqq \ln \abs{x},$ then $g'(x) = \dfrac{1}{x}.$ So

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