THEOREM 1.4: Fundamental Theorem of Calculus, Part 1

If $f(x)$ is continuous over an interval $[a, b]$, and the function $F(x)$ is defined by

$$
F(x)=\int_a^x f(t) d t
$$

then $F^{\prime}(x)=f(x)$ over $[a, b]$

NOTE: constant is BOTTOM bound, variable is UPPER bound.

NOTE: observe difference between variable of integration and variable of function F(x)

Example 2: $\int_3^x \sqrt{x^2 + 45} dx$

NOTE: With a little cleverness, we can use this fact to calculate the value of definite integrals.

Example 3 (proof of FTC2 – optional): $\int_2^5 3x^2-4x dx$.

Let $G(x) = \int_a^x 3t^2-4t dt$ for some constant $a$

Notice that $\int_2^5 3x^2-4x dx = G(5)-G(2)$ (geometric argument – BUT still don’t know how to calculate G(5) or G(2)!!)

Also notice that $G'(x)=3x^2-4x$ by the FTC part 1, so $G(x)$ is an antiderivative of $3x^2-4x$

Let $F(x)$ be any antiderivative of $G(x)$ — give me one! e.g. $F(x)=\frac{3x^4}{4}-\frac{5x^2}{2}$. We know how to calculate $F(2)$ and $F(5)$.

BUT $F(x)$ and $G(x)$ are both antiderivatives of $3x^2-4x$, and we know that these differ only by a constant, so $G(x)=F(x)+C$.

Thus $G(5)=F(5)+C$ and $G(2)=F(2)+C$

so $\int_2^5 3x^2-4x dx = G(5)-G(2)=F(5)+C-(F(2)+C)=F(5)-F(2)$.

THEOREM 1.5: The Fundamental Theorem of Calculus, Part 2

If $f$ is continuous over the interval $[a, b]$ and $F(x)$ is any antiderivative of $f(x)$, then

$$
\int_a^b f(x) d x=F(b)-F(a)
$$

Example 4 (warmup): $\int_2^5 3x^2-4x dx$

GROUP WORK 2

Evaluate the following integrals (using the Fundamental Theorem of Calculus Part 2):

  1. $\int_0^10 x^3+7x-1 dx$
  2. $\int_{-1}^1 e^t dt$
  3. $\int_\pi^{2\pi} -\sin x dx$
  4. $\int_2^16 \sqrt{x} dx$