Calculus Notes

Section 4.4 Fundamental Theorum of Calculus

• this is approximated by a finite number of rectangles to fill the space

• and is exact by taking n (number of rectangles) to infinity by using a limit

Fundamental Theorem of Calculus (Pt. 1)

• If \(f(x)\) is continuous on \([a,b]\) and \(F(x)\) is any antiderivative of \(f(x)\) on \([a,b]\text{,}\) then:

  \begin{equation*} \underbrace{\int_a^b f(x) dx}_{\text{area}} = \underbrace{F(b) - F(a)}_{\text{difference in value of antiderivative}} = \underbrace{F(x) \mid_a^b}_{\text{difference rewritten}} \end{equation*}

  Example 4.4.1.

  Find \(\displaystyle \int_0^3 x dx\text{:}\)

  Solution

  \begin{align*} \int x dx \amp = \frac{1}{2} x^2 + c\\ \int_0^3 dx \amp = \left( \frac{1}{2} x^2 \right) \mid_0^3 \\ \amp = \frac{1}{2} (3)^2 - \frac{1}{2} (0)^2\\ \amp = 4.5 \end{align*}

• With definite integrals, the answer is the net signed area.

  Example 4.4.3.

  Find \(\displaystyle \int_{-1}^1 1 + 2x dx\text{:}\)

  Solution

  \begin{align*} \int 1 + 2x dx \amp = x + 2 (\frac{x^2}{2})\\ \int_{-1}^1 1 + 2x dx \amp = (x + x^2) \mid_{-1}^1\\ \amp = (1 + 1^2) - ((-1) + (-1)^2)\\ \amp = 2 \end{align*}

  Note 4.4.5.

  It is the net signed area because a portion is being subtracted from total.

• You have to be careful with functions that have asymptotes:

  Example 4.4.6.

  Find \(\displaystyle \int_{-1}^1 \frac{1}{x^2} dx \text{:}\)

  Solution

  Note 4.4.7.

  This has a verticle asymptote at \(x = 0\text{.}\)

  Since we cross the asymptote in the integral we cannot evaluate it. Instead, let's find \(\displaystyle \int_1^2 \frac{1}{x^2} dx\text{:}\)

  \begin{align*} \int \frac{1}{x^2} dx \amp = \int x^2 dx = -x^{-1} = -\frac{1}{x}\\ \int_1^2 \frac{1}{x^2} dx \amp = \left( - \frac{1}{x} \right) \mid_1^2 = \left( - \frac{1}{2} \right) - \left( -\frac{1}{1} \right)\\ \amp = - \frac{1}{2} + 1\\ \amp = \frac{1}{2} \end{align*}

• Similarly, we must be careful with functions containing absolute value:

  Example 4.4.10.

  Find \(\displaystyle \int_{-1}^2 \mid x \mid dx\text{:}\)

  Solution

  Note 4.4.11.

  \begin{equation*} \mid x \mid = \begin{cases} -x, \amp x \gt 0\\ x, \amp x \ge 0 \end{cases} \end{equation*}

  \begin{align*} \int_{-1}{2} \mid x \mid dx \amp = \int_{-1}^0 (-x) dx + \int_0^2 x dx \\ \amp = \left( - \frac{x^2}{2} \right) \mid_{-1}^0 + \left( \frac{x^2}{x} \right) \mid_0^2\\ \amp = (0) - \left( - \frac{(-1)^2}{2} \right) + \left( \frac{2^2}{2} \right) - (0)\\ \amp = \frac{1}{2} + 2\\ \amp = 2.5 \end{align*}

• We define the total area under a curve f(x) as:

  \begin{equation*} \text{Total Area} = \int_a^b \mid f(x) \mid dx \end{equation*}
  \begin{equation*} y = sin(x) \end{equation*}

  \begin{equation*} y = \mid sin(x) \mid \end{equation*}