Sequences

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We study the mathematical concept of a sequence.

(Video) Calculus: Single Variable

Note: Don’t worry yet about the Taylor series for $e^x$ which is briefly mentioned at 5:50. We will come to this idea shortly.

Online Texts

Examples

The base of a solid region is bounded by the curves $x = 0$ , $y= 0$ , and $y = \sqrt {1-x^2}$ . The cross sections perpendicular to the $x$ axis are squares. Compute the volume of the region.

Solution:: Lines in the $xy$ -plane which are perpendicular to the $x$ -axis are vertical, so the base of a typical $x$ cross section will extend from $y=0$ to $y = \sqrt {1-x^2}$ . Since each cross section will have area $A(x) = \left ( \sqrt {1-x^2} - 0 \right )^2 = 1-x^2.$ To compute volume, we integrate $dV = A(x) dx$ between $x=0$ and $x=1$ , since these are the most extreme values of $x$ found in our region. Therefore $V = \int _0^1 ( 1 -x^2) dx = \left . x - \frac {x^3}{3} \right |_0^1 = 1 - \frac {1}{3} = \frac {2}{3}.$

PICT

The base of a solid region is bounded by the curves $y = 0$ , $x = \sqrt {y}$ , and $x = 1$ . The cross sections perpendicular to the $y$ -axis are squares. Compute the volume of the region.

Solution:: Lines in the $xy$ -plane which are perpendicular to the $y$ -axis are , so the base of a typical $y$ cross section will extend from the graph $x=\answer {\sqrt {y}}$ to the graph $x = \answer {1}$ . The length of the base is the difference of $x$ -coordinates (since all points on a slice have the same $y$ -coordinate), so the length of the base is $\answer {1 - \sqrt {y}}$ , giving the square an area of $A(y) = \answer {(1-\sqrt {y})^2}$ (note that the answer is a function of $y$ because different $y$ cross sections will generally have different areas). To compute volume, we integrate $dV = A(y) dy$ between $y=0$ and $y=1$ , since these are the most extreme values of $y$ found in our region (note that we can find the upper value $y=1$ by solving for the intersection of the curves $x = \sqrt {y}$ and $x=1$ ). Therefore we integrate $A(y) dy$ to conclude $V = \int _{\answer {0}}^{\answer {1}} \answer {(1-\sqrt {y})^2} dy = \answer {\frac {1}{6}}.$

PICT