definite integrals, which together constitute the Integral Calculus. quality ebook which they do not! x��ZK��6����Qڌ�� 9L2�b�좁=$s�mu[Yr$yz:�>U,R��r;���!h���"Yϯ��NWO�t������)�+�l&����J،q�WF�櫇���D�?=��dD��2�p�L6b6 �z�I���YR��S�YGv�h���x�>�����=Ү{�R��q]=���C��)⻈؃Cy*t"�E=ܥ^��#V�3��@�l�,mn�dBΔ�Ƥ�Y�ϵ՜_��?ߌf��?U]� �'�N���C�6=�������%MES %�h��+�~�#R�6�#W����U�+��w���G�&�r�E�e�dR���X6�F��P ���\�{zۮ%O^���[�͞�QVۦ���y^��~yAc_���i��=UͲ�9�*3)$��$�����Z Have questions or comments? That is, for $$N$$ a positive infinite integer, let, $d x=\frac{b-a}{N},$ and $d y=f(x+d x)-f(x).$ If the shadow of $\sum_{i=1}^{N} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x.$ is the same for any choice of $$N,$$ then we call $$(2.5 .27)$$ the arc length of $$C .$$ Now if $$f$$ is differentiable on an open interval containing $$[a, b],$$ and $$f^{\prime}$$ is continuous on $$[a, b],$$ then $$(2.5 .27)$$ becomes the definite integral of $$\sqrt{1+\left(f^{\prime}(x)\right)^{2}} .$$ That is, the arc length of $$C$$ is given by $L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x.$, Let $$C$$ be the graph of $$f(x)=x^{\frac{3}{2}}$$ over the interval $$[0,1]$$ (see Figure $$2.5 .11)$$ and let $$L$$ be the length of $$C .$$ since $$f^{\prime}(x)=\frac{3}{2} \sqrt{x},$$ we have, $L=\int_{0}^{1} \sqrt{1+\frac{9}{4} x} d x.$ Now $\int \sqrt{x} d x=\frac{2}{3} x^{\frac{3}{2}}+c,$ so we might expect an integral of $$\sqrt{1+\frac{9}{4}}$$ to be $\frac{2}{3}\left(1+\frac{9}{4} x\right)^{\frac{3}{2}}+c.$ However, $\frac{d}{d x} \frac{2}{3}\left(1+\frac{9}{4} x\right)^{\frac{3}{2}}=\frac{9}{4} \sqrt{1+\frac{9}{4} x},$ and so, dividing our original guess by $$\frac{9}{4},$$ we have $\int \sqrt{1+\frac{9}{4}} x d x=\frac{8}{27}\left(1+\frac{9}{4} x\right)^{\frac{3}{2}}+c,$ which may be verified by differentiation. There are basically two types of integrals, Definite and Indefinite. Area Between Curves – In this section we’ll take a look at one of the main applications of definite integrals in this chapter. lol it did not even take me 5 minutes at all! $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:dsloughter", "Area Between Curves" ], $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, 2.4: The Fundamental Theorem of Integrals, 2.6: Some Techniques for Evaluating Integrals. In this section we will look at several examples of applications for definite integrals. Here are a set of practice problems for the Applications of Integrals chapter of the Calculus I notes. /Filter /FlateDecode Chapter 6 : Applications of Integrals. Leibnitz (1646-1716) 288 MATHEMATICS There is a connection, known as the Fundamental Theorem of Calculus, between indefinite integral and definite integral which makes the definite integral as a practical Hence the integral, which is the area of $$R^{+}$$ minus the area of $$R^{-},$$ is $$0 .$$. Let $$P$$ be a pyramid with a square base having corners at $$(1,1,0),(1,-1,0),(-1,-1,0),$$ and $$(-1,1,0)$$ in the $$x y$$ -plane and top vertex at $$(0,0,1)$$ on the $$z$$ -axis. CHAPTER 6 APPLICATIONS OF THE DEFINITE INTEGRAL We may employ the following intuitive method for remembering this limit of sums formula (see Figure 6.5): 1. Given a line, which we will call the $$z$$-axis, let $$V(a, b)$$ be the volume of $$B$$ which lies between planes which are perpendicular to the $$z$$-axis and pass through $$z=a$$ and $$z=b .$$ Clearly, for any \(a.

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