MAT 21B Lecture Notes - Lecture 6: Antiderivative
beigecamel74 and 41 others unlocked
83
MAT 21B Full Course Notes
Verified Note
83 documents
Document Summary
This problem is an example of going from a rate to a quantity. I(cid:374) che(cid:396)(cid:374)ey"s (cid:1006)(cid:1005)b (cid:272)lass, the defi(cid:374)ite i(cid:374)teg(cid:396)al is defi(cid:374)ed as a li(cid:373)it of the (cid:396)ight. Intuitively, you respond 60 miles and this is true in which (cid:888)(cid:882) (cid:1856)=[(cid:888)(cid:882)](cid:2868)(cid:2869) = (cid:2869)(cid:2871)(cid:4666)(cid:883)(cid:4667)(cid:2871) (cid:2869)(cid:2871)(cid:4666)(cid:882)(cid:4667)(cid:2871)= (cid:2869)(cid:2871): example: (cid:2870)(cid:1856)=[(cid:2869)(cid:2871)(cid:2871)](cid:2868)(cid:2869, definite integrals are easy to calculate using the fundamental theorem of. Calculus (ftoc) only if you can find an anti-derivative of the integrand easily: upper limit of integration: 1. =(cid:2869) (cid:2869)(cid:2868: lower limit of integration: 0. Integrand: x2: differential: dx (cid:2869)(cid:2868) variable or symbol that is being used since the same solution is still obtained, be(cid:272)ause defi(cid:374)ite i(cid:374)teg(cid:396)als a(cid:396)e (cid:374)u(cid:373)(cid:271)e(cid:396)s, the va(cid:396)ia(cid:271)le of i(cid:374)teg(cid:396)atio(cid:374) is a (cid:862)du(cid:373)(cid:373)y. (cid:863) The index of sum(cid:373)atio(cid:374) is also a (cid:862)du(cid:373)(cid:373)y. (cid:863) fo(cid:396) i(cid:374)sta(cid:374)(cid:272)e, . If the variable is changed from k to z, then we have . =(cid:883)+(cid:884)+(cid:885)=(cid:888) (cid:2871)=(cid:2869: problem: find an anti-derivative for (cid:1857)2(cid:1856). the anti-derivative must contain (cid:1857)2 in the answer.