MAT 21B Lecture Notes - Lecture 26: Specific Weight, Moment Of Inertia, Riemann Sum
MAT 21B – Lecture 26 – Review of Integral Calculus
I. PART I: Definite Integrals in the Abstract and the FTOC
• The string of symbols
is shorthad for the phrase the defiite
integral of oer the iteral [, ] hih is defied as the number
.
• There are two ways to evaluate
o 1) Evaluating it as the limit of a Right Riemann Sum
.
o 2) Applying the Fundamental Theorem of Calculus (FTOC)
.
• Are the following limits of sums definite integrals?
o a)
. No because definite
integrals have a specific structure. For example, this limit is missing a
factor of
, which would correspond to the differential, dx in the
definite integral
o b)
. No; because
is squared, then
the denominator has a higher degree of polynomial. As a result, the
denominator gets bigger faster as n → ∞ than the numerator. Thus,
the limit of sum → 0. Thus, .
o c)
. Yes, in fact, it is equal to
.
However, this cannot be computed with summation formulas. We do
not know an identity for exponential functions.
• The notation is shorthad for the ati-deriatie of f hile
is shorthad for the defiite itegral of f oer iteral [a, ]
• The Fundamental Theorem of Calculus is a statement about definite
integrals. For example, calculate and show where you use the FTOC.
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Document Summary
Mat 21b lecture 26 review of integral calculus. =(cid:2869: there are two ways to evaluate (cid:1876)(cid:2871)(cid:1856)(cid:1876) (cid:2870)(cid:2869) lim (cid:4666)(cid:883)+(cid:2870) (cid:2869)(cid:4667)(cid:2871) =(cid:2869) = lim (cid:2869) + (cid:2871)(cid:3118) (cid:4666)+(cid:2869)(cid:4667)(cid:2870) + (cid:2871)(cid:3119) (cid:4666)+(cid:2869)(cid:4667)(cid:4666)(cid:2870)+(cid:2869)(cid:4667) (cid:2871) (cid:2874) +(cid:3120)+(cid:2870)(cid:3119)+(cid:3118) (cid:2870)(cid:3118) +(cid:2870)(cid:3119)+(cid:2871)(cid:3118)+ (cid:4667)(cid:2870)=lim (cid:883)+(cid:2871)(cid:3118)+(cid:2871) (cid:4666)(cid:4666)+(cid:2869)(cid:4667)(cid:2870) (cid:2872)(cid:3120) (cid:2870)(cid:3119) (cid:883)+(cid:2869)(cid:2872)=(cid:2872)(cid:2872)+(cid:2874)(cid:2872)+(cid:2872)(cid:2872)+(cid:2869)(cid:2872)=(cid:2869)(cid:2873)(cid:2872) . No because definite (cid:2869)(cid:3118) factor of (cid:3029) (cid:3028) , which would correspond to the differential, dx in the definite integral (cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1856)(cid:1876) (cid:3029)(cid:3028: b) lim (cid:4672)(cid:883)+(cid:2870) (cid:2869) (cid:4673)(cid:2871)(cid:4666)(cid:2870) (cid:2869)(cid:4667)(cid:2870) Yes, in fact, it is equal to (cid:1857)(cid:2870)(cid:2869) (cid:1857)(cid:4666)(cid:2869)+(cid:3118) (cid:3117)(cid:4667) =(cid:2869: the notation (cid:862) (cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1856)(cid:1876)(cid:863) is shortha(cid:374)d for (cid:862)the a(cid:374)ti-deri(cid:448)ati(cid:448)e of f(cid:863) (cid:449)hile (cid:862) (cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1856)(cid:1876) (cid:3029)(cid:3028) integrals. For example, calculate (cid:1876)(cid:1857)(cid:1856)(cid:1876) and show where you use the ftoc. the denominator has a higher degree of polynomial. As a result, the denominator gets bigger faster as n than the numerator. Thus: the fundamental theorem of calculus is a statement about definite (cid:863) is shortha(cid:374)d for (cid:862)the defi(cid:374)ite i(cid:374)tegral of f o(cid:448)er i(cid:374)ter(cid:448)al [a, (cid:271)](cid:863)
