By the Fundamental Theorem of Calculus, Part 1 4 ["* vi(2+1)dt = ["^(2√²+11) de dt 4 - = ( 5 -t² + -t 2 5 4 3 3 1 3 2 5 4 - ²/² · 4¹ ) - ( ²3² · ₁¹² + ²/3 · ₁² ) .42 5 21.7333333333333.

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Im confused where we got some of these number such as the fractions, and was hoping for a walkthrough

 

By the Fundamental Theorem of Calculus, Part 1
4
3
[ * √t (2 + t) dt = [ ^ ( 2 √² + t ³ ) dt
=
=
=
4
3
+
3
2 5
5
21.7333333333333.
2
3
4 2
2 +
3
3 2
+
Transcribed Image Text:By the Fundamental Theorem of Calculus, Part 1 4 3 [ * √t (2 + t) dt = [ ^ ( 2 √² + t ³ ) dt = = = 4 3 + 3 2 5 5 21.7333333333333. 2 3 4 2 2 + 3 3 2 +
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