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 143[ * √t (2 + t) dt = [ ^ ( 2 √² + t ³ ) dt===43+32 5521.7333333333333.234 22 +33 2+

Answered: Image /qna-images/answer/515dbce2-12d9-48a4-bd54-90f586dbd7f1.jpg

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.

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

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

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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Question

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
+

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