For e. and f.: If you have difficulty in making the changes for upper and lower limits,you may temporarily skip them; just work out an anti-derivative first, then plug in thosenumbers, as the Fundamental Theorem of Calculus states, to get the answer.V5/2t2dt #41]е.(1 – 12 )3/23/5f.25a² dt [#44]

∫032t21-t232dt Now, integrate the integral,…

Answered: t2 dt [#41] (1 – t2)3/2

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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For e. and f.: If you have difficulty in making the changes for upper and lower limits,
you may temporarily skip them; just work out an anti-derivative first, then plug in those
numbers, as the Fundamental Theorem of Calculus states, to get the answer.
V5/2
t2
dt #41]
е.
(1 – 12 )3/2
3/5
f.
25a² dt [#44]

Expert Solution

Step 1: (e) Consider the provided integral,

$\int_{0}^{\frac{\sqrt{3}}{2}} (\frac{t^{2}}{{(1 - t^{2})}^{\frac{3}{2}}}) d t$

Now, integrate the integral,

$\begin{array}{rcl} \int_{0}^{\frac{\sqrt{3}}{2}} (\frac{t^{2}}{{(1 - t^{2})}^{\frac{3}{2}}}) d t & = & \int_{0}^{\frac{π}{3}} (\frac{\sin (u) \tan (u)}{\cos (u)}) d u \\ = & \int_{0}^{\frac{π}{3}} (\tan^{2} (u)) d u \\ = & \int_{0}^{\frac{π}{3}} (- 1 + s e c^{2} (u)) d u \\ = & {[- x + \tan u]}_{0}^{\frac{π}{3}} \\ = & - \frac{π}{3} + \sqrt{3} \end{array}$

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