For exercises e. and f.: If you have difficulty in adjusting the upper and lower limits, you may temporarily skip them. Instead, just find an anti-derivative first, and then substitute in those numbers, as indicated by the Fundamental Theorem of Calculus, to obtain the answer.**Exercise e:**\[\int_{0}^{\sqrt{3}/2} \frac{t^2}{(1-t^2)^{3/2}} \, dt \quad [\#41]\]**Exercise f:**\[\int_{0}^{3/5} \sqrt{9 - 25x^2} \, dt \quad [\#44]\]**Diagrams:**In both exercises, there is a right triangle diagram accompanying each integral, with an angle $\theta$ labeled. However, specific details about the triangle sides or further context regarding the angle's significance to the problem are not provided.

∫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 exercises e. and f.: If you have difficulty in adjusting the upper and lower limits, you may temporarily skip them. Instead, just find an anti-derivative first, and then substitute in those numbers, as indicated by the Fundamental Theorem of Calculus, to obtain the answer.

**Exercise e:**

\[
\int_{0}^{\sqrt{3}/2} \frac{t^2}{(1-t^2)^{3/2}} \, dt \quad [\#41]
\]

**Exercise f:**

\[
\int_{0}^{3/5} \sqrt{9 - 25x^2} \, dt \quad [\#44]
\]

**Diagrams:**

In both exercises, there is a right triangle diagram accompanying each integral, with an angle $\theta$ labeled. However, specific details about the triangle sides or further context regarding the angle's significance to the problem are not provided.

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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