Evaluate the following(i) * V16 – x² dx/16 — х2dx(ii) Sx2/4-x2dx

Evaluate the following: (i). ∫0416-x2 dx We have, I=∫0416-x2 dx ⇒I=∫0442-x2 dxWe know that,…

Answered: Evaluate the following i) * V16 – x2 dx…

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

Evaluate the following
(i) * V16 – x² dx
/16 — х2
dx
(ii) S
x2/4-x2
dx

Expert Solution

Step 1

Evaluate the following:

(i). $\int_{0}^{4} \sqrt{16 - x^{2}} d x$

We have, $I = \int_{0}^{4} \sqrt{16 - x^{2}} d x$

$\Rightarrow I = \int_{0}^{4} \sqrt{4^{2} - x^{2}} d x W e k n o w t h a t, \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} s i n^{- 1} \frac{x}{a} + C \Rightarrow I = {[\frac{x}{2} \sqrt{4^{2} - x^{2}} + \frac{16}{2} \sin^{- 1} \frac{x}{4}]}_{0}^{4} \Rightarrow I = [\frac{4}{2} \sqrt{4^{2} - 4^{2}} + 8 \sin^{- 1} \frac{4}{4} - \frac{0}{2} \sqrt{4^{2} - 0^{2}} + 8 \sin^{- 1} \frac{0}{4}] \Rightarrow I = [2 \times 0 + 8 \sin^{- 1} 1 - 0 + 0] \Rightarrow I = 8 \sin^{- 1} 1 \Rightarrow I = 8 \times \frac{π}{2}, \{S i n c e, \sin^{- 1} 1 = \frac{π}{2}\} \Rightarrow I = 4 π H e n c e, \int_{0}^{4} \sqrt{16 - x^{2}} d x = 4 π$