HW08-solutions - kim (tk5895) HW08 Henry (54974) 1 This...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: kim (tk5895) HW08 Henry (54974) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The shaded region in 2 3 2 x y is bounded by the graph of f ( x ) = cos 3 x on [0 , 3 / 2] and the the x-axis. Find the area of this region. 1. area = 3 2. area = 2 3. area = 3 2 4. area = 3 5. area = 3 2 6. area = 2 correct Explanation: The area of the shaded region is given by I = integraldisplay 3 / 2 | cos 3 x | dx which as the graph shows can in turn be writ- ten as I = integraldisplay / 2 cos 3 x dx- integraldisplay 3 / 2 / 2 cos 3 x dx . Since cos 2 x = 1- sin 2 x , we thus see that I = braceleftBig integraldisplay / 2- integraldisplay 3 / 2 / 2 bracerightBig cos x (1- sin 2 x ) dx . To evaluate these integrals, set u = sin x . For then du = cos x dx , in which case integraldisplay / 2 cos x (1- sin 2 x ) dx = integraldisplay 1 (1- u 2 ) du = bracketleftBig u- 1 3 u 3 bracketrightBig 1 = 2 3 , while integraldisplay 3 / 2 / 2 cos x (1- sin 2 x ) dx = integraldisplay- 1 1 (1- u 2 ) du = bracketleftBig u- 1 3 u 3 bracketrightBig- 1 1 =- 4 3 . Consequently, the shaded region has area = 2 . keywords: 002 10.0 points Evaluate the definite integral I = integraldisplay / 4 4 cos x + 5 sin x cos 3 x dx . 1. I = 9 kim (tk5895) HW08 Henry (54974) 2 2. I = 21 4 3. I = 3 2 4. I = 13 2 correct 5. I = 11 4 Explanation: After division we see that 4 cos x + 5 sin x cos 3 x = 4 sec 2 x + 5 tan x sec 2 x = (4 + 5 tan x ) sec 2 x . Thus I = integraldisplay / 4 (4 + 5 tan x ) sec 2 x dx . Let u = tan x ; then du = sec 2 x dx , while x = 0 = u = 0 , x = 4 = u = 1 . In this case I = integraldisplay 1 (4 + 5 u ) du = bracketleftbigg 4 u + 5 2 u 2 bracketrightbigg 1 . Consequently, I = 13 2 . 003 10.0 points Find the value of I = integraldisplay 3 tan 4 x dx . 1. I = 6- 8 3 9 2. I = 4- 2 3 3. I = 4 + 2 3 4. I = 3 3 5. I = 3 correct 6. I = 6 + 8 3 9 Explanation: Since tan 2 x = sec 2 x- 1 , we see that tan 4 x = tan 2 x ( sec 2 x- 1 ) = tan 2 x sec 2 x- tan 2 x . Thus by trig identities yet again, tan 4 x = ( tan 2 x- 1 ) sec 2 x + 1 . In this case, I = integraldisplay 3 bracketleftbig( tan 2 x- 1 ) sec 2 x + 1 bracketrightbig dx = bracketleftbigg 1 3 tan 3 x- tan x + x bracketrightbigg 3 . On the other hand, tan 3 = 3 . Consequently, I = 3 . 004 10.0 points Evaluate the definite integral I = integraldisplay / 4 sin 2 ( 3- 4 cos 2 ) d . kim (tk5895) HW08 Henry (54974) 3 1. I = 3 2. I = 2 3. I = 0 correct 4. I = 1 5. I = 3 2 Explanation: Since cos 2 = 1 2 (1 + cos 2 ) , we see that 3- 4 cos 2 = 3- 2 (1 + cos 2 ) = 1- 2 cos 2 Thus I = 1 2 integraldisplay / 4 sin 2 (2- 4 cos 2 ) d ....
View Full Document

Page1 / 9

HW08-solutions - kim (tk5895) HW08 Henry (54974) 1 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online