🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
Calculus
1
answer
0
watching
210
views
13 Nov 2019

7. Evaluate using integration by parts.

Integral 5sqrt(x) ln sqrt(x) dx

a. 10/3 x^3/2 ln sqrt(x) - 5/3x^3/2 + C

b. 5x^3/2 ln sqrt(x) - 5x^3/2 + C

c. 10/3 x^3/2 ln sqrt(x) + 10/9x^3/2 + C

d. 10/3 x^3/2 ln sqrt(x) - 10/9x^3/2 + C

8. Evaluate using integration by parts.

Integral 5x/sqrt(2x+3) dx

a. 5x(2x+3)^1/2 + 5/3(2x+3)^3/2 + C

b. x(2x+3)^1/2 + 1/3(2x+3)^3/2 + C

c. 5x(2x+3)^1/2 - 5/3(2x+3)^3/2 + C

d. 5x(2x+3)^1/2 - (2x+3)^3/2 + C

Watch
1
answer
0
watching
210
views

For unlimited access to Homework Help, a Homework+ subscription is required.

Jamar Ferry
Jamar FerryLv2
6 Feb 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Share

Related textbook solutions

Calculus
4 Edition,
Rogawski
ISBN: 9781319050733
Single Variable Calculus: Early Transcendentals
4th Edition, 2018
Stewart
ISBN: 9781337687805
CALCULUS:EARLY TRANSCENDENTALS
4 Edition,
Rogawski
ISBN: 9781319050740

Related questions

Answer with solution. Thank you!

  1. d/dx sin^2 (x^2+2x+1)
  2. d/dx [4 tan (3x^3) - 5 csc x]
  3. d/dx sin (2x/x+1)
  4. d/dx (4 sin 3x - x)
  5. d/dx (x cos 5x^2)

1. Evaluate R y(x) dx (that is, the indefinite integral, or the anti-derivative) for the following y(x):

(a) 5x^2/3

(b) 2(x − 1)^−2

(c) sin(2x − 1)

(d) 3e^−2x

(e) (x − 2)^−1

(f) 2x−1/x+1

here is the problem it is already solved and this is the yahooanswer solution and my teachers solution as well. All i need is aGraph and for someone to explain how my teachergot 0,0 and 0,-2 as intresection points.

teachers solution:find all 4 points of intersections: (0,0),(0,-2), (-4sqrt(17sqrt(2)-24), 10sqrt(2)-14),(-4sqrt(17sqrt(2)-24), 10sqrt(2)-14).

yahoo answers:sin(theta) = r^2/4 ==> sin(theta)>= 0 . Hence 0 =< theta =< pi

r = 1 - sin(theta) ; therefore 0 =< r =< 1, because now weknow 0 =< sin(theta) =< 1

Therefore y = r sin(theta) >= 0 so the intersection points are in theupper half plane.

Also, sin(theta) = r^2/4 =< 1/4 < 1/sqrt(2) = sin pi/4 =sin(3pi/4),

so 0 =< theta < pi/4 or 3pi/4 < theta =< pi.

---
At an intersection point, sin(theta) = r^2/4 = 1 - r

r^2 + 4r - 4 = 0

Roots: r = -- 2 +/- 2 sqrt(2); but r >= 0, so r = 2 sqrt(2) - 2= 2(sqrt(2) - 1)

sin(theta) = 1 - r = 3 - 2 sqrt(2)

There is just one value of y, namely

y = r sin(theta) = 2 (sqrt(2) - 1)(3 - 2 sqrt(2) ) = 2 (5 sqrt(2) -7) = 10 sqrt(2) - 14

( or y = 10 (sqrt(2) - 1.4 ) = 0.142135...

However, there are two intersection points, with x^2 = r^2 -y^2

x = +/- sqrt( r^2 - y^2 )

x^2 = r^2 - y^2 = 4 (sin theta) -- y^2 = 4 (3 - 2 sqrt(2)) -- [2(5sqrt(2) - 7)]^2

x^2 = 12 - 8 sqrt(2) -4 [ 99 - 70 sqrt(2)] = 272 sqrt(2) - 384 (=17 x 2^4 sqrt(2) - 3 x 2^7)

x = +/- 4 sqrt(17 sqrt(2) - 24) [ = +/- 0.8161427...]

Weekly leaderboard

Home
Homework Help 3,900,000
Calculus 630,000
Start filling in the gaps now
Log in