Math PLZ help me
 👍 0
 👎 0
 👁 178
Respond to this Question
Similar Questions

Math
Hi, I needed help with this improper integral. lower limit : 2 upper limit : infinity integral of (lnx)/(x+1) dx Thank You so much :D
asked by Bradley on June 17, 2014 
CALCULUS II
Hi, I needed help with this improper integral. lower limit : 2 upper limit : infinity integral of (lnx)/(x+1) dx Thank You so much :D
asked by BRADLEY on June 17, 2014 
12th Calculus
1. Explain why the function f(x)=(x^24)/(x2) is not continous on [0,3]. what kind of discontinuity occurs? 2. use areas to show that integral sign with the upper limit of 3 and a lower limit of 0 (x^24/x2)dx=10.5 3. use the
asked by may on December 10, 2008 
Mathintergration
Show that integrate.[e^(4x)*(cos 3x)] from 0infinity =4/25 I got the answer(without applying limits) as, {(e^(4x) )[3sin(3x)  4cos(3x)]}/25 But when applying the upper limit what is the value of, {(e^(infinity)
asked by Shenaya on July 29, 2017 
calculus
Determine whether the following is convergent of divergent. integral(lower limit=0, upper limit=infinity)of sin(x)sin(x^2)dx Thanks
asked by Hector on July 23, 2008

math
1.integral cos(x)sqrt(sin(x))dx. upper limit pi and lower limit 0 2.integral 1/sqrt(x)(sqrt(x) +1)^2. Upper limit is 4 and lower limit is 1.
asked by Carl on September 10, 2018 
math
Sorry I posted this question earlier with a error. Can anyone help me solve this now? Can someone help me answer this? If a < 5 the define integral [a, 4] 2.4e^(1.4x)dx = 44 Find the value a Define integral = integral sign a =
asked by mark on May 19, 2013 
math
How do you find the definite integral of an absolute function? ∫_0^32x3 lower limit= 0 upper limit = 3
asked by Stephen on February 12, 2009 
math
Evaluate the integral upper limit sqrt11 and lower limit 1 of (s^2 + sqrt 5)/ s^2
asked by vishal on September 10, 2013 
Math
Can someone help me answer this? Determine F ' (x) and F ' (1) if F(x) = integral sign lower limit = 4, upper limit = x 4t^10 (2 t^(1) 0 + 8 ln t) ^4 dt a) F ' (x) = b) F' (1) =
asked by mark on May 19, 2013