Calculus AP
 👍
 👎
 👁
 ℹ️
 🚩

 👍
 👎
 ℹ️
 🚩

 👍
 👎
 ℹ️
 🚩
Respond to this Question
Similar Questions

calculus integrals
Evaluate the integral by making the given substitution. (Use C for the constant of integration. Remember to use absolute values where appropriate.) integral x^5/x^65 dx, u = x6 − 5 I got the answer 1/6ln(x^65)+C but it was

Calc 1
Evaluate the integral by making the given substitution. (Use C for the constant of integration.) ∫ dt/(17)^7, u=17t

Calculus
Integral of sin2xcos3x using Integration by Parts

calc
evaluate the integral: y lny dy i know it's integration by parts but i get confused once you have to do it the second time Leibnitz rule (a.k.a. product rule): d(fg) = f dg + g df y lny dy = d[y^2/2 ln(y)]  y/2 dy > Integral

Stats: Joint Density Function
Let the joint density function of X and Ybe given by: f(x)={kxy^2 for 0

calc: avg value
Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(ba))*inegral of a to b for: f(x) dx f

calculus
1.Evaluate the integral. (Use C for the constant of integration.) integral ln(sqrtx)dx 2. Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the curves about the given axis. y =

calculus2
Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.) ∫√((x^24)/x)dx x=2sec(t)

improper integral
can any one explain how to evaluate this improper integral i.e. the function is not continuous at 0 neither at inf integration of [(e^(sqrt(t)))/sqrt(t)]dt from 0 to infinity the integration part is easy but i want only how to

Calculus Please Help2
Evaluate the integral by using substitution. Use an uppercase "C" for the constant of integration. integral x^5((x^62)^10)dx ty

Integration by Parts
integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=cos(t) i integral sin(t)e^(it)dt= e^(it)cos(t)+i*integral

calculus
How do you find: the Integral of arcsin(1 / (sqrt x^2  1) ) dx ?? (The integral of arcsin of one over the squareroot of x squared minus 1) So far, I've used trig integration to simplify down to arcsin(1/tanb) *secbtanb b is theta