Calculus

posted by .

Given two problems:

Problem A: Tau = Integral(0 to v)
v/(q^2 - v^2) dv
q = constant

Problem B: Tau = Integral(V TO 0)
v/(q^2 - v^2) dv
q = constant

Why is it that, SUBSTITUTION RULE is used in problem A, and QUOTIENT RULE is used in problem B? The two problems are somewhat similar. The only different is their integrals. Problem A, the integration is from 0 to V and problem B, the integration is from V to 0. The answers are shown below. Please help!!

Answer (problem A):
Tau = ln(q^2/(q^2-v^2))

Answer (problem B):
Tau = ln((q^2-v^2)/q^2)

  • Calculus -

    One integral is the negative of the other, since the integrand is the same and only the direction of integration changes. If you think about the answers, and recognize that log a/b = - log b/a, you will see that all that is involved is a sign change there also. Either integration could have been solved by either method.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calc

    how do you start this problem: integral of xe^(-2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious than …
  2. Math/Calculus

    How would I solve the following integral with the substitution rule?
  3. Calculus

    I have two questions, because I'm preparing for a math test on monday. 1. Use the fundamental theorem of calculus to find the derivative: (d/dt) the integral over [0, cos t] of (3/5-(u^2))du I have a feeling I will be able to find …
  4. Integration-Calculus

    How do I integrate this? constant(1- z/[sqrt (z^2 + constant^2)] i know i can take the constant out so I have constant * integral 1- z/[sqrt (z^2 + constant^2)] and I know the integral of z/[sqrt (z^2 + constant^2)] is [ln z^2 + constant^2]/2
  5. calculus

    LEt f and g be continous functions with the following properties i. g(x) = A-f(x) where A is a constant ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx iii. for the integral from 2 to 3 f(x)dx = -3A a find the …
  6. calculus

    LEt f and g be continous functions with the following properties i. g(x) = A-f(x) where A is a constant ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx iii. for the integral from 2 to 3 f(x)dx = -3A a find the …
  7. Physics

    In the following circuit R_1=3 Ohm, R_2=10 Ohm, C=0.003 Farad, V=5 Volts. At time t =0 we connect the power supply to the circuit. At t=0 the capacitor is uncharged. (a) What is the time constant tau (in seconds) to charge up the capacitor?
  8. math

    Note: You can get full credit for this problem by just answering the last question correctly. The initial questions are meant as hints towards the final answer and also allow you the opportunity to get partial credit. Consider the …
  9. Calculus: Integral

    I don't understand how to do this one integral problem that involves secant. I'm asked to find the integral of sec^4 (4x). I'm not really sure how to go about solving this problem.
  10. Calculus Please Help2

    Evaluate the integral by using substitution. Use an upper-case "C" for the constant of integration. integral x^5((x^6-2)^10)dx ty

More Similar Questions