Answers>Maths>IB>Article

What is integration by parts, and how is it useful?

"Integration by parts" is one of several methods in our arsenal that we can use to integrate a function f(x). It involves expressing f(x) as the product of two other functions, which I will call u(x) and v'(x) (where v' = dv/dx is the first derivative of v with respect to x):
 
∫ f(x) dx = ∫ u(x) · v'(x) dx =  u(x) · v(x) - ∫ u'(x) · v(x) dx
 
This formula comes from the product rule for differentiation. If we differentiate the product u(x) · v(x) with respect to x using the product rule, we get the following:
 
d/dx(u(x) · v(x)) = u(x) · v'(x) + u'(x) · v(x)
 
Integrating both sides of this expression with respect to x removes the d/dx on the left hand side, and creates two integrals on the right hand side:
 
u(x) · v(x) = ∫ u(x) · v'(x) dx + ∫ u'(x) · v(x) dx
 
If we subtract the final term from both sides, we arrive at our original formula again:
 
∫ f(x) dx = ∫ u(x) · v'(x) dx =  u(x) · v(x) - ∫ u'(x) · v(x) dx
 
But why would we want to use this method? Well, integration by parts is generally useful in cases such as the one below:
 
∫ Ln(x)/x2 dx
 
At a first glance, we have no idea how to integrate this function. We do, however, know how to integrate 1/x2 (which is our v' in this case), and we know how to differentiate Ln(x) (which is our u). So, using integration by parts:
 
∫ Ln(x)/x2 dx = ∫ Ln(x) · (1/x2) dx = (-1/x) · Ln(x) - ∫ (-1/x) · (1/x) d
-Ln(x)/x + ∫ (1/x2) dx = -Ln(x)/x - 1/x + C
 
(Not forgetting our constant of integration at the end!)

Answered by Alexander J. Maths tutor

2826 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Find the Cartesian equation of plane Π containing the points A(6 , 2 , 1) and B(3, -1, 1) and perpendicular to the plane Π2 (x + 2y - z - 6 = 0).


Solve the equation 8^(x-1) = 6^(3x) . Express your answer in terms of ln 2 and ln3 .


In an arithmetic sequence, the first term is 2, and the fourth term is 14. a) Find the common difference, d. b) Calculate the sum of the first 14 terms, S14.


The velocity of a particle is given by the equation v= 4t+cos4t where t is the time in seconds and v is the velocity in m s ^-1. Find the time t when the particle is no longer accelerating for the interval 0≤t≤2.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences