MYTUTOR SUBJECT ANSWERS

777 views

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) dx  

=  -Ln(x)/x  +  ∫ (1/x2) dx  =  -Ln(x)/x  -  1/x  +  C

 

(Not forgetting our constant of integration at the end!)

Alexander J. IB Physics tutor, A Level Physics tutor, GCSE Physics tu...

2 years ago

Answered by Alexander, an IB Maths tutor with MyTutor


Still stuck? Get one-to-one help from a personally interviewed subject specialist

36 SUBJECT SPECIALISTS

£26 /hr

Tadas T.

Degree: MMathPhil Mathematics and Philosophy (Bachelors) - Oxford, St Anne's College University

Subjects offered:Maths, Further Mathematics + 3 more

Maths
Further Mathematics
.MAT.
-Personal Statements-
-Oxbridge Preparation-

“University of Oxford Maths and Philosophy student happy to help students learn and stay motivated!”

£26 /hr

Michelangelo M.

Degree: MMath Mathematics with Placement (Masters) - Bath University

Subjects offered:Maths, Italian+ 4 more

Maths
Italian
Further Mathematics
Computing
.STEP.
.MAT.

“Mathematics student at the University of Bath, willing to help you love this subject and improve your grades.”

£36 /hr

James G.

Degree: Mathematical Physics (Doctorate) - Nottingham University

Subjects offered:Maths, Physics+ 2 more

Maths
Physics
Further Mathematics
.STEP.

“Currently a 3rd year PhD student in Mathematical Physics. I'm very passionate about teaching as well as my subject area. Look forward to hearing from you.”

About the author

Alexander J. IB Physics tutor, A Level Physics tutor, GCSE Physics tu...
£26 /hr

Alexander J.

Degree: Astrophysics (Masters) - Cambridge University

Subjects offered:Maths, Physics

Maths
Physics

“5-star experienced and enthusiastic tutor with a first-class MSci from the University of Cambridge”

You may also like...

Posts by Alexander

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

What is Olbers' paradox?

Other IB Maths questions

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

Differentiate x^3 + y^4 = 34 using implicit differentiation

Factorise z^3+1 into a linear and quadratic factor. Let y=(1+i√3)/2. Show that y is a cube root of -1. Show that y^2=y-1. Find the value of (1-y)^6.

An arithmetic sequence has third term 56 and 15th term 92. What is the common difference and the first term?

View IB Maths tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok