Let f(x)=xln(x)-x. Find f'(x). Hence or otherwise, evaluate the integral of ln(x^3) between 1 and e.

We use the product rule with u=x and v=ln(x) (so u'=1 andv'=1/x) to differentiate xln(x) to ln(x)+1, and -x just differentiates to -1, hence we have. f'(x)=ln(x).
Now note that ln(x^3)=3ln(x) using properties of logarithms.Hence, we are just integrating 3ln(x). We know, from the first part, that ln(x) will integrate to xln(x)-x, and we require 3 lots of this, so 3ln(x) integrates to 3xln(x)-3x. Plugging in 1 and e into this formula, we get that the integral of ln(x^3) between 1 and e is (3eln(e)-3e)-(3ln(1)-3) = (3e-3e)-(30-3) = -3 (since ln(e)=1 and ln(1)=0).

MJ
Answered by Matthew J. Maths tutor

3068 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has the equation y=12+3x^4. Find dy/dx.


(1.) f(x)=x^3+3x^2-2x+15. (a.) find the differential of f(x) (b.) hence find the gradient of f(x) when x=6 (c.) is f(x) increasing or decreasing at this point?


The graph with equation y= x^3 - 6x^2 + 11x - 6 intersects the x axis at 1, find the other 2 points at which the graph intersects the x axis


The curve C has the parametric equations x=4t+3 and y+ 4t +8 +5/(2t). Find the value of dy/dx at the point on curve C where t=2.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning