f(x)=ln(x). Find the area underneath the curve f(x) between 1 and 2.

We cannot dirrectly intergrate ln(x), so instead we intergrate 1ln(x) using intergration by parts.
The formula for intergration by parts is: ∫ (u
dv/dx) dx = uv − ∫ vdu/dx dx .
We let u=ln(x) so that du/dx=1/xWe let dv/dx=1 so that v=x
We put those values into the formula and we get ∫ ln(x) dx = x
ln(x) - ∫ (x1/x )dx∫ ln(x) dx = xln(x) - ∫1 dx∫ ln(x) dx = xln(x)-x + c
Finding the area under the curve between 1 and 2. ∫21 ln(x) dx = [x
ln(x)-x]2121 ln(x) dx = 2ln(2)-2-(1ln(1)-1) ∫21 ln(x) dx = 2*ln(2)-1

Related Further Mathematics A Level answers

All answers ▸

A=[5k,3k-1;-3,k+1] where k is a real constant. Given that A is singular, find all the possible values of k.


Expand (1+x)^3. Express (1+i)^3 in the form a+bi. Hence, or otherwise, verify that x = 1+i satisfies the equation: x^3+2*x-4i = 0.


Evaluate the following product of two complex numbers: (3+4i)*(2-5i)


What is sin(x)/x for x =0?


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences