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: ∫ (udv/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 = xln(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 = [xln(x)-x]21 ∫21 ln(x) dx = 2ln(2)-2-(1ln(1)-1) ∫21 ln(x) dx = 2*ln(2)-1
AT
Answered by Angus T. • Further Mathematics tutor
3559 Views
See similar Further Mathematics A Level tutorsRelated Further Mathematics A Level answers
All answers ▸The infinite series C and S are defined C = a*cos(x) + a^2*cos(2x) + a^3*cos(3x) + ..., and S = a*sin(x) + a^2*sin(2x) + a^3*sin(3x) + ... where a is a real number and |a| < 1. By considering C+iS, show that S = a*sin(x)/(1 - 2a*cos(x) + a^2), and find C.
Integrate x^2sin(x) between -pi and pi
Given z=cosx+isinx, show cosx=1/2(z+1/z)
How do I sketch accurate graphs for rational functions in a short amount of time? (I.e. A step by step guide of sketching graphs)