How do you evaluate the definite integral #int logx dx# from #[2,4]#?

Question

Scarlett Allison · Accepted Answer

#I=6log2-2/ln10#. See details below

#I=intlogxdx# Lets do it by parts #u=logx# and #dv=dx# With this we have #du=1/(xln10)dx# and #v=x# #I=xlogx-int1/ln10dx=xlogx-1/ln10x=F(x)# #F(4)-F(2)=4log4-4/(ln10)-2log2+2/ln10=4log4-2log2-2/ln10=4log2^2-2log2-2/ln10=6log2-2/ln10#

Axel Alvarez · Answer

undefined To evaluate the definite integral ∫log(x) dx from x = 2 to x = 4, you can use the properties of logarithms and integration techniques.

First, rewrite the integral as follows:

∫log(x) dx = ∫ln(x) dx

Then, use integration by parts:

Let u = ln(x) and dv = dx
Then, du = (1/x) dx and v = x

Apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values:

= xln(x) - ∫x*(1/x) dx
= xln(x) - ∫dx
= xln(x) - x

Now, evaluate the integral from 2 to 4:

∫[2,4] log(x) dx = [4ln(4) - 4] - [2ln(2) - 2]
= (4ln(4) - 4) - (2ln(2) - 2)
= (4*ln(4) - 4) - (2*ln(2) - 2)

Calculate the values:

= (4*ln(4) - 4) - (2*ln(2) - 2)
≈ (4*1.386 - 4) - (2*0.693 - 2)
≈ (5.544 - 4) - (1.386 - 2)
≈ (1.544) - (-0.614)
≈ 2.158

Therefore, the value of the definite integral ∫log(x) dx from x = 2 to x = 4 is approximately 2.158.