Question

Find the area of the region between the curve y= 2ln(x) and the horizontal axis for 1<= x <= 4

Answer 1

Answer:

5.09 units

Step-by-step explanation:

Given equation

$y=2\ln x=f(x)$ in the interval $1\le x\le 4$

So we integrate $y$ in the given interaval

$\int f(x)=2\int\limits^4_1 {\ln x}dx$

Let us integrate $\ln x$ first.

let

$u=\ln x, dv=dx$

$du=\dfrac{1}{x}, v=x$

$\int\ln x dx=udv$

Using integration by parts we get

$uv-\int vdu$

$=x\ln x-\int x\dfrac{1}{x}dx$

$=x\ln x-dx$

$=x\ln x-x+C$

So here

$\int f(x)=2\int\limits^4_1 {\ln x}dx\\ =2(x\ln x-x)_1^4\\ =2[(4\ln 4-4)-(1\ln 1-1)]\\ =2[4\ln 4-4+1]\\ =5.09\ units$

The area of the the region between the curve and horizontal axis is 5.09 units.