[tex]A=\frac{1}{2}\ln17 = 1.417[/tex]
Step-by-step explanation:
The area A under the curve can be written as
[tex]\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2}[/tex]
To evaluate the integral, let
[tex]u = 1+4x^2 \Rightarrow du = 8xdx\:\text{or}\:\frac{1}{2}du = 4xdx[/tex]
so the integral becomes
[tex]\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\int\!\dfrac{du}{u} = \dfrac{1}{2}\ln |u|[/tex]
or
[tex]\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\ln |1+4x^2|[/tex]
Putting in the limits of integration, our area becomes
[tex]\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\left.\ln |1+4x^2|\right|_0^2[/tex]
[tex]\;\;\;\;= \frac{1}{2}[\ln (1+16) - \ln (1)][/tex]
[tex]\;\;\;\;=\frac{1}{2}\ln17[/tex]
Note: [tex]\ln 1 = 0[/tex]
