Question

How would I solve this

Answer 1

$A=\frac{1}{2}\ln17 = 1.417$

Step-by-step explanation:

The area A under the curve can be written as

$\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2}$

To evaluate the integral, let

$u = 1+4x^2 \Rightarrow du = 8xdx\:\text{or}\:\frac{1}{2}du = 4xdx$

so the integral becomes

$\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\int\!\dfrac{du}{u} = \dfrac{1}{2}\ln |u|$

or

$\displaystyle \int\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\ln |1+4x^2|$

Putting in the limits of integration, our area becomes

$\displaystyle A = \int_0^2\!\dfrac{4x\:dx}{1+4x^2} = \dfrac{1}{2}\left.\ln |1+4x^2|\right|_0^2$

$\;\;\;\;= \frac{1}{2}[\ln (1+16) - \ln (1)]$

$\;\;\;\;=\frac{1}{2}\ln17$

Note: $\ln 1 = 0$