# What is the net area between f(x) = sqrt(x^2+2x+1)  and the x-axis over x in [2, 4 ]?

Jul 14, 2017

$8$ sq. units

#### Explanation:

Actually we can do this one without the need of Calculus, as follows.

$f \left(x\right) = \sqrt{{x}^{2} + 2 x + 1}$

$= \sqrt{{\left(x + 1\right)}^{2}}$

$= \left\mid x + 1 \right\mid$

Since we are only concerned with $x \in \left[2 , 4\right]$ we now only have to consider a straight line of slope 1 and $y -$ intercept 1.

The area between $f \left(x\right)$ and the $x -$axis $x \in \left[2 , 4\right]$ can be separated into:

(i) a rectangle of sides $\left(4 - 2\right)$ and $\left(2 + 1 - 0\right)$
and
(ii) a right triangle of sides $\left(4 - 2\right)$ and $\left(4 + 1 - \left(2 + 1 - 0\right)\right)$

Area of rectangle = $2 \times 3 = 6$

Area of right triangle = $\frac{1}{2} \times 2 \times 2 = 2$

Required area $= 6 + 2 = 8$ sq. units