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

Question

Emilia Aguilar · Accepted Answer

#[(1/2-5/6sqrt5+1/12sqrt5)-(8-34/3sqrt17+17/60sqrt17)]=36.38# areal units, nearly. Please do not attempt to edit this answer. I would review my answer, for bugs, if any, myself, after some time.

The integrated area is in y-negative #Q_4# and is negative. The end points for the curved boundary are A at #(1, -(sqrt5-1))=(1. -1.236)# and B at #(4,=4(sqrt17-1))=(4, =12.402)#. The numerical area ( suppressing negative sign ) #=- int y dx#, with #y = x(1-sqrt(4x+1))# and x, from 1 to 4 #=-int(x-xsqrt(4x+1)) dx#, for the limits 1 and 4 #=-[x^2/2-(2/3)(1/4)int x d(4x+1)^(3/2)]#, for x = 1 to 4 #=-[x^2/2-1/6x(4x+1)^(3/2)+1/6int(4x+1)^(3/2) dx], for x = 1 to 4 #=-[x^2/2-1/6x(4x+1)^(3/2)+1/6(2/5)(1/4)(4x+1)^(5/2) ],# between x = 1 and 4 #=-[(8-34/3sqrt17+17/60sqrt17)-(1/2-5/6sqrt5+1/12sqrt5)]# graph{x(1-sqrt(4x+1)) [0, 50, -25, 0]}

Ezekiel Alexander · Answer

undefined To find the net area between $ f(x) = x - x\sqrt{4x + 1} $ and the x-axis over $ x $ in the interval $[1, 4]$, you first need to integrate the function within the given interval, and then take the absolute value of the result since the function dips below the x-axis. The integral will be:

$$ \int_{1}^{4} \left| x - x\sqrt{4x + 1} \right| \, dx $$

You can break this integral into two parts since the function changes its behavior at $ x = \frac{1}{4} $. Therefore, the integral becomes:

$$ \int_{1}^{4} (x - x\sqrt{4x + 1}) \, dx + \int_{\frac{1}{4}}^{4} (x\sqrt{4x + 1} - x) \, dx $$

After integrating each part, you should get the net area between the function and the x-axis over $ x $ in the interval $[1, 4]$.