To find the area under the curve \(y = 2\sqrt{3x+4} - x\) from \(x = 0\) to \(x = 4\), we need to integrate the function with respect to \(x\).
The integral is:
\(\int (2\sqrt{3x+4} - x) \, dx\)
First, integrate \(2\sqrt{3x+4}\):
Let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{1}{3} \, du\).
\(\int 2\sqrt{u} \, \frac{1}{3} \, du = \frac{2}{3} \int u^{0.5} \, du\)
\(= \frac{2}{3} \cdot \frac{2}{3} u^{1.5} = \frac{4}{9} (3x+4)^{1.5}\)
Next, integrate \(-x\):
\(\int -x \, dx = -\frac{1}{2} x^2\)
Thus, the integral of the function is:
\(\frac{4}{9} (3x+4)^{1.5} - \frac{1}{2} x^2\)
Evaluate from \(x = 0\) to \(x = 4\):
\(\left[ \frac{4}{9} (3(4)+4)^{1.5} - \frac{1}{2} (4)^2 \right] - \left[ \frac{4}{9} (3(0)+4)^{1.5} - \frac{1}{2} (0)^2 \right]\)
\(= \left[ \frac{4}{9} (16)^{1.5} - 8 \right] - \left[ \frac{4}{9} (4)^{1.5} \right]\)
\(= \left[ \frac{256}{9} - 8 \right] - \left[ \frac{32}{9} \right]\)
\(= \frac{256}{9} - 8 - \frac{32}{9}\)
\(= \frac{256}{9} - \frac{72}{9} - \frac{32}{9}\)
\(= \frac{152}{9}\)
