Concept:Given:y = \(\sqrt{(3x + 4)}\)Area = \(\int \limits_0^4 \sqrt{(3x + 4)} dx\)Let 3x + 4 = tOn differentiating we get,3dx + 0 = dt3dx = dtdx = \(\frac{dt}{3}\)When x = 0 then t = 4When x = 4, then t = 16Now the function is,\(\frac{1}{3}\int \limits_4^{16} \sqrt t dt\)Calculation:? \(\frac{1}{3}\rm \left [t^{\frac{1}{2} +1}\over\frac{1}{2} + 1\right]_{4}^{16}\)? \(\frac{1}{3}\rm \left [\frac{2}{3}t^\frac{3}{2}\right]_{4}^{16}\) ? \(\frac{1}{3} \times \frac{2}{3} [(16)^\frac{3}{2} - (4)^\frac{3}{2}]\)? \(\frac{1}{3} \times \frac{2}{3} [(4^2)^\frac{3}{2} - (2^2)^\frac{3}{2}]\)? \(\frac{2}{9} [(4)^3 - (2)^3]\)? \(\frac{2}{9} [64 - 8]\)? \(\frac{2}{9} \times 56\)? \(\frac{112}{9} sq. units\)? The area under the curve y = \(\sqrt{(3x + 4)}\) above the x - axis and between the lines x = 0 and x = 4 is \(\frac{112}{9} sq. units\)