Question

If the area of the region bounded by the curve $y=a\sqrt{x}+bx,x$-axis and the lines x = 0 and x = 4 is 8 sq. units, then the value of 2a + 3b is _____________.

Answer 1

Given: area of the region bounded by the curve $y=a\sqrt{x}+bx,x$-axis and the lines x = 0 and x = 4 is 8 sq. units

We know,
the required area of the region = ${\int }_0^{4}ydx$
Thus,
$$\begin{gathered} 8=\left|{\int }_0^4\left(a\sqrt{x}+bx\right)dx\right| \\ \Rightarrow \pm 8={\left(a\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}+b\frac{x^2}{2}\right)}_0^4 \\ \Rightarrow \pm 8={\left(\frac{2a}{3}{x}^{\frac{3}{2}}+\frac{b}{2}{x}^2\right)}_0^4 \\ \Rightarrow \pm 8=\left(\frac{2a}{3}{4}^{\frac{3}{2}}+\frac{b}{2}{4}^2\right)-0 \\ \Rightarrow \pm 8=\left(\frac{2a}{3}\times 2^3+\frac{b}{2}\times 16\right) \\ \Rightarrow \pm 8=\left(\frac{16a}{3}+8b\right) \\ \Rightarrow \pm 8=\left(\frac{16a+24b}{3}\right) \\ \Rightarrow \pm 24=\left(16a+24b\right) \\ \Rightarrow \pm 24=8\left(2a+3b\right) \\ \Rightarrow \pm 3=\left(2a+3b\right) \\ \mathrm{Thus},2a+2b=\pm 3 \end{gathered}$$


Hence, the value of 2a + 3b is $\underline{\pm 3}.$