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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 _____________.

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Solution

## 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, $8=\left|{\int }_{0}^{4}\left(a\sqrt{x}+bx\right)dx\right|\phantom{\rule{0ex}{0ex}}⇒±8={\left(a\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}+b\frac{{x}^{2}}{2}\right)}_{0}^{4}\phantom{\rule{0ex}{0ex}}⇒±8={\left(\frac{2a}{3}{x}^{\frac{3}{2}}+\frac{b}{2}{x}^{2}\right)}_{0}^{4}\phantom{\rule{0ex}{0ex}}⇒±8=\left(\frac{2a}{3}{4}^{\frac{3}{2}}+\frac{b}{2}{4}^{2}\right)-0\phantom{\rule{0ex}{0ex}}⇒±8=\left(\frac{2a}{3}×{2}^{3}+\frac{b}{2}×16\right)\phantom{\rule{0ex}{0ex}}⇒±8=\left(\frac{16a}{3}+8b\right)\phantom{\rule{0ex}{0ex}}⇒±8=\left(\frac{16a+24b}{3}\right)\phantom{\rule{0ex}{0ex}}⇒±24=\left(16a+24b\right)\phantom{\rule{0ex}{0ex}}⇒±24=8\left(2a+3b\right)\phantom{\rule{0ex}{0ex}}⇒±3=\left(2a+3b\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},2a+2b=±3$ Hence, the value of 2a + 3b is $\overline{)±3}.$

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